23.Ray Optics

Ray Optics

* Light • Lens " Reflection of Light .. Refraction Through a Prism • Spherical Mirrors ,. Dispersion " Refraction " Optical Instruments * Total Internal Reflection ~· Resolving Power of a Microscop~~ .. Refraction from a Spherical Surface " Resolvi~g Power of Telescope

------------------,~.~------------------·

t> light The light is that form of energy which makes objects visible to our

eyes. The branch of physics which deals with nature of light, its sources, properties, measurement, effects and vision is called "optics". For the sake of convenience, study of optics is generally divided into two parts namely (i) geometrical optics or ray optics, and (ii) wave optics. This chapter deals with the geometrical optics.

*> Reflection of Light When a beam of light is incident on a polished interface, it is thrown

back in same medium. This phenomenon is called reflection. In reflection the frequency, speed and wavelength do not change, but

a phase change may occur depending on the nature of reflecting surface. Experimentally it is found that the rays corresponding to the incident

and reflected waves make equal angles with the normal to the surface . Thus, the two laws of reflection can be summarized as under.

1. Li = Lr 2. Incitlent ray, reflected ray and normal lie on the same plane.

886 Chapter 23 • Ray Optics

J":

>bs0·t-,

Normal

.(\iil Note · The above two laws of reflection can b.e appli.ed to the reflecting surfaces · wh}ch a(e not ev.en horizontal.

9rh r(g,.

1t811e·ction from Plane Mirror (Surface} In case of reflection from plane surface such as plane mirror (i) The image is always erect, virtual and of exactly the

same size as the object. The image is formed as much behind the mirror as the object is in from of it.

(ii) The image is laterally inverted. (iii) If keeping the incident ray fixed, the plane mirror

is rotilted through an angle e, the reflected ray turns through double the angle ie, 28 in that very direction.

(iv) If the object is fixed and the mirror moves relative to the · ,object with a speed v, the image moves with a speed 2v

~elative to the object. (v) If the mirror is fixed and the object moves relative to the

mirror with a speed v, the image also moves with the same speed .. y relative to the mirror.

(vi) Deviation suffered by a light ray incident at an angle i is given by 8 = (180 - 2i) ,

(vii) If there are two mirrors inclined at an angle 8, the total number of images formed for an object kept between

the two is equal to 2

87t or ( ~7t - 1} which ever is odd.

(viii) The minimum size of a mirror required to see the full image of a person, is half the height of the person.

(ix) If a plane mirror is rotated about an axis perpendicular to plane of mirror then reflected ray image do not rotate.

Instance 1 The minimum size of the mirror fixed on the wall of a room in which an observer at the centre of room may see the full image of the wall of heightJLbehind him is

(a) ~ (b) ~ 3 2

(c) 2h (d) h 3

lnterp.ret From ~ 0' M1M2 and O'AB

M 1M 2 =-x-h 2x +y

h

,

Size of mirror, M M - hx 1 z-(2x+y)

If

Instance 2 Find velocity of image when object and mirror both are moving toward each other with velocity 2 ms-1 and 3 ms-1

respectively . (a) 8ms-1

(c) -5ms-1

Interpret Here

(b) --Bms-1

(d) 5ms-1

Vo-VM=-(Vl-VM)

=> ( +2ms-1)- ( -3ms-1) =~vi+ ( -3)

Instance 3 Two plane mirrors are inclined at 30° as shown in figure. A light ray is incident at angle 45°. Find total deviation produced by combination of mirror after two successive reflection.

(a) 60° i (c) 50° i

(b) 58° i (d) 68° i

Interpret Deviation at mirror M1,81 = 180°-2 x 45°= 90° i

Deviation at mirror M2 ,82 = 180°-2 x 15° = 150°t

Total deviation 8 = 82 -81 = 150°- 90° = 60° i

• Spherical Mirrors Mirrors having their reflecting surface spherical are called

spherical mirrors. Spherical mirrors are of two types

(i) Concave mirror If reflection takes place from the inner surface, the mirror is called concave [Fig. (a)].

(ii) Convex mirror If reflection takes place from the outer surface , the mirror is called convex [Fig. (b)] .

A Incident

light ---+

Incident light

-----...--....-=: P----.. +ve --=p~------

---+ - c

(a) Concave mirror (b) Convex mirror

Definitions of Some Terms Related to Spherical Mirrors Centre and radius of

curvature The centre of curvature and radius of curvature of a mirror are the centre and radius of the sphere of which the mirror is a part. In the given figure, AC is the radius of curvature and C, the centre of curvature.

Pole Pole of the spherical mirror is mid point of its reflecting surface. In figure it is shown by P.

c

Principal ,axis The principal axis of a spherical mirror is the line joining the pole and centre of curvature. In the figure PC is principal axis.

Principal focus Principal focus is a point on the principal axis of the mirror at which the light rays coming parallel to principal axis actually meet after reflection or appear to meet.

c

(a) Concave mirror (b) Convex mirror

0

For concave mirror focus is infront of the mirror, while for convex mirror focus is behind the mirror. Focus of concave mirror is real, while focus of convex mirror is virtual.

Focal length The distance between pole and focus of a spherical mirror is called its focal length. It is represented byf

ie, f = !!_ 2

Table 23.1 Image Formation by Concave Mirror

1. At infinity

2. Between infinity and C

Chapter 23 • Ray Optics 887

Ray Tracing In geometrical optics, to locate the image of an object.

Tracing of a ray as it reflects or refracts, is very important. 1. A ray going through centre of curvature is reflected back

along the same direction. Concave Convex

~· c

2. A ray parallel to principal axis is reflected through the focus, and vice-versa. Also, mutually parallel rays. !\{!]1t . reflection intersect on the focal plane.

F

.E)

3. The light corning through the focus of rriirror or coming towards focus, becomes parallel to principal axis.

;)

F

Sign Convention for Mirrors

M'

According to the sign convention (i) Origin should be placed at the pole (P).

(ii) All distances should be measured from the pole (P). (iii) Object distance is denoted by u, image distance by v,

focal length by f and radius of curvature by R. (iv) Distance measured in the direction of incident ray are

taken as positive while in the direction opposite of incident ray are taken negative.

p

p

Real inverted, very small [m < < -1], at F

t-

Real, inverted, diminished (m < -1) between F and C


3. AtC

4. Between F and C

5. Atf

6 . Between F and P

Table 23.2 Image Formation by Convex Mirror

1. At infinity

c \

M'

)

2. In front of mirror

M'

M

M'

M

M'

c

Real, inverted, equal in size [m = -1] at C

Real, inverted and very large (m > -1) between 2F and iiliini.ty

\

Real, inverted, very large [m ~ ( -oo)] at infinity

Virtual, erect, large in size (m > + 1) behind the mirror

Virtual, erect, very small (0 < m < < + 1) at F

Virtual, erect, diminished (m < + 1) between PandF

/

I I \

Formula and Magnification for Spherical Mirrors Mirror fonnula

1 1 1 -=-+-! v u

where symbol possess their usual meanings.

Lateral magnification

I v f f - v m=-=-=--=--

0 u f-u ·f

where I = size of image perpendicular to principal axis 0 = size of object perpendicular to principal ruds.

Axial magnification 2 - "

max = - ~~ = :: = ( f ~ J = ( f ~! r Areal magnification

m = Ar =::.:_=(-! )2

=(f - v)2

ar Ao u2 f- u . f where , A1 = area of image

A0 = area of object.

Instance 4 An object of length 2.5 em is placed at l.5f from a concave mirror; where f is the focal length of the mirror. The length of the object is perpendicular to the principal a·ds. Find the length of the image. Is the image erect or inverted?

(a) -5 em (b) 5 em (c) 6 em (d, --6 em

Interpret The focal length F = -f and u = -l.5f

1 1--f • 0 F

1----- 1.5 f ------;~

Chapter 23 •· Ray Optics 889

we have, 1 1 1 1 1 1 -+-=-or--+-=--u v F -l.Sf v f

1 1 1 1 -=----=--v l.Sf f 3f

Now, ni = -~"" _ _l_[_ = -2 u l.Sf

h or __£ = -2 or h2 = -2h1 = -5 em.

hl The image is 5 em long. The minus sign shows that it is inverted.

Instance 5 ·A concave mirror of focal length 10 em and a convex mirror of focal length 15 em are placed facing each other 40 em apart. A point object isplaced between the mirrors, on their common axis and 15 em from the concave mirror. Find the position of the image produced by the successive reflections, first at concave mirror and then at convex mirror.

(a) 6 em (b) +10 em (c) + 15 em (d) +30 (.7n

· Interpret According to given problem, for concave mirror.

So,

f= -10 M

u = ~ 15 em andf = - 10 em

.!.+-1-=-1- ie, v = -30em

v -15 -10 ie, concave mirror will form real, inverted and enlarged

image 11 of object 0 at a distanci! 30 em from it, ie, at a distance 40-30 = 10 em from convex mirror.

For convex mirror the image 11 will act as an object and so for it u = -10 em andf = + 15 em.

.!.+-1-=_.!._ ie,v= +6Cin v -10 15

So, final image 12 is formed at a distance 6 em behind the convex mirror and is virtual as shown in figure.

lntext Que,~!lo~,. -2a.l -M~'"" ,

(i); Does the mirror fommla hold g?od for a plane mirrgr? .· , , ······ . ..· _._. __ _. :,: , ..... (iit An object :is'placed between two · plane parallel mirrors. Why do tpe t images ;~t fainter . · · (iii)' Why are inir:rors used in search-lights parabolic and not concave spherical? , ; ·)>

qy~ If yo~ were driving a car, what typ~ ofmirtoi would .you prefer to use (9_r;ob~~~il:affic .at ""''"''"•Adr?

w> Refraction When light passes from one medium, say air, to another

medium, say glass, a part is reflected back into the first medium andthe restpa~,;, ·., . : .. .. ;. ·n-''.': _,·: ;\;)assesintu the second medium, it either bends towards the normal or away from the normal. This phenomenon is known as refraction.

laws of Refraction (Snell's law) (i) If meditim 1 is a vacuum (or in practice air) we refer 1J.l2

as the absolute rE>fra i'! h•" ir><f• ·: •)f medium 2 and denote it by J.l/ or simply J.l (if no other medium is there).

(ii) Now, we can write Snell's law as, J.l sin i = constant For two media, 111 sin i1 = J.lz sin i2

... (i)

... (ii)


2

(iii) Snell's can be written as.

sin i1 v1 A.1 l-12 1112 =-.-.-=-=-=-

Sllll2 v2 A.2 111

Here, v1 is the speed of light in medium 1 and v2 in medium 2. Similarly 1..1 and A-2 are the corresponding wavelengths.

;1 > ;2

V2 < VI

l-12 > l-11

"-2 < "-1

Rarer

Denser 2

ll :> < u l

"-2 > "-1

Denser

Rarer

If 1-1-2 ::> 1-1-1 then v1 > v2 and A-1 > 1..2 , ie, in a rarer m~dium, speed and hence, wavelength of light is more .

(iv) In general, speed of light in any medium is less than its speed in vacuum. It is convenient to define refractive index 11- of a medium as.

Speed of light in vacuum c 11 = Speed of light in medium v

Instance 6 Light is incident from air on oil at an angle of30°. After moving through oil-1, oil-2, and glass it enters water. If the refractive indices of glass and water are 1.5 and 1.3, respectively, find the angle which the ray makes with normal in water.

Air

Oil-1

Oil-2

0 -1 ( 1 J sm -2 .6

(b) sin-1 (_]__J 2.6

(a)

(c) sin-1 (-

1 J 3.6

Interpret As we know

(d) sin-1(2 .6)

1-l sin i = (constant)

=> !l-air sin ~air) = 1-l-glass sin r (glass)

0 0 !l-air 0 0

sm~glass) =--Sin lair 1-l-glass

Again, ll·glass sin iglass = llwater sin r water

From Eqs. (i) and (ii) sin 30 = 1.3 sin r

0 1 1 smr=--=-,

, 2xl.3 2.6 r=.sin-

1(-

1 J 2.6

... (i)

... (ii)

Instance 7 A ray of light is incident on a transparent glass slab of refractive index 1.62. If the reflected and refracted rays are mutually perpendicular, what is the angle of incidence?

(a) 58.3° (b) 85 .3° (c) 60° (d) 65°

Interpret Let the angle of incidence, angle of reflection and an,le of refraction be i , rand r', respectively. Now, as per the question 90° -r + 90°- r' = 90° => r' = 90°- i (because i = r) In case of reflection according to Snell's law, 1 sin i = 1-l sin r ' or sin i = 1-l sin (90° - i)

=> tan i = 1-l or i = tan-1[!1-] = tan-1(1.62) = 58.3°

Instance 8 Refractive index of glass with respect to water is 1.125. If the absolute refractive index of glass is 1.5, find ,the absolute index of water.

(a) 1.33 (b) 2.33 (c) 0.33 (d) 0.44

Interpret Here, the refractive index of glass with respect to water ie, wll-g = 1.125 and absolute refractive index of glass llg = 1.5. We know that

Apparent Shift of an Object due to Refraction Due to bending of light at the interface of two different

media, the image formation due to refraction creates an illusion of shifting of the object position.

Consider an object 0 in medium. After refraction, the ray at the interface bends. The bent ray when it falls on our eyes, is perceived as corning from I.

For nearly normal incident rays, 81 and 82 will be very small.

Rarer

J.lz

. · AB tan 81 = sm 81 = -=-':-:-:-':__-:-:----:---.-=-:-----"':'--'---:----:--Object distance from the refracting surface

Similarly,

. AB sm82 = -------------,------

Image distance from the refracting surface

sin 81 _ 11

_ 112 :::} AB 1

AB _ 112 sin82 - 1 2 - 111 OB BI - 11

1

BI Apparent depth J..lz -= ·=-OB Real depth J..L1

So, Shift = Real depth- Apparent depth = Real depth ( 1 - ~~ J Case I

If ).ll < ).lz

Shift becomes negative , image distance > object distance, ieJ image is farther from the refracting surface.

Case II If ).ll > ).lz.

Shift becomes positive, image distance < object distance, ie image is closer to the refracting surface.

Case III If ).lz = 1 or ).!1 ~ ).l

Shift = Real depth ( 1 - ~ J Instance 9 A fish in an aquarium, approaches the left wall at a rate of 3 ms-1, and observes a fly approaching it at 8 m.~-1 . If the refractive index of water is (4/ 3), find the actual velodty of the fly.

~X ).lX

(a) 3.75 ms-1 (b) 2.75ms-1

(c) 0.75 ms-1 (d) 4.75 ms-1

Interpret For the fish, appa.rent distance of the fly from the wall of the aquarium is ).lX. If x is actual distance, then apparent

velocity will be d(J.!X) dt

(v.pp)fly = !l vfly

Now, the fish observes the velocity of the fly to be 8 ms-1•

Therefore, apparent relative velocity = 8 ms-1

Vfish + Cvapp)fly = 8 ms-1 ~ 3 + ).l vfly = 8

vf!y = 5 x ~ = 3.75ms-1

4

Instance 10 A layer of oil 3 em thick is flowing on a layer of coloured water 5 em thick. Refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36/7 em. What is the refractive index of oil?

(a) 1.4 (c) 3

Interpret

or

or

(b) 2.4 (d) 2

Apparent depth (AD) = .EL + 2_ Ill ' 112

,

36 5 3 -=--+-7 5/3 J..lz

.2_= 36 -3= 15 J..lz 7 7

7 J..lz =- = 1.4

5


v~ Total Internal Reflection Whenever a ray of light goes from a denser medium to a rarer

medium it bends away from the normal. As angle of incidence in denser medium increases, angle of refraction also increases in rarer medium.

The angle of incidence in denser medium for whi.ch the angle of refraction in rarer medium is 90° is called the critical angle (C) .

=>

=>

sin C = ~Lrarer = ~ sin 90° J..ldenser J..ld

sinC=~ J..ld

c = sin-1 (~:J

Now, if the angle of incidence in the rarer medium is greater than the critical angle (C), then the ray instead of suffering refraction is reflected back in the same (denser) medium.

This phenomenon is called total internal reflection. For total internal reflection to take place following set of conditions must be obeyed.

(i) The ray must travel from denser medium to rarer medium.

(ii) The angle of incidence i must be greater than critical angle C.

Instance 11 An isotropic point source (bulb) is placed at a depth h below the water surface. A floating opaque disc is placed on the surface of water, so that the bulb is not visible from the surface. What is the minimum radius of the disc?

Take refractive index ofwater=).l.

Interpret As shown in figure, light from bulb will not emerge out of the water if at the edge of disc. · 1

i>C

sin i >sin C ... (i)

Now, if R is the radius of disc and h is the depth of bulb from it

. . R smt=-,===

Jiz +hz

and . c 1 sm =-J..l

So, Eq. (i) becomes

or

R 1 --===>-~R2 +hz J..l

h R>--

~J..l-1


• Refraction from a Spherical Surface Spherical surfaces are of two types (i) Convex

(ii) Concave

()

1

~ I 2 • ······;;,··· ··· ·· ········ ··· ·· ·:················

1E' 0 I 2 ·- -·· ·· ·····-p· · ---------- ------ -- --- •--·-·······

For both surfaces refraction formula is given by

~-~= Jllz-1 v u R

11-12 is refractive index of second medium with respect to first. If 1-11 and 1-lz are refractive indices of first and second medium with respect to air, then,

llz _ J-11 = 1-lz ·· 1-11 v u R

instance 12 A linear object of length 4 em is placed at 30 em from the plane surface of hemispherical glass of radius 10 em. The hemispherical glass is surrounded by water. Find the final position and size of the image:

(a) 5.3 em (c) 5 em

(b) 4.3 em (d) 2.3 em

;nterpret 4 3 For 1" surface 1-11 = -,J-12 =- ,u = -20cm,

3 2 and R = +lOcm,

8"

5.3cm

A"

1----- v' ----~

Using 1-lz _ J-11 = (J.lz -1-11) v u R

==> (3/ 2) - ( 4/ 3) = (3/2- 4/ 3) v (-20) 10

==> v = -30 em

Using A' B' = ~ => A' B' = ( 4/3) ( - 30) AB JlzU ( 4cm) (3/2) ( -20)

==> A'B' = 5.3 em A'B' behaves as the object for plane surface

3 4 ' Jl1 = 2,Jlz=j and R=oo,u=-40

==> llz_ = llz v' u'

==> ( 4/3) = (3/2) v' ( -40)

Solving it we will get, v' = -35.4cm

N . A' B'' (Jl1 v')

ow usmg --= -, A' B' (Jl2u')

A"B" = (3/2)(-35.4) => A"B" = 5.3cm (5 .3) ( 4/3)( -40)

=>

The final images in all the above cases are shown in figure .

~'~ Lens Lens is a transparent medium bounded by two curved

surfaces. Lenses are of two types 1. Convex or convergent lens 2. Concave or divergent lens

1. Convex or Convergent Lens The traqsparent medium bounded by two bulging surfaces

is called convex lens . It is ofthree types (as shown) .

(a) Double-convex lens

(b) Plano-convex (c) Concavo-convex lens lens

2. Concave or Divergent Lens The transparent medium bounded by two hollow surfaces is

called concave lens. It is of three types (as shown).

(a) Double-concave (b) Plano-concave (c) Concavo-concave lens lens lens

Some Definitions Relating Lenses Optical centre The optical centre is a point within or

outside the lens, at which incident rays refract without deviation in its path.

s

Principal axis The straight line passing through the optical centre of lens is called principal axis of lens.

1 Principal focus Lens has two principal foci.

(i) First principal focus It is a point on the principal axis of lens, the rays starting from which (convex lens) or appear to converge at which (concave lens) become parallel to principal axis after refraction.

··.· .. ... -::· .. _... F1

(ii) Second principal ,focus It is the point on the principal axis at which the rays coming parallel to the principal axis converge (convex lens) or appear to diverge (concave lens) after refraction from the lens.

Both the foci of convex lens are real while that of concave lens are virtual.

Focal length The distance between focus and optical centre of lens is called focal length of lens.

Table 23.3 Formation of Image by a Convex Lens

1. At infinity

2. Beyond 2F1

Chapter 23 • Ray Optics 8~~ l,i

Laws of Formation of Images by Lens ' ·' (i) The rays corning parallel to principal axis of lens pas~

through the focus after refraction. · · · (ii) The rays corning from the focus ofltilS go parallel tot~

principal a~is of lens after refraction. ': (iii) The rays of light passing through optical centre go

straight after refraction without changing their path.

Lens Maker's Formula If R1 and R2 are the radii of curvature of first and seconq

refracting surfaces of a thin lens with optical centre C of foc~l length/ and refractive index 1 ~2 then according to Lens Maker'$ formula

c u-----1...._--v

..!..=(1~2-1)(_..!__ _ _..!__] f Rl R2

~ ..!..=(!l-l)(_..!__ _ _..!__J f R1 R2

where, 1 ~2 = ~ is refractive index of material of lens with' respect to surrounding medium. .

Thin lens formula is

1 1 1 -= - --! v u

At the principal focus (F2) or in the focal plane

Between F 2 and 2F 2

Real, inverted and extremely diminished

Real, inverted and diminished


· 4. Between F1 and 2Fl

5. At F1

6. Between F1 and optical centre

Formation of Image by Concave Lens The image formed is always virtual, erect and diminished

and lies between the lens and F2 for all positions of the · object.

Instance 13 The focal length of convex lens is 10 em in air. Find its focal length in water. (Given, llg. = 3/ 2 and llw = 4/ 3)

(a) 10 em (b) 20 em (c) 30 em (d) 40 em

Interpret -1-=CI!g -1)(2__2_J ... (i) fair Rl Rz

and 1 (llg X 1 1 J ff'ater = llw -

1 Rl - Ri

..• (ii)

Beyond 2F2

At infinity

On the same side as the object

Real, inverted and of same size as the object

Real, inverted and magnified

Real, inverted and highly magnified .

Virtual, erect and magnified

Dividing Eq. (i) by (ii), we get

fwater ( llg -1 J "fair = llg f ll w -1

Substituting the values, (3 / 2-1)

fwater = ( 312 )fair ---1 4/ 3

= 4 fair = 4 X 10 = 40cm

Instance 14 An object is placed at a distance of 10 em to the left on the axis of a convex lens L1 of focal length 20 em. A second convex lens L2 of focal length 10 em is placed co-axially to the right of the lens L1 at a distance of 5 em from it. Find the position of the final image and its magnification.

(a) 163. em on the right of the second lens, 3.33 3 .

(b) 163. em on the right of the second lens, 1.33 3

(c) 163. em on the right of the first lens, 1.33 3

(d) None of the above

Interpret Here, for 1" lens,

=>

=>

u1 = -10 em f1 = 20 em

1 1 1 ---=-vl u1 !1

1 1 1 -=---vl 20 10

v1 =-20cm

L1 L2

~ o· , o;:;; · 0 2 '! o \ '! !

';.: \ i \! v ,1,. ·1'2

Scm

ie, the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens.

1 1 1 For 2nd lens, ---=-

Vz u2 fz

Here, u2 = -(20 + S),J2 = 10cm

1 1 1 50 2 -+-=- => v2 =-=16-cm v2 25 10 3 3

ie, final image is at a distance o; 163_ em on the right of the second lens. , 3

T)le magnification of the image is given by;

m = 2:1_ Vz = 20 _29_ = _± = 1.33 u1 u2 10 3x25 3

Magnification of Lens The lateral, transverse or linear magnificauon produced by

a lens is defined by Height of image I

m= =-Height of object 0

A real image II' of an object 00' formed by a convex lens is shown in figure.

Height of image = _!£__ = ~·: ' ' '' Height of object 00' u

Substituting v and u with proper sign, ) ,; . . If -I v

'. 00' 0 -u I v or -=m=-0 u

Thus, v , m=-u


Important Features

1. Power of lens P = --1-

f(inm)

=> P= 100

j(incm) Power of convex lens is positive and of concave lens is negative .

2. If distance of an object from first focus of lens is a 1 and distance of image from second focus is a2, then its focal length.

f = ~ala2 . This is Newton's formula.

3 . If two or more lenses are placed in contact, then equivalent focal length of the combination.

1 1 1 n 1 -=-+-+ .. ·= L.-f !1 fz i=l h

Power of combination n

p =Pl +Pz + ... = L,P; i=1

Magnification of combination n

M = m1 xm2 x .. . =11m i=l

4. If two lenses offocallengthsf1 andf2 are separated by a distance x, then its equivalent focal length

1 1 1 X -=-+ - --F !1 fz fdz

Power of combination,

P = P1 + Pz- x P1P2 Total magnification remains unchanged ie,

m=m1 xm2

5. If a lens is made of a number of layers of different refractive indices, then number of images of an object formed by the lens is equal to number of different media.

6. Cutting of a lens (i) If a symmetrical convex

lens of focal length f is cut into two parts along its optic axis, then focal length of each part (a plano convex lens) is 2f. However, if the two parts are joined as shown in figure, the focal length of combination is again f.

2(, 2f f f

(a) (b) (c) (d)

I


(ii) If a symmetrical convex lens of focal length f is cut into two parts along the principal axis, then focal length of each part remains changed at f . If these two parts are joined with curved ends on one side focal

length of the combination is f_ . But on joining two 2

parts in opposite sense the net focal length becomes ~ (or net power = 0).

(a) (b) (c) (d)

Table 23.4 Difference between Lens and Mirror

1. Convex lens + ve + ve

2. Concave mirror - ve + ve

3. Concave lens - ve -ve

4. Convex mirror + ve -ve

,

7. Silvering of a lens (i) Let a plano-convex lens is having a curved surface of

radius of curvature R and has refractive index J.l· if its plane surface is silvered, it behaves as a concave mirror of focal length.

R f =- 2(~-t- 1)

(ii) If the curved surface of plano-convex lens is silvered then it behaves as a concave mirror of focal length.

R f =- 2!l

(iii) If one surface of a symmetrical double convex lens (R1 = R2 = R) is silvered, then the lens behaves as a concave mirror of focal length

f = R 2(2J.l-l)

8 . The tabular difference between lens and mirror is given in table.

converging

converging

diverging

diverging

Instance 15 A convergent lens of 6 D is combined with a diverging lens of -2 D. Find the power and focal length of the combination.

(a) 26 em (c) 30 em

(b) 20 em (d) 25 em

Interpret Here, P1 = 6 D, P2 = -2 D Using the formula, P = P1 + P2 = 6- 2 = 4 D

f = liP= 1/4 m = 25 em.

Is the ratio of frequencies o_f ultraviolet rays and ....... ,,a""' Can convergent Why

!I) Refraction Through a Prism A prism is a homogeneous,

transparent medium bounded by two plane surfaces inclined at an angle A with each other. These surfaces are called as refracting surfaces and the angle between them is called angle ofprismA.

p

A

Figure shows the 0 R refraction of monochromatic light through a prism. Here' i and e represent the angle of incidence and angle of emergence respectively, r1 and r2 are two angles of refraction. If J.l is the refractive index of the material of the prism, then

sini sine J.l=--=-

sinrl sinr2

The angle between the incident ray and the emergent ray is lmown as the angle of deviation o. For refraction through a

· prism it is found that i + e =A + o and r1 + r2 = A

Minimum Deviation It is found that the angle

of deviation o varies with the angle of incidence i of the ray incident on the first refracting face of the prism. The variation is shown in Om figure and for one angle of incidence it has a minimum value omm. At this value

i = e It therefore, follows that

A r=-

2

L---~---:i-="-e----.i1

r1 = r2

Further at o1 = om;= (i + 0 -A


Instance 16 A convex lens of 10 em focollength is combined with a concave lens of 6 em focollength. Find the focollength of the combination.

(a) -15 em (b) 15 em (c) 10 em (d) -10 em

Interpret Here,f1 = 10 cm,f2 = -6 em, F =?

Use the formula .!. = ..!._+..!._ = _.!__.!_ = _ _.!_ ' F !1 ! 2 10 6 15

or

F = -15 em

. A+om !=--

2

sm --· [A+om) smt 2

J.!=-.- or J.l= A smr sin-

2

For thin prism, om" {ll l)A.

<Jl T

Instance 17 The angle of minimum deviation for a glass prism with J.l = .J3 equals the refracting angle of the prism. What is the angle of the prism?

Interpret

Using,

we have

or 2

. A A sinA Slll -· COS -

.J3=--= 2 2

sin(1) sin(1)

A .J3 COS-=-

2 2 A - = 30° or A = 60° 2 .

~~ .Dispersion

ut

Dispersion of light is the phenomenon of splitting of white light into its constituent wavelengths on passing through a dispersive medium, eg, prism. Cause of dispersion is the variation of refractive index of prism with wavelengJ:h. As A.v < A.R, hence, J.lv > llR and consequently Ov > Ow


Angular Dispersion It is the angular separation between the fwo extreme rays.

Angular dispersion 8 = 8v - 8R

= C11v - 11R )A

Dispersive Power The dispersive power of a prism material is measured by the ratio of angular dispersion to the mean deviation suffered by light beam.

:. Dispersive power

ro- Ov -oR _ 1-lv -1-lR --0--~,

where 11 is the mean value of refractive index of prism. The dispersive power of a prism depends only on its material

and is independent of angle of prism, angle of incidence or size of the prism.

Dispersive power is a unitless and dimensionless term. Dispersive power of a flint glass prism is more than that of

a crown glass.

Dispersion without Deviation (Direct Vision Prism) 1. To produce dispersion witho~t mean deviation we use a

combination of two prisms of different materials such that

A'= -(l-l.-1JA !-l-1

2. Net dispersion caused = C11v- 11R) A + C11' v - 11' R)A' = (11- 1)A (m - m') = 8 (m - m')

Deviation without Dispersion (Achromatic Prism) 1. To produce deviation

without dispersion we use a combination of two prisms of different materiaPs such that ~-

A'=- [!-lv -1-lR] ·A [!-l~ -!-l~]

2. Resultant ~eviation produced =·O[ 1-:]

Instance 18 Find the dispersion produced by a thin prism of 18° having refracting index for red light = 1.56 and refractive index for violet light = 1.68.

(a) 2.16° (c) 3.16°

(b) 1.16°(d) 2.10°

Interpret We know that dispersion produced by a thin prism e = C!-lv -11R)A

Here, 1-lv = 1.68,~-tR = 1.56 and A = 18°

e.= (1.68 -1.56) x 18° = 2.16°

Instance 19 Calculate the dispersive power for crown glass from the given data

(a) 0.01639 (c) 0.05639

1-lv = 1.523 and 1-LR = 1.5145

(b) 1.05639 (d) 2.05639

Interpret Here, 1-lv = 1.523 and 1-lR = 1.14S.

M fr . . d l.S23+1.S145 ean re actlve m ex, 11 = l.S187S

2 Dispersive power ro is given by,

ro = !lv -llR = l.S23 -l.S14S 0 _01639 (!l-1) (l.S187S-1)

Instance 20 A prism of crown glass with refracting angle of so and mean refractive index = 1.51 is combined with a fiint glass prism of refractive index = 1.6S to produce no deviation. Find the angle of fiint glass .

(a) 3.92° (b) 4.68° (c) 5.32° (d) 7.28°

Interpret Let A' be the angle of flint glass prism. Here, A = so and J-l = 1.51 for crown glass prism.

8 = (!l-1)A= (l.S1-1)xS = 2.SS0

Deviation produced by flint glass o' = (!!' - 1)A' = (1.6S -1)A' = 0.6SA'

For no deviation, 8' = 8 or 0.65A' = 2.55

> Optical Instruments

A'= 2.S5 = 3.92o 0.65

Optical instrument is a device which is made from proper combination of mirrors, prisms and lenses. The principle of working of optical instruments depends on laws of reflection and refraction of light.

Microscope It is an optical instrument which forms a magnified image

of a small nearby object and thus, increases the visual angle subtended by the image at the eye so that the object is seen to be bigger and distinct.

) .. l ~

(i) Simple microscope A simple microscope is a convex lens of short focal length which is fixed in a frame provided with handle . . ·c··j ',

Magnification of simple microscope (a) When final image is formed at least distance of

distinct vision,

D M=1+-

f

(b) For relaxed eye, M = D f

where D = least distance of distinct vision.

(ii) Compound microscope Figure shows a simplified version of a compound microscope. It consists of two converging lenses arranged coaxially. The one facing the object is called objective and the one close to eye is called eye piece. The objective has a smaller aperture and smaller focal length than those of the eye piece.

h

Magnification of compound microscope (a) For relaxed eye

M~ = _ vo (E-J Uo fe

In this position, length of microscope

L~ = Vo + fe

(b) When final image is formed at least distance of distinct VlSlOn.

MD =~(1+ DJ Uo f e

Length of microscope,

L0 =v0 +Ue

Telescope

v 0

= distance of first image from object lens. U

0 = distance of object from objective lens.

f. = focal length of eye piece.

Telescope is an optical instrument which increases, the visual angle at the eye by forming the image of a distant object at the least distance of distinct vision, so that the object is seen distinct and bigger.

(i) Astronomical telescope It consists of two converging lenses placed coaxially. The one facing the distant object is called the objective and has a large aperture and large focal length. The other is called the eye-piece, as the eye is placed closed to it. The eye-piece tube can slide within the objective tube, so that the separation between the objective and the eye-piece may be varied.

'


Object

Magnification of astronomical telescope

(a) For relaxed eye, M = _ fo ~ fe

In this position, length of telescope

L~ =fo +fe

(b) When final image is formed at least distance of distinct vision

M = _ fo ( 1 + fe J D fe D

Length of telescope

LD =fo+ue f o = focal length of objective lens f e = focal length of eye piece

(ii) Terrestrial telescope In an astronomical telescope, the final image is inverted with respect to the object. To remove this difficulty, a convex lens of focallengthf is included between the objective and the eye-piece in such a way that the focal plane of the objective is a distance 2f away from this lens.

B

1~~~~~ A

~-r0~2r---er-~

Magnification of terrestrial telescope

(a) For relaxed eye, M~ = 1 In this position, length of telescope

L~ = f o + 4f + f e (b) When final image is formed at least distance of distinct

vision,

M = fo (1+ fe J D fe D .

Length of telescope , n

LD = fo + 4 f + de fo = focal length of objective lens f e = focal length of eye peice

Galilean telescope A simple model of Galilean telescope is shown in figure. A convergent lens is used as the objective and a divergent lens as the eye-piece.

900 Chapter 23 * Ray Optics

~-------- fo ______ ___..,

8

1 A

Magnification of Galilean telescope

(a) For relaxed eye, M~ = fo fe

In this position, length of telescope L, = fo- fe

(b) When final image is formed at least distance of distinct

vision M = fa (1 - fe J ' D fe D

Length of telescope

Lv = fo -u,

Instance 21 An object is seen through a simple microscope of focal length 20 em. Find the angular magnification produced if the image is formed at 30 em from the lens.

(a) 2.08 (c) 3.08

Interpret Given,f = + 20 em

(b) 2.05 (d) 1.5

and v = - 30 em ,

Using thefonnula, .!._.!.=.!. wehave, v u f

1 1 1

-30 -U0

20 u0 = 12 em

'{<The >angular magnification, M = .!!_ = 25 = 2.08 uo 12

Instance 22 A galilean telescope is 27 em long when focussed to form an image at infinity. If the objective has a focal length of 30 em, what is the focal length of the eye piece?

(a) 3 em (b) -3 em (c) 2 em (d) -2 em

Interpret Given,fo = + 30 em. Objective Length of telescope is given

27cm. Therefore, u, = + 3 em. For the final image at infinity,

the intermediate image should lie

Eye-piece

at first focus of eye piece of the I-- 27 em --l~ Galilean telescope. 1---~- 30 em •I

fe=-3cm

Resolving Power of a Microscope Resolving p;Jwer of e. microscope is defined as the reciprocal

of the least separation between two close objects, so that they appear just separated, when seen through the rnicroscope.

The least separation between two objects, so that they appear just separated is given by

d=--"'-2J.1sin8

where fl is che refractive index of the medium between the objective of the microscope and the object. This distance is called limit of resoh~tion of the microscope.

Resolving power of a micro5mpe = .!_ = 2J.1 sine d lc

8 = half angle of the cone of light from the point object, fl sin 8 = numerical aperture

.... de. _____ _____ _ 0

1. Resolving power of microscope increases with increase in the value of the refractive index of the meaium between objective and object that's why oil immersion objective microscopes are used to achieve high resolving power.

2. The resolving power of microscope increases, with decrease in the value of the wavelength of the light used to illuminate the object, so microscopes using ult~aviolet light for illu(Tlinating the objects are used to achieve high resolving power. These are called ultra microscopes. Higher resolving power is obtained in electron microscope.

> Resolving Power of Telescope Resolving power of telescope is defined as the reciprocal

of the smallest angular separation between two distant objects, so that they appear just separated, when seen through the telescope.

The smallest angular separation between two objects, so that they appear just separated is found to be

de= 1.221c D

where D is the diameter of objective

D Resolving power of telescope 1.22/c

lntext Questions 23.3 en does a ray fucid~nf on _a prism deviate away from the base?

"ngs) observed sometimes round the sun or moon? f gla.~s for lights ?f yellow, green and red colours are f!y. llg and flr respectively. Rearrange 'thes~ fv~l~71: _;

e position of a object relative to a biconvex lens so that it behaves like magnifying lens? erted, ~ill it serve as a microscope?


Chapter Compendium

are o\:1eyed at every reflectin,g surface.

ofsy:mmetrical spherical surfaces, are

two (i) convex, and (ii) concave. which the reflection takes place at the bulged

a convex mirror and the mirror in which place at the depressed surface is called a

.Ll ~'"wouL rays, the refracted ray and the normal surface separating the two media-all lies in one

law For any two media the ratio of the sine of the of incidence to the sine of the angle of refraction

constant for a light beam of a particular frequency, ie

T t

L--..-~...,...--~~

C is the critical angle ,

,, '·

10. Refraction at Sphe,rical Surfaces,

(i) For a spheric1il surface, f.Lz . ..., f.LI = Hz - f.L1

(ii) Magnification, m = 111::_ flzU

(iii) When the object is relation can be obtainted vv<<.:muL)(.UI

relation becomes

_ f.L2 +~t1 = !11 -J.Lz u v R

11. Thin Lens

1 1 1 12. Thin lens fonnula .,--- =

v u f

13. Linear magnification of a lens

14. Power ofLens (i) Power of a lens = (1/J) where f is the focal

the lens in meters with proper sign . .. (ii) Power of combination of lenses.P ':"

where P1, P2, P3, ••• are powers o{co (iii) Magnification produced by equiValent lens.uf = m~ ;5 ,m2

15. Prism (i) Refracting angle

(ii) A + o = i + e


" I

s ..

l '

l ·~ "l •

,

I

Illustrative Example 1 A beaker contammg liquid is placed on a table underneath a microscope which can be moved along a vertical scale. The microscope is focussed, through the liquid, on a mark on the table and the reading on the scale is a. It is next focussed on the upper surface of the liquid and the reading is b. More liquid is added and the observations are repeated, the corresponding readings being c and d. The refractive index of the liquid is

d-b (a)

d-c-b+a

b-d (b)

d -c-b+a

d-c-b+a (c) d-b

(d) d+c-b-a d-b

Solution The real depth = (Refractive index) apparent depth ::::>In first case, the real depth h1 = ).!(b- a).

Similarly, in the second case, the real depth h2 = J.!(d- c) Since, h2 ~ h1 the difference of the depths = h2 - h1

= J.!(d- c- b + a) Since, the liquid is added in second case, h2 - h1 = d - b

d-b ll= d-c-b+a

Example 2 A ray of light from a denser medium strikes a rarer medium at angle of incidence i. The reflected and refracted rays make an angle of90° with each other. The angles of reflection and refraction are r and r', respectively. The critical angle is

(a) sin- 1 (tan r)

(b) sin-1 (cot i)

(c) tan-1 (sin r)

(d) tan-1 (sin i)

Solution Applying Snell's law for refraction,

sini = 112

sinr' 111

From the given condition, r + r' = 90° sin r' =cos r

Solution of Eqs. (i) and (ii) yields, sin i = 112 cosr 111

According to the Law for refraction

i = r ,

... (i)

... (ii)

.. . (iii)

... (iv)

/.12

Using Eqs.(iii) and (iv), we obtain sini = 112

cosi 111

tani = 1-lz Ill

Since, at the time of total internal reflection,

sinec = 1-lz, using Eq.(v) we obtain ec = sin-1(tanr) . Ill

... (v)

Example 3 A given ray of light suffers minimum deviation in an equilateral prism R Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer

(a) greater deviations (b) no deviation (c) same deviation as before (d) total internal reflection

Solution No deviations occur on interfaces 2 and 3 as there is no change in medium. However, deviation at interface 4 is same as it was on interface 2 with only prism P.

Example 4 The sun (diameter D) subtends an angle of e rad at the pole of a concave mirror of focal length f The diameter of the image of the sun formed by the mirror is

(a) f6 (b) f26!D (c) 2f6 (d) De

Solution Since, the sun is at very large distance, u = oo


1 1 1 ~ -+-=-

00 v f ==> v=f If the diameter of the image be d,

T d 2

.J_

d/2 --=a:::> d = (2a)v

v :::> d=(2a)v Putting, 2a =6 and v = f, we obtain d = f6

Example 5 A ray enters a glass sphere of refractive index fl = ../3 at an angle of incidence 60° a ray is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is

(a) 50° (c) 90°

(b) 60° (d) 40°

Solution Refraction at P

sin 60° = ../3 . sinr1

sinr1 = (1/2) ==> r1 = 30° .l Since r2 = r;

r2 = 30° sinr2 1

Refraction at Q = ---sini2 - ../3

Putting r2 = 30°, we obtain i2 = 60° Reflection at Q, r; = r2 = 30°

Example 6 A soldier directs a laser beam on an enemy by reflecting the beam from a mirror. If the mirror is rotated by an angle 8, by what angle will reflected beam rotate?

(a) 8/2 (b) 8 (c) 28 (d) None of these

Solution Let MPM2 be the initial position of the mirror. The mirror is rotated through an angle 8 to the position M't OM2.PO is the incident light. OQ on the initial reflected ray and OQ' is the reflected ray after rotating the mirror by angle 8. If i = initial incidence angle, then L.POQ = 2i and L.POQ' = L.PON' +L.N'OQ = 2i- 26.

Q'

p Q

:. The reflecte~ beam rotates through an angle 28.

Example 7 When an object is at distances of u1 and u2 from the poles of a concave mirror, images of the same size are formed. The focal length of the mirror is

(a) lu1 +uzl (b) lu1 -uzl

(r) lul ;uzl

(d) lu1 ;u21

Solution One image will be real and the other will be virtual. ___ Since they are of the same size, one will have magnification m

and the other -m.

1 1 1 -+-=-ul u1m f

or 2_(1 + ]:_) = _!_ _ u1 m f

... (i)

1 1 1 ----=-and Uz u2m f

2_(1-]:_)=.!. u2 m f

... (ii) or

From Eqs. (i) and (ii), we get

ul +uz = 2 f

f = ul +uz 2

or

Example 8 A ray of light passes through four transparent media with refractive indices J..lJJ J..lz, J..t3 and J..l4 as shown in the adjacent figure . The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

(a) J..l1 = J..l3 (b) J..l2 = J..l4

(c) J..l4 = J..l1 (d) J..l2 = J..l3

Solution Considering Snell's law, J..l sin 8 = constant, and i1 = 4 (given).

Since,

sin i1 = flz sm 12

sin i2 111 ' sin i3

= 113 sini3 = 114

112 '-sin i4 113

sini1 = 114

sin i4 111

i1 = i4, therefore J..t4 = J..t1•

Example 9 Light is incident normally on face AB of a prisTTJ. as shown in the figure. A liquid of refractive index J.1 is placed on face AC of the prism. The prism is made of glass of refractive index 3/2.

Tile limits of fl for which total internal reflection takes place on

faceAC is

3 (a) !J.> -

4

8

3J3 (b) ll <--

4

(d) .)3 !J.<2

Solution Critical angle between glass and liquid face is

sine = 312 = -~ c ll 21J.

Angle of incidence at face AC is 60° ie, i = 60° For total internal reflection to take place

... (i)

i > ec ----------------------------------------- -----or sin i >sin ec _-:_-_:::.-_-_-_:::::.-_·:::.-_-_-_:::.-_-:_-_:· . 600 3 or sm >--

or

or

2}l

.)3 3 ->-

2 2}l

}l> .)3

Example 10 Two identical glass (J.18=3!2) equiconvex lenses of focal length fare kept in contact. .The space between the two lenses is filled with water Cflw = 4/3). The focal length of the combination is

(a) f

·(c) 4f 3

(b) f_ 2

(d) 3f 4

8

~olution Let R be the radius of curvature of each surface. Then

1 ( 1 1 J 7=CL5 - 1) "R+"R . For the water lens

; . =(i- 1)( -*+*) =M -7 l ;. =- 3~

,


. 1 1 1 1 Nowusmg, -=-+-+- wehave

Ff1fzf3 1 1 1 1 -=-+-+-F f f f'

2 2 4 =---=-

! 3f 3f

F= 3f 4 ····

Example U A concave lens jQrms the image of an object such that the distance berween the object. and image is 10 em ana the magnification produced is 1/4. The focal (ength of the lens will be

(a) 8.6 em (b) 6.2 em

(c) 10 em (d) 4.4 em

Solution Concave lens fom1S the virtual image of a real object. So let

u=-4xandv=-x Then 3x = 10 ern

10 or x =-ern

3 40 10

u=--cmand v=--cm 3 3

S b . . . 1 -3 3

u stltutmg m -=-+! 10 40

or f=-~ or f = - 4.4 ern

0

3x

4x

Example 12 A microscope has an objective of focal length 1.5 em and eye piece of focal length 2.5 em. If the distance between objective and eye-piece is 25 em, what is the approximate value of magnification produced for relaxed eye? ·

(a) 75 (b) 110 (c) 140 (d) 25

Solution Length of the tube is L = V0 +f •

Now applying

V0

= L- F.= ~5 - 2.5 = 22.5 ern

.!.. - .!.. = .!.. we have Vo Uo fo

1 1 1 ----=-22.5 U0 1.5

luolz1.6 ern

IMI=vox!!_ uo f.

=(22.5) (~)-\ 1.6 2.5

z140.

Chapter Practice

Exercise I

Photometry 1. What is the ratio of luminous intensity ot two sources,

which produce shadows of equal intensities at ci1stance 25 em and 50 em from the photometer screen? (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1

2. The time required for making a print a distance of 0.25 m from a 60 W lamp is 5 s. If the distance is increased to 40 em, the time required in second to make a similar print is (a) 3.1 (c) 12.8

(b) 8 (d) 16

3. A source is at 4m height above the centre of a circular table of a circular table of radius 3m. The ratio of illuminance at 0 and P will be

, S

4m

PF---3-m_...Jo

64 (a) 125

(c) 1

(b) 125 64

(d) 16 25

4. A lamp of 250 candela power is hanging at a distance of 6 m from a wall. The illuminance at a point on the wall at a minimum distance from lamp will be (a) 9.64lux (b) 4.69 lux (c) 6.94lux (d) None ofthese

5. An electric bulb illuminates a plane surface . The intensity of illumination on the surface at a point 2 m away from the bulb 5 x 10- 4 phot (lumen cm-2) . The line joining the bulb to the point makes an angle of 60° with the normal to the surface. The intensity of the bulb in candela (candle power) is ' (a) 40 x 10- 4

(c) 40vG (b) 40 (d) 20

6. A lamp is hanging at a height of 40 em from the centre of the table. If its height is increased by 10 em, the illuminance of the lamp will decrease by ,

(a) 10% (c) 27%

(b) 20% (d) 36%

7. In a photometer, two sources of light when placed at 30 em and 50 em respectively produce shadows of equal intensities. Their candle powers are in the ratio of

(a) _.2._ (b) 16 25 25 3 5

(c) - (d) 5 3

8. A book can be read if it is placed at a distance of 50 em from a source of 1 cd. At what distance should the book placed if the source is of 16 cd? (a) 8 m (b) 4 m (c) 2m (d) 1 m

9. In a grease spot photometer, light from a lamp with dirty chimney is exactly balanced by a point source distance 10 em from the grease spot. On dearing the dirty chimney, the point source is moved 2 em to obtain a balance again. Then the percentage of light absorbed by the dirty chimney is nearly (a) 64% (b) 36%

(d) 56% (c) 44%

10. Two point sources A and B of luminous intensities 1 cd and 16 cd respectively are placed 10.0 em apart. A grease spot screen is placed between the two sources. For the grease spot to become indistinguishable from both the sides, it should be placed at (a) 80 em from 16 cd lamp and 20 em from 1 cd (b) 20 em from the 16 cd and 80 em from 1 cd

400 100 (c) - - em from 16 cd and - em from 1 cd

3 3

(d) 100

em from 16 cd and 400

em from 1 cd 3 3

11. A point source of light moves in a straight line parallel to a plane table. Consider a small portion of the table directly below the line of mowment of the source. The illuminance at this portion varies w1tn this d1stance r from the source as ~-

(a) 1

r

1 (b) oc -;z

1 1 (c) oc ~ (d) =-;A

12. As the wavelength is increased from violet to red, the luminosity

(a) continuously increases (b) continuously decreases (c) increases then decreases (d) decreases then increases

Reflection of Light 13. A dentist has a small mirror of focal length 16 mm. He

views the cavity in the tooth of a patient by holding the mirror at a distance of 8 mm from the cavity. The magnification is (a) 1 (c) 2

(b) 1.5 (d) 3

14. Given width of aperture = 3 mm and 1c = 500 nm. For what distance ray optics is good approximation? (a) 18m (b) 18 mm (c) 18A (d) 18 light years

15. The separation between the screen and a plane mirror is 2r. An isotopic point source of light is placed exactly mid way between the mirror and the screen. Assume that mirror reflects 100% of incident light. Then the ratio of illuminance on the screen with and without the mirror is (a) 10 : 1 (b) 2 : 1 (c) 10 : 9 (d) 9 : 1

16. An object is placed asymmetrically between two plane mirrors inclined at an angle of 72°. The number of images formed is (a) 5 (b) 4 (c) 2 (d) infinite

17. From a spherical mirror, the graph of 1/v versus 1/u is 1 1 v v

(a) (b) ~ 0 Ti 0 u

1 v

F (c) (d)

0 Ti 0

18. For a convex mirror, the variation of u versus v is given by

v v

(a) (b)

0 u 0 u

v v

(c) ~ (d) ~ 0 u 0 , u


19. A fish is a little away below the surface of a lake. If the critical angle is 49°, then the fish could see things above water surface within an angular range of 8° where

Air Water

e 49°

(a) 8=49°

"(c) 8=24_!_ 0

4

(b) e = 98°

(d) 8=90°

20. A car is fitted with a convex mirror of focal length 20 em. A second car 2 m broad and 1.6 m height is 6 em away from the first car. The position of the second car as seen in the mirror or the first car is (a) 19.35 em (b) 17.45 em (c) 21.48 em (d) 15.49 em

21. A convex mirror forms an image one-fourth the size of the object. If object is at a distance of 0.5 m from the mirror, the focal length of mirror is (a) 0.17 m (b) -1.5 m (c) 0.4 m (d) - 0.4 m

22. A person 6 feet in length can see his full size erect image in a mirror 2 feet in height. This mirror has to be (a) plane or convex (b) plane or concave (c) necessarily convex (d) necessarily concave

23. A point object is placed at a distance of 30 em from a convex mirror of a focal length 30 em. The image will form at

24.

25.

(a) infinity (b) pole (c) 15 em behind the mirror (d) no image will be formed

An object is placed at a distance of 10 em from a concave mirror of radius of curvature 0.6 m. WJ:tich ofthe following statements is incorrect? (a) The image is formed at a distance for 15 em from the

mirror. (b) The image formed is real. (c) The image is 0.5 times the size of the object. (d) The image is 1.5 times the size of the object.

With a concave mirror, an object is placed at a distance x1

from the principal focus, on the principal axis. The image is formed at a distance x2 from the principal focus. The focal length of the mirror is _

(b) xl +xz 2

(d) Jx1x2

26. A man has a concave shaving mirror of focal length 0.2 m. How far should the mirror be held from his face in order to give an image of two fold magnification? (a) 0.1 m (b) 0.2 m (c) 0.3 m (d) 0.4 m


2 7. To focal length of a concave mirror is 12 em. Where should an object of length 4 em be placed so that an image 1 em

. long is formed? (a) 48 em (c) 60 em

(b) 3 em (d) 15 em

28. The focal length of a concave mirror is 20 em. Where an object must be placed to form an image magnified two times when the image is real? (a) 30 em from the mirror (b) 10 em from the mirror (c) 20 em from the mirror (d) 15 em from the mirror

29. A spherical mirror forms diminished virtual image of magnification 1/3. Focal length is 18 em. The distance of the object is (a) 18 em (c) 48 em

(b) 36 em (d) infinite

30. Sun subtends an angle of 0 .5° at the centre of curvature of a concave mirror of radius of curvature 15 m. The diameter of the image of the sun formed by the mirror is (a) 8 .55 em (b) 7.55 em (c) 6.55 em (d) 5.55 em

31. A convex mirror and a concave mirror has radii of curvature 10 em each are placed 15 em apart facing each other. An object is placed midway between them. If the reflection first takes place in the concave mirror and then in convex mirror. the position of the final image is (a) on the pole of the convex mirror (b) on the pole of the concave mirror (c) at a distance of 10 em from convex mirror (d) at a distance of 5 em from concave mirror

32. An object 5 em tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 em. The size of the image is (a) 0.11 em (c) 0.55 em

(b) 0.50 em (d) 0.60 em

33. A convex mirror of radius of curvature 1.6 m has an object placed at a distance of 1 m from it. The image is formed at a distance of (a) 8/13 min front of the mirror (b) 8/13 m behind the mirror (c) 4/9 min front of the mirror (d) 4/9 m behind the mirror

34. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is equal to

(a) b( u;ffz

(c) b(u;f)

(b) b(_L)1/2 u-f (d) b(f~u)

35. When an object is kept at a distance of 30 em from a concave mirror, the image is formed at a distance of 10 em. If the object is moved with a speed of 9 ms-1

, the speed with which images moves is (a) 0.1 ms-1 (b) 1 ms-1

(c) 3 ms-1 (d) 9 ms-1

,

36. Two plane mirrors are inclined to each other at an angle 8. A ray of light is reflected first at one mirror and then at the other. The total deviation of the ray is (a) 28 (b) 240°- 28 (c) 360°-28 (d) 180°-8

37. A plane mirror is approaching you at 10 cms-1. Your image shall approach you will a speed of (a) + 10 cms-1

(b) - 10 cms-1

(c) + 20 cms-1

(d) - 20 cms-1

38. A candle is placed before a thick plane mirror. When looked obliquely in the mirror, a number of images are seen from the surfaces of the plane mirror. Then (a) first image is brightest (b) second image is brightest (c) third image is brightest (d) all images beyond second are brighter

39. An object is approaching a plane mirror at 10 cms-1• A stationary observer sees the image. At what speed will the image approach the stationary observer? · (a) 10 cms-1 (b) 5 cms-1

(c) 20 cms-1 (d) 15 cms-1

40. A small object is placed 10 em in front of a plane mirror. If you stand behind the object, 30 em from the mirror and look at its image, for what distance must you focus your eyes? (a) 20 em (b) 60 em (c) 80 em (d) 40 em

41. When a convergent beam of light is incident on a plane mirror, the image formed is (a) upright and real (b) upright and virtual (c) inverted and virtual (d) inverted and real

42. It is necessary to illuminate the bottom of a well by reflected solar beam when the light is incident at an angle of a = 40° to the vertical. At what angle p to the horizontal should a plane mirror be placed? (a) 70° (b) 20° (c) 50° (d) 40°

43. The sun (diameter d) subtends an angle 8 radian at the pole of a concave mirror of focal length f . The diameter of the image of sun formed by mirror is

(a) 8f

(b) ~ f 2

(c) 28 f

(d) ~ f n

44. A plane mirror reflects a pencil of light to form a real image. Then the pencil of light incident on the mirror is

(a) parallel (b) convergent

(c) divergent (d) Any of these

45. A spherical mirror forms an image of magnification 3. The object distance, if focal length of mirror is 24 em, may be (a) 32 em, 24 em (b) 32 em, 16 em W32~o~ ~16~~~

/

/

/

Refraction of Light 46. How will the image formed by a convex lens be

affected, if the central portion of the lens is wrapped • in blank paper, as shown in the figure . (a) No image will be formed (b) Full image will be formed but is less bright (c) Full image will be formed but without the central

portion (d) 1\vo images will be formed, one due to each exposed

ha~ ·

47. The distance v of the real image formed by a convex lens is measured for various object distance u. A graph is plotted between v and u . Which one of the following graphs is correct?

v v

\ (a) (b)

u u

v

'~ (c) (d)

u u

48. If the space between the lenses in the lens combination shows were filled with water, what would happen to the focal length and power of the lens combination?

; /

/ / Focal Length

(a) Decreased

I (b) Decreased (c) Increased (d) Increased

Power increased unchanged unchanged decreased

49. 1\vo convex lenses placed in contact form the image of a distant object at P. If the lens B is moved to the right, the image will

A B

(a) move to the left (b) move to the right (c) remain ,at P

p

(d) move either to the left or right, depending upon focal , length of the lenses


50. A layered lens as shown in figure is made of two types of transparent materials indicated by different shades. A point object is placed on its axis. The object will form

(a) 1 image (c) 3 images

(b) 2 images (d) 9 images

51. As shown in figure position of an images I of an object 0 formed by lens. This is possible if

oLJ (a) a convex lens is placed to the left of 0 . (b) a concave lens is placed to the left of 0 (c) a convex lens is placed between 0 and I (d) a concave lens is placed to the right of I

52. The relation between n1 and n2 if the behaviour of light ray is as shown in the figure.

~ens

(a) n2 > n1 (b) n1 >> n2 (c) n1 > n2 (d) n1 = n2

53. A convex lens A of focal length 20 em and a concave lens B of focal length 56 em are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves Bas a parallel beam, tnen distanced in em will be (a) 25 (c) 30

(b) 36 (d) 50

54. As shown in figure, the liquids L1, L2 and L3 have refractive indices 1.55, 1.50 and 1.20 respectively. Therefore , the arrangement corresponds to

(a) biconvex lens (b) biconcave lens (c) concavo-convex lens (d) convexo-concave lens

55. A thin double convex lens has radii of curvature each of magnitude 40 ern and is made of glass with fl = 1.65. The focal length of the lens in nearly (a) 30 ern (b) 31 ern (c) 40 ern (d) 41 ern


56. A convex lens of focal length f produces a virtual image n times the size of the object. Then the distance of the object from the lens is (a) (n- 1) f (b) (n + 1) f

(c) (n~1Jt

(d) (n:1)f 57. A concave lens of focal length 20 em produces an image

half in size of the real object. The distance of the real object is (a) 20 em (b) 30 em (c) 10 em (d) 60 em

58. An object 15 em high is placed 10 em from the optical centre of a thin lens. Its image is formed 25 em from the optical centre on the same side of the lens as the object. The height of the image is (a) 2.5 em (b) 0 .2 em (c) 16.7 em (d) 37.5 em

59. One surface of a lens is convex and the other is concave. If the radii of curvature are r 1 and r 2 respectively, the lens will be convex, if (a) r 1 > r 2 (b) r 1 = r 2

(c) r 1 < r 2 (d) r 1 = 1/r2

60. A lens forms a virtual image 4 em away from it when an object is placed 10 em away from it. The lens is a .... .lens offocallength ... (a) concave, 6.67 em (b) concave, 2.86 em (c) convex, 2.86 em (d) May be concave or convex, 6.67 em

61. A convex lens of focal length I_ m forms a real, inverted 3

image twice in size of the object. The distance of the

object form the lens is (a) 0.5 m (c) 0.33 m

(b) 0.166 m (d) 1m

62. A object is placed at a distance offj2 from a convex lens of focal length f. The image will be (a) at one of the foci, virtual and double its size (b) is greater than 1.5 but less than 2.0 (c) at 2f, virtual and erect (d) None of the above

63. Consider an equiconvex lens of radius of curvature R and focal length f. Iff> R, the refractive index I! of the material of the lens (a) is greater than zero but less than 1.5 (b) is greater than 1.5 but less than 2.0 (c) is greater than one but less than 1.5 (d) None of the above

64. A convex lens forms an image of an object placed 20 em away from it at a distance of 20 em on the other side of the lens . If the object is moved 5 em towards the-lens, the

, image will move (a) 5 em towards the lens (b) 5 em away from the lens (c) 10 em towards the lens (d) 10 crrf away from the lens

65. A convex lens is placed in contact with a mirror as shown in figure. If the space between them is filled with water, its power will

(a) decrease (b) increase (c) remain unchanged (d) increase or decrease depending on the focal length

66. The power of a thin convex lens Cang = 1.5) is + 5.0 D. When it is placed in a liquid of refractive index ani, then it behaves as a concave lens of focal length 100 em. The refractive index of the liquid ani will be (a) 5/3 (b) 4/3

(c) F3 (d) 5/4 67. A concave lens with unequal radii of curvature made

of glass C~-t8 = 1.5 ) has a focal length of 40 em. If it is immersed in a liquid of refractive index 1-ti = 2, then (a) it behave like a convex lens of 80 em focal length (b) it behave like a concave lens of 20 em focal length (c) its focal length becomes 60 em (d) nothing can be said

68. An object is put at a distance of 5 em from the first focus of a convex lens of focal length 10 em. If a real image is formed, its distance from the lens will be (a) 15 em (b) 20 em (c) 25 em (d) 30 em

69. A virtual image twice as long as the object is formed by a convex lens when the object is 10 em away from it. A real image twice as long as the object will be formed when it is placed at a distance .... from the lens (a) 40 em (b) 30 em (c) 20 em (d) 15 em

70. An achromatic convergent doublet of two lenses in contact has a power of+ 2 D . The convex lens has power +5 D. What is the ratio of the dispersive powers of the convergent and divergent lenses? · (a) 2 : 5 (b) 3 : 5 (c) 5 : 2 (d) 5 : 3

71. If i~Lj represents refractive index when a light ray goes from medium ito medium j, then the product 2~-t1 x 3 ~-t2 x 4~-t3 is equal to

(a) 3~1 (b) 3 ~2

(c) 1

1~4

72. What. is the relation between refractive indices ~-t1 , ~-t2 , and ~-t3 if the behaviour of light rays is as shown in figure.

(a) ~3 < ~2.~2 = ~1

(c) ~3<~2 < ~1

(b) ~2 < ~1.~2 = ~3

(d) ~3 > ~2 > ~1

73. Monochromatic light of wavelength A.1 travelling in medium of refractive index n1 enters a denser medium of refractive index n2• The wavelength in the second medium is

(a) A1 ( :: J (b) A1 ( ~: J (c) 1.1 (d) A1( nz~n1 J

74. A bucket contains some transparent liquid and its depth is 40 em. On looking from above, the bottom appears to be raised up by 8 em. The refractive index of the liquid is (a) 5/4 (b) 5 (c) 4/5 (d) 8/5

75. What is the angle of incidence for an equilateral prism of refractive index /3 so that the ray is parallel to the base inside the prism? (a) 30° (c) 60°

(b) 45° (d) Either 30° or 60°

76. A lens of refractive index n is put in a liquid of refractive index n'. If focal length of lens in air isf, its focal length in liquid will be

(a) fn'(n -1) n'-n

n'(n-1) (c) ___:_ _ __:_ j(n'-n)

(b) J(n' -n) n'(n -1)

fn'n (d)

n-n'

77. ::~f Jl:~ :::~~~nili: ~ ;;;; iiii:~::i;J critical angle 8c. If thin layer of l : water (1-1 = ~ J is now poured on '

the glass air surface, the angle at which the ray emerges into air at the water-air surface is (a) 60° (b) 45° (c) 90° (d) 180°

78. Light is incident from a medium X at an angle of incidence i and is refracted into a medium Y at angle of. refraction r. The graph sin i versus sin r is shown in figure. Which of the following conclusions would fit the situation?

1. Speed of light in medium Y is .J3 times that in

d 0.2

me iumX. sin r 2. Speed of light in medium

Y is 1/ .J3 times that in medium X.

3. Total internal reflection will occur above a certain i value.

(a) 2 and 3 (b) 1 and 3 (c) 2 only (d) 3 only

sin i

0.4

79. When a glass slab is placed on a cross made on a sheet, the cross appears raised by 1 em. The thickness of the glass is 3 em. The critical angle for glass is

(a) sin-1(0.33) (b) sin-:-1(0.5)

(c) sin-1( .67) (d) sin-1(-h/2)


80. Monochromatic light of frequency 5 x 1014 Hz travelling in vacuum enters a medium of refractive index 1.5. It wavelength in the medium is · (a) 4000 A (b) 5000A (c) 6000A (d) 5500A

Optical Instruments 81. The magnifying power of a telescope is 9. When it is

adjusted for parallel rays, the distance between the objective and the eye-piece is found to be 20 em. The focal lengths of the lenses are (a) 18 em, 2 em (b) 11 em, 9 em (c) 10 em, 10 em (d) 15 em, 5 em

82. In compound microscope, magnifying power is 95 and

the distance of object from objective lens is -1- em.

1 3.8 The focal length of objective lens is - em. What is the

4 magnification of eye piece? (a) 5 (b) 10 (c) 100 (d) 200

83. The focal lengths of the objective and eyelenses of a microscope are 1.6 em and 2.5 em respectively. The distance between the two lenses is 21.7 em. If the final image is formed at infinity, the distance between the object and the objective lens is (a) 1.8 em (b) 1.70 em (c) 1.65 em (d) 1.75 em

84. Two points, separated by a distance of 0.1 mm, can just be inspected on a microscope when light of wavelength 6000 A is used. If the light of wavelength 4800 A is used, the limit of resolution is (a) 0.8 mm (b) 0.08 mm (c) 0.1 mm (d) 0 .04 mm

85. The diameter of moon is 3.5 x 103 km and its distance from the earth is 3.8 x 105 km. The focal length of the objective and eye-piece are 4 m and 10 em respectively. The diameter of the image of the moon will be approximately (a) 2° (b) 21° (c) 40° (d) 50°

86. With diaphragm of the camera lens set atf/2, the correct exposure time is 1/ 100 s. Then with diaphragm set atf/8, the correct exposure time is (a) l/100 s (b) l / 400 s (c) 1/200 s (d) 16/100 s

87. An object is viewed through a compound microscope and appears in focus when it is 5 mm away from the objective lens. When a sheet of transparent material 3 mm thick is placed between the objective and the microscope, the objective lens has to be moved 1 mm to bring the object back into the focus . The refractive index of the transparent material is (a) 1.5 (c) 1.8

(b) 1.6 (d) 2.0

88. A hypermetropic person having near point at a distance of 0.75 m puts on spectacles of power 2.5 D. The near point now is at (a) 0.75 m (c) 0.26 em

(b) 0.83 m (d) 0.26 m


89. An astronomical telescope has a converging eye-piece of focal length 5 ern and objective of focal length 80 em. When the final image is formed at the least distance of distinct vision (25 em), the separation between the two lenses is (a) 75.0 em (b) 80.0 em (c) 84.2 em (d) 85.0 em

90. The focallengthofobjective and eye lensof:mastronomical telescope are respectively 2 m and 5 em. Final image is formed at (1) least distance of distinct vision (2) infinity. Magnifying powers in two cases will be (a) - 48,-40 (b) - 40, 48 (c) -40, + 48 (d) -48, + 40

91. A man's near point is 0.5 m and far point is 3 m. Power spectacle lanses repaired for (i) reading purposes

(ii) seeing distant objects, respectively. (a) -2 D and + 3D (b) + 2 D and- 3 D (c) + 2 D and- 0.33 D (d) -2 D and + 0.33 D

92. A hypermetropic person has to use a lens of power + 5 D to normalise his vision. The near point of the hypermetropic eye is (a) 1m (c) 0 .5 m

(b) 1.5 m (d) 0.66 m

93. A compound microscope has an objective and eye-piece as thin lenses of focal lengths 1 em and 5 em respectively. The distance between the objective and the eye-piece is 20 em. The distance at which the object must be placed infront of the objective if the final image is located at 25 em from the eye-piece, is numerically (u) 95/6 em ' (b) 5 em (c) 95/ 89 em (d) 25/6 em

94. The focal length of the objective and the eye-piece of a microscope are 4 mm and 25 mm respectively. If the final image is formed at infinity and the length of the tube is 16 em, then the magnifying power of microscope will be (a) -337.5 (b) -3.75 (c) 3.375 (d) 33. 75

95. A simple microscope consists of a concave lens of power - 1 OD and a convex lens of power + 20D in contact. If the image is formed at infinity , then the magnifying power CD= 25 em is (a) 2.5 (c) 2.0

(b) 3 .5 (d) 3.0

96. The magnifying power of an astronomical telescope is 10 and the focal length of its eye-piece is 20 em. The focal length of its objective will be (a) 200 em (b) 2 em (c) 0 .5 em (d) 0 .5 x 10-2 em

Prism 97. For a prism, its refractive index is cos A. Then minimum

2 angle of deviation is (a) 180° -A (b) 180°- 2A

(c) 90° - A (d) A 2

98. The refractive index of a prism for a monochromatic wave is J2 and its refracting angle is 60°. For minimum deviation, the angle of incidence will be

99.

100.

101.

(a) 30° (c) 60° It is desired to make a converging achromatic combination of mean focal length 50 em by using two lenses of materials A and B. If the dispersive powers of A and B are in ratio 1: 2, the focal lengths of the convex ar.d the concave lenses are respectively (a) 25 em and 50 em (b) 50 em and 25 em (c) 50 em and 100 em (d) 100 em and 50 em

Two parallel light rays are incident at one surface of a prism of refractive index 1.5 as shown in figure. The angle between the emergent rays is nearly (a) 19° (b) 37° (c) 45° (d) 49°

The refractive index of the material of a prism is J2 and the angle of prism is 30°. One of its refracting faces is polished. The incident beam of light will retrace back for angle of incidence (a) 0° (c) 60°

(b) 45° (d) 90°

102. The cross-section of a glass prism has the form of an isoceles triangle. One of the refracting faces is silvered. A ray of light falls normally on the other refracting face . After being reflected twice, it emerges through the base of the prism perpendicular to it. The angles of the prism are (a) 54°, 54°, no (b) no, no, 36° (c) 45°, 45°, 90° (d) 57°, 57°, 76°

103. Parallel beam containing light of'), = 400 nm and 500 nm is incident on a prism as shown in figure. The refractive index fl of the prism is given by the relation

1-!C"-) = 1.20 + o.s x ~o-14 I.

Which of the following statement is correct? (a) Light on= 400 nm undergoes

total internal reflection. (b) Light of A. = 500 nm undergoes

total internal reflection.

10

,'

(c) Neither of the two wavelengths undergoes total internal reflection.

(d) Both wavelengths undergoes total internal reflection.

104. The maximum refractive index of a prism which permits the passage of light through it, when the refracting angle of the prism is 90°, is

(a) J3

(c) J3 2

(b) J2

(d) ~ 105. An object is placed 30 em to the left of a diverging lens

whose focal length is of magnitude 20 em. Which one of the following correctly states the nature and position of the virtual image formed?

Nature of image (a) inverted, enlarged (b) erect, diminished (c) inverted, enlarged (d) erect, diminished (e) inverted, enlarged

Distance from lens 60 em to the right 12 em to the left 60 em to the left 12 em to the right 12 em to the left


Exercise II

Only One Correct Option 1. A ray oflight passes through four transparent medium with

refractive indices J.11, J.12, J.13 and J.14 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB. we must have

1!3

(a) J.11 == J.12 (b) J.12 == J.l3 (c) J.13 == J.l4 (d) J.13 == J.11

2. A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing (a) a convex mirror of suitable focal length (b) a concave mirror of suitable focal length (c) a convex lens of focal length less than 0.25 m (d) a concave lens of suitable focal length

3. A ray of light passes through an equilateral prism such that the angle of incidence and the angle of emergence are both equal to 3/ 4 th of the angle of prism. The angle of minimum deviation is (a) 15° (b) 30° (c) 45° (d) 60°

4. A lens of focallengthf projects i'rt:tiines magnified image of an object on .a scnen. the distance of the screen from the lens is

(a) (m~ 1) (b) (m: 1)

(c) f (m -1) (d) f (m + 1)

5. A thin equiconvex lens of refractive index 3/ 2 and radius

of curvature 30m is put in water (refractive index = ~). Its focal length is (a) 0.15 m (b) 0.30 m (c) 0.45 m (d) 1.20 m

6. A 16 em long image of an object is formed by a convex lens on a screen. On moving the lens towards the screen, without changing the positions of the object and the screen, a 9 em long image is formed again on the screen. The size of the object is (a) 9 em (b) 11 em (c) 12 em (d) 13 em

7. TWo lenses, one concave and the other convex of same power' a:re placed such that their principal axes coincide. If the separation between the lenses is x, then (a) real image is formed for X== o<&tily (b) real image is formed for all vafues of x (c) system will behave like a glass-~late- for x = 0 (d) virtual image is formed for a'liSV.alues of x other than

. .Jr. zero ' · 8. A ray of light falls on a transparent glass sl~b with

refractive index (relative to air) of 1.62. The angle of incidence for which the reflected and refracted rays are mutually perpendicular is ·

(a) tan-1(1.62) (c) cos-1(1.62)

(b) sin-1(1.62) (d) None of these

9. A double convex lens made out of glass (refractive index J.1 == 1.5) has both radii of curvature of magnitudes 20 em. Incident light rays parallel to the axis of this lens will converge at a distance d such that

20 (a) d = 10 em (b) d =-em

3 (c) d == 40 em (d) d == 20 em

10. A 4 em thick layer of ,--,...,---,-----water covers a 6 em thick glass slab. A coin is placed at the bottom of the slab and is being observed from the air side along the normal to the surface. Find _the

4cm

apparent position of the coin from (a) 7.0 em (b) 8.0 em (c) 10 em (d) 5 em

Coin

11. The light takes in travelling a distance of 500 m in water. Given that J.1 for water is 4/ 3 and the velocity of light in vacuum is 3 x 1010 cms-1 • Calculate equivalent optical path. (a) 566.64 m (b) 666.64 m (c) 586.45 m (d) 576.64 m p

12. How many images are formed by the lens shown, if an object is kept on its axis?

(a) 1 (c) 3

1!1

(b) 2 (d) 4

13. For a optical arrangement shown in the figure. Find the position and nature of image.

(a) 32 em (c) 6 em

(b) 0.6 em (d) 0.5 em

u

14. A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image plane is 1.8 m. The magnification of the image fonned by one Of the half lens is 2. Find the focal-length of the lens and separation between the two halves.

I :

I


(a) 0.1 m (c) 0.9 m

(b) 0.4 m (d) 1m

15. The refractive indices of the crown glass for blue and red light are 1.51 and 1.49 respectively and those of the flint glass area 1. 77 and 1. 73 respectively. An isosceles prism of angle 6° is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light. (i) Determine the angle of the flint glass prism. (ii) Calculate the net dispersion of the combined system. (a) -4°, 0.04°, (b) 4°, 0 .04 (c) 5°, 0.04 (d) -5, 0 .04°

16. A plano-convex lens has a thickness of 4 em. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 em. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of

uJ L the centre of the plane face is found to be 25/ 8 em. Find the focal length of the lens. Assume thickness to be negligible (a) 85 em (b) 59 em (c) 75 em (d) 7.5 em

17. A convex lens of focallengthf is placed some where in between an object and a streen. The distance between object and screen is x. If numerical value of magnification produced by lens is m, focal length of lens is

to w, 2r br(a)

mx (b)

(m-1)2

mx

(m+1f

(m+ 1)2

(c) x (m-1f

(d) X m m

18. A ray oflight from a denser medium strikes a rarer medium at angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angles of

2j f::.J:.!eflection and refraction are r and r' respectively. The

critical angle is (a) sin-1(tanr') (b) sin-1(tanr)

(c) tan-1(tanr')

19. If eye is kept at a depth h inside water of refractive 21 2

[{ : d d . d "d th h d" f h . l 1, m ex an VIewe outs1 e, en t e 1ameter o t e crrc e ,r ' through which the outer objects become visible, will be

(a) _h_ (b) _h_ ~~2 -1 ~~2+1

ri!gn9J'. . 2 h h suld 'c) ~ 2 (d) G

~ -1 vw 20. A ray of light is incident at 60° on one face of a prism

whi.ch has angle 30°. The angle between the emergent ray and incident ray is 30°. What is the angle between the ray and the face from. which its emerges? (a) 0° (b) 30° (c) 60° (d) 90°

21. An object is kept at a distance of 16 em from a thin lens and the image formed is real. If the object is kept at a distance of 6 em from the same lens, the image formed is virtual. If the sizes of the images formed are equal the focal length of the lens will be (a) 21 em · (b) 11 em (c) 15 em (d) 17 em

22. P is a point on the axis of a concave mirror. The image of P formed by the mirror, coincides with P. A rectangular glass slab of thickness t and refractive index f.l is now introduced between P and the mirror. For the image of P to coincide with P again, the mirror must be moved

(a) towardsPby (~-1)t (b) awayfromPby (~-1)t

(c) towardsPby t (1-~ J

(d) awayfromPby {1 -~ J 23. A point object is placed at a distance of '25 em from

a convex lens of focal length 20 em. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness t is (a) 15 em (c) 10 em

(b) 5 em (d) 20 em

24. A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different materials of refractive indices f11 and f.l2 and R is the radius of curvature of the curved surface of the lenses, then focal length of the combination is

(a) R (b) R 2(~1 + ~2) 2(~1- ~2)

R 2R (c) (d)

(~1 - ~2 ) (~2 -~1) 25. One of the refracting surfaces of a prism of angle 30° is

silvered. A ray of light incident at an angle of 60° retraces its path. The refractive index of the material of prism is

(a) ~ (b) 3/2

(c) 2 (d) ifi 26. The focal length of objective and eye-piece of a microscope

are 1 em and 5 em respectively. If the magnifying power for relaxed eye is· 45, then length of the tube is (a) 9 em (b) 15 em (c) 12 em (d) 6 em

27. A glass prism ABC (refractive index 1.5) , immersed in water (refractive index 4/3) . A ray of light is incident normally on face AB. If it is totally reflected at face AC then

-:_-------. c _-_-_-_-_-_-_-_-_-_-.:: .::_-_-_-.

(a) . e 8 (b) . e 2 Sill 2':- Sill 2':-

9 3

sine= F3 (d) 2 .

9 8

(c) - <Sill < -

2 3 9 28. A plane mirror is placed at the bottom of a tank containing

a liquid of refractive index fl · P is a small object at a height h above the mirror. An observer 0 -vertically above P outside the liquid sees P and its image in a mirror. The apparent distance between these two will be

29.

(a) 2 fl h

(c) _3!!_ J.L-1

o. I I I I I

Light takes t 1 second to travel a distance x in vacuum and the same light takes t2 second totravel10xcminamedium. Critical angle for corresponding medium will be

(a) sin-1 ( 1~1tz J

(c) sm -- sm --. -1(10t1 J (d)• . -1( t1 J t2 10t2

30. A plano convex lens of (f = 20 em) is silvered at plane surface. New fwill be (a) 20 em (c) 30 em

(b) 40 em (d) 10 em

May have More than One Correct Option 31~ A point object is at 30 em from a convex glass lens

(lls =%) of focal length 20 em. The final image of object

will be formed at infinity if (a) another concave lens of focal length 60 em is placed

in contact with the previous leJ:lS (b) another convex lens of focal length 60 em is placed at

a distance of 30 em from the first lens (c) the whole system is immersed in a liquid of refractive

index 4/3 ,. · (d) the .whole system is immersed i:p. fi liquid of refractive

index 9;8 ... u.

32. For which of the pairs of u and f for a mirror image is smaller in size. (a) u = - 10 em, f = 20 em (b) u = - 20 em, f = - 30 em (c) u =- 45 em, f = -10 em (d) u =- 60 em, f = 30 em

33. There are three optical media, 1, 2 and 3 with their refractive indices fll j> f.!z > f.!3. (TIR- total internal reflection)


(a) When a ray of light travels from 3 to 1 no TIR will take place \

(b) Critical angle between 1 and 2 is less than the critical angle between 1 and 3

(c) Critical angle between 1 and 2 is more than the critical angle between 1 and 3

(d) Chances ofTIR are more when ray oflight travels from 1 to 3 compare to the case when it travel from 1 to 2

34. Parallel rays oflight are falling on convex spherical surface of radius of curvature R = 20 em as shown. Refractive index of the medium is fl = 1.5 . After refraction from the spherical surface parallel rays

~ (a) actually meet at some point (b) appears to meet after extending the refracted rays

backwards (c) meet (or appears to meet) at a distance of 30 em from

the spherical surface (d) meet (or appears to meet) at a distance of 60 em from

the spherical surface 35. A ray of light travelling in a transparent medium falls on

a surface separating the medium from air, at an angle of incidence of 45°. The ray undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible values of n from the following (a) 1.3 (b) 1.4 (c) 1.5 (d) 1.6

.'\f

Comprehension Based Questions Passage I

I

The power of a convex lens depends on the radius of curvature and refractive index of lens material a'nd is given by

Pa = Ca llg -1) ( ~1 - ~2 } when lens is in air.

The refractive index of material is roughly.given by JH B ~

Jl =A+ /.} (Cauchy's formula). 1

If a lens is dipped in a liquid, its power is changed and is given by

~ = C;J.tg -1)( ~1 - ~J· J:;

36. A lens (f.!g=l.5) in air has a power + 2 D. When l~n~fs dipped in water of refractive index 4/3, its powe~ will become/remain 1

(a) + 4 D (b) + 8 D "' (c)+ 2D (d) -4D

3 7. The refractive index of glass for yellow light of wavelength 6000 A is 1.5 . Then the refractive index of glass fo1 blue light A = 4000 A will become/remain (a) 1.5 (b) less than 1.5 (c) greater than 1.5 (d) information is insufficient

38. If a hollow convex shaped glass is filled with water and surrounding is glass. The lens will act as (a) convex lens (b) concave lens (c) glass plate (d) convex mirror

\


Passage II Total internal refl'ection is the phenomenon of reflection of light into denser medium at the interface of denser medium with a rarer medium. Light must travel from denser to rarer and angle of incidence in denser medium must be greater than critical angle (C) for the pair of media in contact. We can show that

1 11 =sine

·39. Critical angle for water air interface is 48.6°. What is the refractive index of water? (a) 1 (b) 3/2 (c) 4/3 (d) 3/ 4

40. Light is travelling from air to water at Li = 50°, which is greater than critical angle for air water interface. What fraction of light will be totally reflected? (a) 100% (b) 50% (c) None (d) Cannot say

41. Critical angle for glass air interface where J.! of glass is 3/ 2 is (a) 41.8° (b) 60° (c) 30° (d) 44.3°

42. Critical angle for air water interface for violet colour is 49°. Its value for red colour would be (a) 49° (b) 50° (c) 48° (d) cannot say

43. A point source of light is held at a depth h below the surface of water. If Cis critical angle of air water interface, the diameter of circle of light coming from water surface would be (a) 2 h tan C (c) h sin C

Assertion and Reason

(b) h tan C (d) h/ sin C

Directions Question No . 44 to 50 are Assertion-Reason type. Each of these contains two Statements : Statement I (Assertion), Statement II (Reason). Each of these questions also has. four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a) , (b), (c) and (d) given below: (a) If both Assertion and Reason are true and the Reason is

correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not

correct explanation of the Assertion. (c) If Assertion is true but Reason is false . (d) If Assertion is false but the Reason is true.

0 44. Assertion Convergent lens property of converging remain same in all mediums. Reason Property of lens whether the ray is diverging or converging depends on the surrounding medium.

45. Assertion A concave mirror and convex lens both have ' the same focal length in air. When they are submerged in

water, they will still have the same focal length. Reason The refractive index of water is greater than the refractive index of air.

46. Assertion The focal length of the objective of the telescope is larger than that of eye-piece Reason The resolving power of telescope increases when the aperture of objective is small.

~

47. Assertion A short sighted person cannot see objects clearly when placed beyond 50 em. He should use a concave lens of power 2 D. Reason Concave lens should form image of an object at infinity placed at a distance of 50 em.

48. Assertion By roughening the surface of a glass sheet its transparency can be reduced. Reason Glass sheet with rough surface absorbs more light.

49. Assertion The refractive index of diamond is /6 and

that of liquid is .J3 . If the light travels from diamond to the liquid, it will initially reflected when the angle of incidence is 30°.

Reason J.! = - .-1

- where J.! is the refrective index of smC

diamond with respect to liquid.

50. Assertion The colour of the green flower seen through red glass appears to be dark. Reason Red glass transmits only red light.

Previous Year's questions 51. The distance between an object and a divergent lens ism

times the focal length of the lens. The linear magnification

52.

produced by the lens is (WB JEE 2009) (a) m (b) 1/m

1 (d)

m+1 (c) (m + 1)

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through origin and making an angle of 45° with the x-axis meets the experimental curve at P. The coo~dinate of P will be (AIEEE 2009)

(a) ( f.f) (b) (j,j)

(c) (4j, 4f) (d) (2/, 2/)

53. A thin lens of ( J.! 1.5) of focal length + 10 em is immersed in water (J.! = 1.33). The new focal length is

(DCE 2009) (a) 20 em (c) 48 em

(b) 40 em (d) 12 em

54. A telescope consists of two thin lenses of focal lengths 0.3 m and 3 em respectively. It is focused on moon which subtends an angle of 0.5° at the objective. Then, the angle subtended at the eye by the final image will be

(a) 5° (c) 0.5°

(b) .0.25° (d) 0.35°

(UP SEE 2009)

55. A ray oflight passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence

and the latter is equal to ~ the angle of prism. The angle of deviation is

4 (UP SEE 2009)

(a) 25° (bJ--:50° (c) 45° (d) 35°

56. Two plane mirrors are inclined at an angle e. It is found that a ray incident on one mirror at any angle is rendered parallel to itself after reflection from both the mirrors. The value of 8 is (UP SEE 2009) (a) 30° (b) 60° (c) 90° (d) 120°

57. When a ray of light enters a glass slab from air (UP SEE 2009}

(a) its wavelength decreases (b) its wavelength increases (c) its frequency increases (d) Neither its wavelength nor its frequency changes

58. Critical angle of light passing from glass to water is minimum for (UP SEE 2009) (a) red colour (b) green colour (c) yellow colour (d) violet colour

59. In a spectrometer experiment, "as wavelength of a spectral line increases, deviation also increases." This is true

(Karnataka CET 2008) (a) in grating spectrum (b) in prism spectrum (c) Both in grating as well as prism spectrum (d) Neither in prism spectrum nor in grating spectrum

60. The focal length of a biconvex lens of refractive index 1.5 is 0.06 m. Radius of curvature are in the ratio 1 : 2. Then radii of curvatures of two lens surface are

(a) 0.045m, 0.09 m (c) 0.04m, 0.08 m

(Karnataka CET 2006} (b) 0.09 m, 0.18 m (d) 0.06m, 0.12 m

61. A person inside water llw = ' 4/3 sees the setting sun at about (Karnataka CET 2008)

(a) 42° to the horizontal (b) 48° to the horizontal (c) 42° to the vertical (d) 24° to the vertical

62. A ray of light is incident on a glass slab of thickness t, at an angle i, r is the angle of refraction in the glass plate. Distance travelled in the glass plate is

(a) t cos r (c) t/cos r

(Karnataka CET 2008)

(b) ttanr (d) t/sin r

63. A right angled hollow prism is filled with a liquid. A ray of light entering the prism grazing one face emerges out grazing other face. Refractive index of the liquid is

(a) 1. 73 (c) 1.54

(b) 1.33 (d) 1.41

(Karnataka CET 2008)

64. The radius of curvature of the convex face of a plano-convex lens is 15 em and the refractive index of the material is 1.4. Then the power of the lens in diopter is

(Kerala CET 2008)

(a) 1.6 (c) 2.6

(b) 1.66 (d) 2.66

65. The wavelength of red light from He-Ne laser is 633 nm in air but 474 nm in the aqueous humor inside the eye ball. The speed of red light through the aqueous humor is

(Kerala CET 2008) (a) 3 x 108 ms-1 (b) .1.34 x 108 ms-1

(c) 2.25 x 108 ms-1 (d) 2.5 x 108 ms-1


66. If the angle of minimum deviation is 60° for an equilateral prism, then the refractive index of the material of the prism is (Kerala CET 2008)

(a) 1.41 (b) 1.5 (c) 1.6 (d) 1.33

67. A student measures the focal length of a convex lens by putting an object pin at a distance u from the lens and measuring the distance v of the image pin. The graph between u and v plotted by the student should look like

(AIEEE 2008)

) v(cm) v(cm)

t I t (a) (b)

" 0 - 0 -u(cm) u(cm)

(c) 'IT~

(d)

) tm) 0 - 0 -u(cm) u(cm)

68. An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by (AIEEE 2008) (a) a screw gauge provided on the microscope (b) a vernier scale provided on the microscope (c) a standard laboratory scale (d) a metre scale provided on the microscope

69. Given allg = 3/2 and allw = 4/3. There is an equiconvex lens with radius of each surface equal to 20 em. There is air in the object space and water in the image space. The focal length of lens is (DCE 2008) (a) 80 em (b) 40 em (c) 20cm (d) 10 em

70. An observer looks at a tree of height 15m with a telescope of magnifying power 10. To him the tree appears

(BVP Engg. 2008) (a) 10 times taller (b) 15 times taller (c) 10 times nearer (d) 15 times nearer

71. The resolution limit of the eye is 1 min. At a distance x km from the eye, two persons stand with a lateral separation of 3m. For the two persons to be just resolved by the naked eye, x should be (BVP Engg. 2008) (a) 10 km (b) 15 km (c) 20 km (d) 30 km

72. A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 tb produce dispersion without deviation. The angle of the prism P 2 is

(a) 5.33° (c) 3°

(BVP Engg. 2008)

73. The image formed by a concave mirror (UP SEE 2008) (a) is always real (b) is always virtual (c) is certainly real if the object is virtual (d) is certainly virtual

I I

.9t8 Chapter 23 • Ray Optics

~~~A lens made of glass whose index of refraction is 1.60 has · a focal length of + 20 em in air. Its focal length in water,

', whose refractive index is 1.33, will be (UP SEE 2008)

(a) three times longer than in air (b) two times longer than in air (c) same as in air (d) None of the above

(~5 . ·Magnification at least distance of distinct vision of a (B) ·simple microscope having its focal length 5 em is ( ::>) (Kerala CET 2007) (L) (a) 2 (b) 4 (d) (c) 5 (d) 6 (B) (e) 7 (' .. , 76. lL (B)

(d)

The limit of resolution of an optical instrument arises on account of (Gujarat CET 2007) (a) reflection (b) diffraction (c) polarization (d) interference

77. The length of deviation for a glass prism is equal to its refracting angle. The refractive index of glass is 1.5 . Then the angle of prism is (J & K CET 2007)

(E) (a) 2 cos-1 (3/ 4) (b) sin-1 (3/ 4) li') (c) 2 sin-1 (3/ 2) (d) cos-1 (3/2)

9(. ~. A convex lens is placed between object and a screen. ::> J 'The size of object is 3 em and an image of height 9 em is '3

1 obtained on the screen. When the lens is displaced to a new lB) position, what will be the size of image on the screen?

(Orissa JEE 2007) (a) 2 em (c) 4 em

(b) 6 em (d) 1 em

79. A glass slab (J.! = 1.5) of thickness 6 em is placed over a paper. What is the shift in the letters? (DCE 2007) (a) 4 em (b) 2 em

·cc) 1 em (d) None of these

80. In a laboratory four convex lenses L1, L2, L3 and L4 of focal lengths 2, 4, 6 and 8 em, respectively are available. Two of these lenses form a telescope of length 10 em and magnifying power 4. The objective and eye lenses are respectively (UP SEE 2007) (a) L2, L3 (b) L1, L4 (c) L1, L2 (d) L4 , L1

,

------

81. A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principle axis. If the power of the original lens is 4 D, the power of a cut lens will be (UP SEE 2007) (a) 2 D (b)- 3D (c) 4D (d) 5D

82. Focal length of objective and eye piece of telescope are 200 em and 4 em respectively. What is the length of telescope for normal adjustment? (DCE 2006)

(a) 196 em (b) 204 em (c) 250 em (d) 225 em

83. The refractive index of glass is 1.520 for .Jed light and 1.525 for blue light. Let D1 and D2 be the angles of minimum deviation for red light and blue light respectively in a prism of this glass. Then {A IEEE 2006)

(a) D1 > D2

(b) D1 = D2

(c) D1 = D2

(d) D1 can be less than or greater than 9-epending upon the angle of the prism

84. A point object is placed at the centre of a glass sphere of radius 6 em and refractive index 1.5. The distance of the virtual image from the surface of sphere is (BHU 2006)

(a) 2 em (b) 4 em (c) 6 em (d) 12 em

85. The minimum diameter of the cross-section of cone of light with surface of water due to luminous point 8 ern below the surface is (J.! = 1.33) (BVP Engg. 2006)

(a) 18.24 em (b) 20.24 em (c) 25 em (d) 30 em

86. Light of certain colour has 200 waves to the millimeter in air. What will be the wavelength of this light in medium of refractive index 1.25? (BVP Eng g. 2006)

(a) 1ooo A Cb) 2ooo A Cc) 3ooo A Cd) 4ooo A

87. A light moves from denser to rarer medium. Which of the following is correct? (UP SEE 2006)

(a) Energy increases (b) Frequency increases (c) Phase changes by 90° (d) Velocity increases

Chapter 23 • Ray Optics I .~919

Answers Exercise I

1. (a) 2. (c) 3. (b) 4. (c) 5 . · (b) 6. (d) • 7. (a) 8. (c) 9. (b) 10. ~)

11. (c) 12. (c) 13. (c) 14. (a) 15. (c) 16. (a) 17. (b) 18. (c) 19. (b) 20. (a)

2;1. (a) 22. (c) 23. (c) 24. (a) 25. (d) 26. (a) 27. (c) 28. (a) 29. (b) 30. (c)

31. (a) 32. (c) 33. (d) 34. (d) 35. (b) 36. (c) 37. (c) 38. (b) 39. (a) 40. (d)

41. (a) 42. (a) 43. (a) 44. (b) 45. (b) 46. (b) 47. (d) 48. (d) 49. (b) 50. (b)

51. (d) 52. (a) 53. (b) 54. (c) 55. (b) 56. (c) 57. (a) 58. (d) 59. (c) 60. (a)

61. (a) 62. (a) 63. (c) 64. (d) 65. (a) 66. (a) 67. (a) 68. (d) 69. (b) 70. \h) 71. (c) 72. (a) 73. (a) 74. (a) 75. (c) 76. (a) 77. (c) 78. (c) 79. (c) 80. (a)

81. (a) 82. (a) 83. (d) 84. (b) 85. (b) 86. (d) 87. (a) 88. (d) 89. (c) 90. (a)

91. (c) 92. (a) 93. (c) 94. (a) 95. (a) 96. (a) 97. (b) 98. (b) 99. (a) 100. (b)

101. (b) 102. (b) 103. (a) 104. (b) 105. (b) \

Exercise II

1. (a) 2. (c) 3. (b) 4. (d) 5. (d) 6. (c) 7. (b,~) 8. (a) 9. (d) . 10. (a)

11. (b) 12. (a) 13. (b) 14. (b) 15. (a) 16. (c) 17. (a) 18. (b) 19. (c) 20. (d)

21. (b) 22. (c) 23. (a) 24. (c) ' 25. (b) 26. (b) 27. (a) 28. (b) 29. (c) 30. ~~ 31. (a,d) 32. (a,c,d) 33. (a,c,d) 34. (a,d) 35. (c,d) 36. (b) 37. (c) 38. (b) 39. (c) 40. · c)

41. (a) 42. (c) 43. (a) 44. (d) 45. (d) 46. (d) 47. (a) 48. (c) 49. (a) 50. (a)

51. (d) 52. (d) 53. (b) 54. (a) 55. (b) 56. (c) 57. (a) 58. (d) 59. (d) 60. (a)

61. (a) 62. (c) 63. (d) 64. (d) 65. '(c) 66. (e) 67. (d) 68. (b) 69. (b) 70. (c)

71. (a) 72. (c) 73. (c) 74. (a) 75. (d) 76. (b) 77. (a) 78. (d) 79. (b) 80. (d)

81. (a) 82. (b) 83. (a) 84. (c) 85. (a) 86. (d) 87. (d)

,

920 _ ~hapter 23 • Ray Optics

Hints & Solutions til{ l•

w 9't. ~.AU p- Exercise I

E=~=-I-=~ e 3 . o r2 (4)2 16

r h

E = Icos8 = I x (415) p r ' 2 C5i

41 125

E0 I 125 125 :. -=-X-=-

Ep 16 4I 64 4. At minimwn distance, incidence is normal. Therefore,

I 250 E=-z=-2 =6.94 lux.

r 6

5. 5 x 1o- 4 Icos60°

200x200 or I= 5 x 1o- 4 x 4 x 104 x 2 = 40 cd

6. r 40 x 40 16 -=---=-I 50 x 50 25

1-E.= 1 - 16 =_2_ I 25 25

I-I' 9 or --x 100=- x 100=36o/o

I 25

7. When the screen is equally illuminated, E1 = E2

or !J_ = .!1_ or !J_ = r/ = 30 x 30 = _2_ r1

2 r] I 2 ri 50 x SO 25

1 16 8. 502 =d2

or d2 = (50i x 16

or d=50 x 4cm=200cm=2m. 9. r 1 = 10 em, r 2 = 8 em

I1 _ 64 1

I1 _1

64 I

2 - 100 ' - I

2 - - 100

or I 2 -I1 = 36

or

I 2 o 100

Iz-IJ x 100 '=36o/o I2

1 16 10. x z (IOO-x)2

or 1 4 x 100-x

or Sx =-100 or X= 20 em ,

1 or E cx: -r 3

12. The variation of relative luminosity with wavelength is shown here .

1.0

Relative luminosity

400 700 Wavelength (in nm) -

Note that 360

is odd and object line asymmetrically. .8

1 1 1 1 1 1 17. -+-=-or-=--+-

u v f v u f Now, compare withy = mx + c Therefore graph is a straight line having negative slope.

18. Think in terms of rectangular hyperbola.

19. The angular range is clearly twice the critical angle .

1 1 1 1 31 20. -+--=-or-=-

v -600 20 v 600

or 600 v =-em= 19.35

31

21. _!_=_!_ f-u 4

f

f -(-0.5)

or 4f=f+0.5 or 3/=0.5

or f 0.5

=-m=0.17m 3

22. The image is erect and diminished. So, the mirror in necessarily convex.

1 1 1 23. -+-=-

-30 v 30

or 1 2 1 v 30 15

or v = 15 em

24. f =-0

·6

= -0.3m = -30cm 2

25.

1 1 1 -+-=-v -10 -30

1 1 1 3-1 v 10 30 30

30 or v =- em= 15 em

2

v 15 m = -- = -- = 1.5

u -10 Object lies between principal focus and pole. So, the image is virtual and erect.

1 1 1 -+-=-u v f

1 1 1 --+--=-f-xl f-x2 f

or f-x2 + f-x1 _ 1 Cf-xl)(f-x2) -~

or f2

- fx2- fx1 + X1X2 = 2/2 - f(xl + X2)

or f 2 = x1x2 or f = Jx1x2 This is Newton's mirror formula .

- f 26. m---

or

or or

f-u

27 m =_f___ . f-u

or

28. m =_f___ f-u

,

2 = -0.2 -0.2-u

0.2 2=-- or 0.4+2u=0.2

0.2+u

2u = 0.2-0.4 = -0.2 u = -O.lm

JlW c,,

1 -12 ._n ;l; --=---=---

4 -12-u 12rou

12 + u = -48 or u = _:66\:m.

- 20 -2=---

-20-u

or or


-2=~ 20+u

20+u = -10 u =-30cm.

29. Clearly, the given mirror is a convex mirror.

m =_f___ f-u

1 18 -=--3 18-u

or 3x18 = 18-u or u=-2 x 18 cmoru =-36crn.

1 1t 30. 8=-x-rad

2 180

diameter of image = 8

focal length or diameter of image

1 1t 15 =-x-x-x100cm=6 55cm 2 180 2 . .

31. For concave mirror

15 u=--cm v="' 2 ' .

10 f =--em= -Scm 2

1 1 1 1 1 =---=-----

v f u -5 -15/2. 1 2 -1

=--+-=-5 15 15

or v = -15 em

Clearly; the position of the final image is on the pole of the convex mirror.

32. f = -10 ern, 0 = 5 em, u = - 100 em, I = ?

!__ =_f___ 0 f-u

-10 -10 I= xs =-·xs ern

-10-(-100) 90 = -0.55 em

33. f=~m=0.8m,u=-1m 2

_!_=_1 __ __!__ = 10 + 1 = 18 = 2_ v 0.8 -1 8 8 4

4 or v=-m

9

34. Length of image = (-f-)b f-u

1 1 1 35'. -+-=-

u v f _du_dv=O

u2 v2

dv du or -7=~

or dv v2 du 10x10 _1 · • _1 -=---=----x9rns_ =-lms dt u2 dt 30x30 ·


36. Total deviation

37.

38.

·39.

40.

41.

But

or

= (180°-2a) + (1'80°-2~)

= 360°-2(a+~)

90°-a+90°-~+8 = 180°

8=a+~

:. Total deviation = 360°-28

O• •I

I-X x-1

O• I •I

1--x-v-t-x-v--.j

As is clear from figure, the new distance is 2x - 2v. The distance of image from object is reduced by an amount 2v in orie second.

Th~ first images is due to r,eflection from the front surface ie unpolished surface of the mirror. So, only a small fraction is the incident light energy is reflected. The second image is due to reflection from polished surface. So, a major portion of light is reflected. Thus, the second image is the brightest.

10 em 1---1

10 em ~I

I I' 0 0'

~~X---!Iiiii--Y-Observer

1--- x+y-10 ---1

As is clear from figure the distance of image with reference to observer reduces by 10 em in one second.

Clearly, the distance of image from observer is 40 em.

0 ._1_0_e_m-+!i--em-ef

~ • 30 em --~JM Observer

,

42. Clearly, i + r = i + i = 140° or i = 70°

Clearly, plane mirror makes an angle of 20° with vertical and 70° with horizontal.

43. !_ = l_ d u

or

44.

d

d I = - f or I = 8f.

u

45. m=_j_ f-u

. If m = + 3, then

-24 3=--

-24-u

or -24- u = -8 or u + 24 = 8 .

or u = 8 -24 em= -16 em If m = -3 , then

-3 =--=3..±_ -24·-U

u+24 = -8

or u = -32cm.

46. Only the light-gathering power is reduced.

47. Think in terms of rectangular hyperbola.

48. P=(!l-1)(2__2_) Rl Rz

J..l decreases, p QJ.e.Jreases andf incre~:~ses. 49. Power of the system decreases due to separatidn between

the lenses. So, the focal length increases.

50. Note that two refractive indices are involved. Therefore, two images will be formed.

51. Diminished erect image is produced only by a concave lens.

52. When convex lens is surrounded by denser medium, it behaves like a diverging lens.

i

I I I

I

/

53. P=_!_+_!_-~=0 A !2 fd2

1 1 d .. fl+f2=fd2

1 1 d 20-56 = 20 (-56)

56-20 d 20x56 -20x56

d = -36 em

54. As seen from a rarer medium (L2 or L3) the interface L1 L2

is concave and L2 L3 is convex. The divergence produced by concave surface is much smaller than the convergence due to the convex surface. Hence, the arrangement corresponds to concave-convex lens.

55. ..!.. = (1.65 -1)(.3._) f 40

56.

1 0.65 or -=--

or

n=-'-f+u

! 20

20 !=-em 0.65

= 30.77 em "' 31 em

f+u=L n

or u = ~- f = C ~ n )'

or u =-( n~1 )!.tul= n~1 f

57. m=-'f+u

58.

59.

1 -20 . -=---or -20+u=-40 2 -20+u

or u=-40+20 or u=-20cm

I v 0 u

I -25 -=-15 ~10

I= 15 x 2.5 em= 3'7!5 em

..!.. = (~t-1)(_!_- _!_) f rl r2

'· r··""'

For lens to be concave, (_!_ _ _!_)·~ >O c, rl r2 . J.)

. •: . 1 1 :T<\:~ . " or ->- orr

1 <r

2 .rh ,.

rl r2 ~'n ·

1 1 1 60. y=~--;]

' ·2.. 1 1 1 ,-=---! -4 -10


1 1 1 or -=---

! 10 4 1 2-5 3

or -=--=--! 20 20

20 or f = -- em= --6.67 em

3 The negative sign indicates that the lens is concave.

1

61. f -

m=--=-2=-3-f+u 1

- +U 3

2 1 ---2u=-

3 3 1 2

or -2u = - +- = 1 3 3

1 or u = --m = -0.5m

2 1 1 1

62. ----=-v -f/2 f

1 1 2 1 -=---=--v f f f

or v = -f v -f Again, m=-=--=2 u -f/ 2

Clearly, the image is virtual and double the size.

63. ..!_ = C~t-1)(3_) or f = _R_ f R 2(1!. -1)

Now,f> R R

--->R 2(~t-1)

1 or --- > 1 or 2(!1-1) < 1

2(!1-1)

or ~t-1 < % or 11 < ( 1 +%)

or 11 < 1.5. 64. Clearly, 2f = 20 em orf = 10 em

Now, u = -15 em, v = ? F = 10 em

1 1 1

v -15 10

[ 1 1 1 1 1 or -+-=-or-=---

v 15 10 v 10 15

1 3-2 1 or -=--=- orv = 30 em

v 30 30 The change in image distance is (30- 20) em ie, 10 em.

65. The power of the given system is a combination of the positive power of the convex lens, negative power of the plano-concave lens of water and zero power of the plane mirror. Clearly, the power of the system decreases.

66. 5 = (1.5 -1)( *) ... (i) ... (ii)

924 I Chapter 23 • Ray Optics

Dividing Eq. (i) by Eq. (ii), we get _5 = 0.5n

1.5 -n or -7.5+5n=0.5n or -7.5=-4.5n

75 5 or n=-=-

45 3

67. _ __l.= (l.5-1)(~-~J 40 R1 R2

1 1 1

R1 R2 20

No~ _!_=(~-1](-~J ' f 2 20

or J =- 0;5 (- 210 J 1 1

or - =- or f = 80 em f 80 \

It behave like a convex lens of focal length 80 em.

1 1 1 68. ---=-

v u f

or

or

or

69. m=-ff+u

Now,

or

Again,

or

1 1 1 v -15 10

1 1 1 -=--+-v 15 10

1 -2+3 -=--v 30 v = 30 em

+2=_1__f-10

2f- 20 = f or f = 20 em 20

-2=--- or-40-2u = 20 20+u

-2u = 20 + 40 or u = - 30 em

70. The condition for achromatism is ffit P1 + WzPz = 0

ffit P1 = -wzPz

=> W1 = _ Pz Wz P1

Now, P1 + P2 = 2D

or 5 + P2 = 2 or P2 = -3D

ffit -3 3 -=--=-(1)2 5 5

_ !11 llz _ !13 _ _ 1 71. zll1 X3 llz X4 113 -- x--- -4 !11 --

llz 113 !14 1!14

72. The central ray goes undeviated. So, J..lz = J..t1. Also, J..t3 < J..tz·

c VA A 73. n1 =-=-=-

v1 vA1 A1 c VA A

nz "'-=-=-v2 vA2 A2

76. ]-=(i-1)(~1- ~J ;1 = (~. - 1 )( ~1- ~J

D. 'd' / 1 (n -1)n' lVl mg, - = _:____________:___

f n-n'

or fl = fn'~n -1) n -n

77. llg sinec = 111 sin 90° or llg sin ec = 1 When water is poured,

llw sin r = lls sin ec or llw sin r = 1 Again, lla sine = llw sin r or lla sine= 1 or sine = 1 or e = 90° sinr 1

78. -.-:- = tan30°= r;::; smz v3

sini r;::; r;::; or - .-=v3 or !l=v3

smr So, speed of light in Y is .J3 times less.

79. !1=1=3(1-~) 1 1 1 1 2 2

or 1-- = -,--- or - = 1--=- or 11 = -11 3 ·' 11 3 3 3

1 3 Now,--=

sinic 2 . . 2 . . _1(2J

or sm zc = 3 or zc = sm ~ 3 or ic = sin-1(0.67).

[

80. Wavelength in vacuum,

A = 3

X 108

X 1010 A= 0.6 X 104 A Sx1014

= 60ooA 'A

Now, 11=~

or 'A' = ~ = 6000 A=4oooA

M =fo . 81. f e

11 1.5

9 = fo or f. = 9f fe o 'Je

Also, L = fo + fe or 20 = fo + fe or 20 = 9 fe + f. or 20 = 10 fe or fe = 2 em :. f 0 =9x2cm=18cm

. . 1 1 1 82. FortheobJectlve, vo- _

11 3_8

=11 4

or _!_+ 3.8 = 4 or_!_= 0.2 = .!_

or 5

Now: M =--= -19 ' 0 1

3.8 Again, M = M0 X Me

95 -95 =-19 xMeor Me=-=5

19 83. L = v 0 + fe -==- v 0 = L - f. '

or V0 =19.2cm

1 1 1 19.2 uo 1.6

1 10 10 or --::::::---

uo 16 192

1 120-10 100 or -

uo 192 192

192 or u =--cm=-1.75cm 0 110

84. 'A2 = 4800 = 0_8 'A1 6000

5

New resolution limit = 0.8 x 0.1 mm = 0.08 mm

a= 3.5x103

rad = 3.5 x 180° 85. 3.8x10s 3.8x100 7t

86.

3.5x180x7° 38x100x22

Also M = fo = 400 = 40 ' f. 10

!2 tcx:

d2

~= 40x35x180x7° = 2l.lo"' 21o. 3Sx100x22 '

j_ means that the diameter of aperture is L. d ' 2

Chapter 23 • Ray Optics I 925

1 !2 Now: -ex:--

' 100 / f J2

l2 1

or - cx:4 100

!2 Again, t ex: ( f J or t ex: 64 Dividing Eq. (ii) by Eq. (i),

64 16 100t=-=16 or t=-s

4 100

87. Shift =t(1-~)

1=3(1-.! ) or .!=1-.!_ Jl 3 11

1 1 2 3 or - = 1--=- or 11 =- = 1.5

Jl 3 3 2 1 1 1

88. !=--;;--;;_

or

or

1 1 1 100 25 2.5=---- or-=----

-0.75 u u 75 10 1 4 5 1 -8-15 23 -=-- - -or-=--=--u 3 2 u 6 6

6 u = --m = -0.26m

23 · ; feD 89. Separat10n = 10 +-F--

J e +D

= 80 + Sx25 = 80 + 125 5+25 30

= 84.16 em= 84.2 em.

90. (1) M =-J0 (1 + f. J fe D

M=- 200(1 +~) 5 25

M = -40 ( 1+~ J = -40 x % = -48

(2) M = fo = - 200 = -40 fe 5

91. For reading purposes u =- 25 em, v =-50 cm,J = ? 1 1 1 1 1 1 -=---= -+-=-.. f v u 50 25 so p = 100 = +2D

f For distinct vision,

f' = distance of far point = -3m

P = _!_ = _ _!D =- 0.33 D. f' 3

1 1 92. f=-=-m=20cm

p 5 1 1 1

Now, ---=v u 20

... (i)

... (ii)

I

926 I Chapter 23 • Ray Optics

1 1 1 1 1 1 or ---=-or-= - - -

v -25 20 v 20 25 1 5-4 1 1

or ~ = 100 or ~ = 100

or d = lOOcm =1m

93. For the eye-piece ve = -25 cm,f, = 5 em

1 1 1 ---=--25 ue 5

1 1 1 1 -1-5 or -=----or-=---

ue 25 5 ue 25 25

or u =--e 6

Now, v = L-lu I= 20-25

o e 6 120-25 95 ---em=- em

6 6 1 1 1 1 6

Now, ----=- or -=--1 95 / 6 uo 1 uo 95

or

94. When final image is formed at oo ,

M- ~:(~)- ~(~) Now, v;; = 16- fe = 16-2.S= 13.5 em

M= 13 ·5 x~=-337.5 -0.4 2.5 .

95. Power of combination P = P1 +P2

96.

97.

For image at infinity

= + 20- 10 = + 100

F= _!_=_!_m=10cm p 10

M = !?_ = 25

= 2.5 F 10

M = fo ,10 = fo ,f0

= 200cm. f. 20

--Tij 2 ~= . A

sm -2

· . (A+om) sm --A 2

cot- = ---'-------'-2 . A sm -

2

A . (A+om 1 cos2 = sm - 2- J

. A ' ·. A or

sm- sm -2 2

or . ( 1t A ) . (A + om ) sm 2- 2 =sm -2-

or Jt A A om ---=-+-2 2 2 2

or ~~A= om , 2 2

98.

7t-2A =om 2 2

. . om = 180°- 2A

~=sin(~) .A

sm -2

or

or

sin J2 = --'1..---...=-----.J.

1 · . (60°+om) J2 =sm 2

. 45o . (60o+om) sm =sm 2

om= 30°

. A+om l=--

2

= 60 + 30 = 90 = 450 2 2

COt 1 99. -=-

100.

~ 2

Now, !1 =-COt = _.! fz ~ 2

or !2 =-2!1

1 1 1 Now, -=-+-

F !1 fz 1 1 1 -=-+--so fl -2!1

or 50f= -2+1 =-1--2!1 2f1

or 2j1 = SO or f1 = 25 em

Again, f2 = -2 x 25cm = -SOcm

following arguments lead us easily to the right choice. (i) Angle between any two lines is the same as the angle

between their perpendiculars. .. i=3Q0

(ii) 1 sin30°

1.5 sinr

. l.S 0 75 or smr=-= . 2

-' or r =48.6°

(iii) 9 = r- i = 18,6° Required angle = 2 x 18.6 = 37.2°

101. A = r 1 + r 2

102.

103.

30° = r1 + 0°

or r 1 = 30°

Now: sin i = J2 ' sin30°

. . {;:;2 1 or sml = v.t.X-2

. . 1 or Slnl= J2

or i = 45°

From ,:\,.abc, A+ 90° + (90°- i) = 180°

or i =A Now, complementary angle at point d, 8 = 2 i

e = 2A Only option (b) satisfies this.

-1 20+ 0.8 x 1o-14 ll1 - · (400 x 10-9 ) 2

or

or

or

or

or

-1 20+ 0.8x1o-14 Ill- · 400x400x10 18

0.8 Ill= 1.20+-

16

Ill = 1.20 + 0.05

' . ' Ill = 1.25 . . 1 0 8 Slnl =--= .

c 1.25

2.

,

or

Again,

or

or

or

or


ll2 = 1.232

sini =-1-=0.81

c 1.232

ic = sin-1 0.81 = 54.26°

Now, sin 8 = 0 .8 or 8 = 53 .13° This angle is clearly greater than critical angle corresponding to wavelength 400 run. So, light of 400 nm wavelength undergoes total internal reflection.

104. For passing the ray from prism,

. . A )l < cosec 2

(90°) )l <cosec Z

ll < J2 llmax = J2

105. When an object is placed between 2f andf (focal length) of the diverging lens, the image is virtual, erect and diminished as shown in the graph. To calculate the distance of the image from the lens, we apply

1 1 1 1 1 1 -=---=>--=---! v u -20 v 30

V= (20) (30)

20+30 = - 12 em (to the left of the diverging lens.)

121. Chapter 23 • Ray Optics

Exercise II

1. As there is no deflection between medium 1 and 2. Therefore, J.1.1 = J.1.2

2. Imaae can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens. Let u = 2/ + x, then

1 1 1 -+-=-u v f

1 1 1 => --+-=-

2/+x v f 1 1 1 f+x

=> -=----= v f 2f +x j(2f +x)

j(2f + x) => v= .

f+x It is aiven that u + v = 1.0 m

· 2f +x+ j(Zf +x) = (2f +x)[1+-1-] < l.Om f+x ,f+x

(2j +X)2 1 0 or < . m f+x

or (2f +x)2 < (f +x) This will be true only whenf < 0.25 m.

3. Given, A = 60°

i = i' =I' A = 45° 4

i+i'=A+o or 90°= 60° +8

() = 30° A

N()te that i = i' is the condition for minimum deviation.

Hence, 8 = 30° = <>min . 4. Image will be real.

We know that 1 1 1 -=---! v u

2'_=1-2'_ f u

2'_=1+m f.

~ v= j(m+1)

. s. !~(.&.-1x]__]_J · f 1!2 R1 R2

( :. u is negative)

or 7=(~-1)(0~3) 1 1 2

or -=-x~ orf= 1.20m f 8 0.3

6. y=~y1 xh =~16x9 =4x3=12 em.

111 x 11x 7. -=-+---- =---+----z

f !1 !2 fd2 !1 !1 !1 1 X

=> f > 0 for every x. -=-f !12

For X=O,J=oo __..

Hence, for x = 0, system will behave like a glass plate. 8. We know that

or

11= s~ni and i+r=90° smr

r = 90°- i sini .

11 = = tant sin(90°-i)

or i = tan-1 (~-t) = tan-1(1.62).

1 2 2 1 9. Used= f; where,-= C~-t-1)- = (1.5-1)x- =-

f R 20 20 ·=> f= 20 em

10. Using equation, the total apparent shift is

or s=4(1--1

)+6(1--1

) 4/3 3/2

= 3.0 em Thus, h = h1 + h2 - s = 4 + 6- 3

= 7.0 em. 11. We know that

velocity of light in vacuum 11 = velocity of light in water

4 3x1010

3 velocity of light in water

=>velocity of light in water = 2.25 x 1010 cms-1

Time taken = 500 X

100 = 2.22 X 1 o-6 s · 2.25x1010

Equivalent optical path = Jl x distance travelled in water

= ix500 = 666.64m 3

I i

I

I

12. Only one converging point is found by this lens. Therefore, only one image is formed .

13. According to Cartesian sign convention u = - 40 em, R = - 20 em l-11 = 1, 112 = 1.33

Applying equation for refraction through spherical surface, we get

f..lz _ f..l1 = f..lz - f..l1 v u R

1.33 1 1.33 -----=--

v -40 -20

After solving, v = - 32 em.

The magnification is m = hz = f..lJ v h1 f..lzU

h2 = _ __ 1(32) 1 1.33(-40)

or h2 = 0.6cm

The positive sign shows that the image is erect.

14. This is a modified displacement method problem.

a+d 2 Here, a = 1.8m and --=-

a -d 1 Solving we get d = 0.6 m

a2 -dz !=~

=0.4-m.

15. (i) When angle of prism is 'Small and angle of incidence is also small, the deviation is given by 8 = (f.1-1)A. Net deviation by the two prisms is zero. So,

(ii)

or

Flint

Crown

81 +82 =0

(~L 1 -1)A1 + (f.12 -1)A2 = 0.

Here, 111 and 112 are the refractive indices for crown and flint glasses respectively.

H 1.51 +1.4g 1.5 and ence, f..l1 = 2

- 1. 77 + 1. 73 -1 75 f..l2- 2 - .

This givesA2 =- 4°

Hence, angle of t1int glass prism is 4°. Negative sign shows that flint glass prism is inverted with respect to the crown glass prism.

Net dispersion due to the two prisms is

= (f..l~ - f..l r1 )Al + (f..lbz - f..lr2 )A2

= (1.51-1.4g) (6°) + (1. 77 -1.73) ( -4°) = -0.04°.

:. Net dispersion is- 0.04° t


16. As shown in Fig. (a)

·T3: 2 ·TO ~ 0 (a) (b)

In this case refraction of the rays starting from t0 takes d

place from a plane surface. So, we can use dapp = actual

or 3=± f..l 4

or J.1=:3

f..l

As shown in Fig. (b). In this case refraction takes place from a spherical surface. Hence, applying

f..l2 _ f..l1 = f..lz - f..l1 v u R

we have, 1 4/3 1-4 / 3

--=---(-25 / 8) -4 -R

1 1 8 1 or -=---=-

3R 3 25 75 . . R = 25 em. Now, to find the focal length we will use the lens maker formula

17. Here, x = u + v f f -v

m=--=--f+u f

and image is real, magnification is negative. f -(m+1)f

-m=-- u= f+u' m

f-v From -m=--~v=(m+1)f

f Put in Eq.(i)

- (m + 1) f ( 1)f x= + m+ m

Solving, we get, f = mx 2 (m+1)

1 1 sini 18. sin C=-=----

f..l sinr' I sini sinr'

As is clear as shown in figure LCBD=goo

goo -r +goo -r' = goo or

sini :. sin C = _ ___:__ __ sini sin(g0°-r) cosr

sini . ( . ) =--. =tant ·: t=r COSt

C = sin-1(tani) = sin-\tanr).

i r Den~er

B

Rarer


19. Let r be the radius of circle through which other objects become visible. The rays of light must be incident at critical angle C.

. C 1 r sm =~= ~rz+hz

J..lzrz = rz + hz

(J..lz -1)rz = hz

Diameter

h r=---

~J..lz -1 2h

2r= ~J..lz-1

20. Here, i1 = 60° ,A= 30°,8 = 30°

As i1 + i2 -= A+ 8,

i2 = 0

Hence, angle between the ray and the face from which it emerges = 90°-0° = 90°.

21. As a convex lens alone can form a real images as well as a virtual image, therefore, the lens in the present question is a convex lens. Let f be the focal length of the lens and m be the magnification produced.

In the first case, when image is real,

As

u = -16cm, v = (m x 16)cm

1 1 1 ---=-v u f

1 1 1 1 16 --+-=-or 1+-=-16m 16 f m f

In the second case, when image is virtual,

u = -6cm, v = ( -6m)cm

1 1 1 From---= -

v u f 1 1 1 1 6

--+-=-or 1--=--6m 6 f m f

Add Eq. (i) and Eq. (ii) we have

22 22 2=- or f=-=11 em.

f 2

.. (i)

.. (ii)

22. When a slab of thickness tis introduced between P and the mirror, the apparent position of P shifts towards the

mirror by ( t- ~} Hence, the mirror must be moved in

the same direction through the same distance.

23. In the situation given, the image will be formed at infinity, if the object is at focus of the lens ie, at 20 em from the lens. Hence, shift in position of object

X = 25- 20 = ( 1 - ~ J t

5 = (1 - J:__) t 1.5

t = 15 em.

,

24. The combination of two lenses is

As 1 1 1 - =-+-F f1 fz

' -

_!_ = (J..l1 -1)(2_ + _!_J+ (J..l2 -1)(J:__- _!_J F oo R -R oo

- Jll -1 J..lz -1 ---+--R R

_!_ = Jl1- J..lz F R

or F=-R-. Jl1- J..lz

25. As is clear from figure , A = 30°, i1 = 60° As the ray retraces its path on reflection at the silvered face , therefore,

i2 = 0, r2 = 0

As r1 +r2 =A r1 + 0 = 30°

or r1 = 30°

J..l= sini1 = sin60° = .J3;2 =.J3. sinr1 sin30° 1/2

26. For the relaxed eye, magnifying power is

M=-~£_ " uo fe

-45 =-VoX 25' Vo = 9. uo 5 uo

For objective lens, image is real. v

Vo=+vo,Uo= - ;.

Given, f a = 1 em.

1 1 1 From---=-

vo Uo fa

1 9 1 -+-=-;v

0 =10cm.

vo vo 1

Length of the tube = V 0 + f, = 10 + 5 = 15 em.

2 7. For total internal reflection at AC face

sini:::: J.lw J.lg

sine::::-4-3xl.5

. e s sm ::::-9

28. Note that image formation by a mirror does not depend on the medium. As P is at a height h above the mirror, image of P will be at a depth h below the mirror.

If d is depth of liquid in the tank, apparent depth of P, d-h

Xl=--J..l

:_ Apparent distance between P and its image

d+h d-h 2h =Xz-Xl =-----=-

J..l J..l J..l

I

x lOx c=-v=--

tl ' tz 29.

sin C' = _.!_ = ~ = 10 x x £L

11 c t2 x

e, . _1 (10t1J =Sill --tz

.. . (i)

30. As _.!_ = (!l- 1) (__.!__- __.!__) f R1 Rz

:. __.!__=(1.5-l)(I__I_) 20 = R

1 -1 -=- R=-10cm 20 2R'

Refraction from rarer to denser medium

_&+llz =llz-111, where u==,v=f u v R

0+~= 1.5 -l =__.!__ f=30cm. f 10 20'

31. Final image is formed at infinity if the combined focal length of the two lenses (in contact) becomes 30 em

1 1 1 or -=-+-

30 20 f ie, when another concave lens of focal length 60 em is kept in contact with the first lens. Similarly, let J..l be the refractive index of a liquid in which focal length of the given lens becomes 30 em. Then

210 =(%-1 )(.~1 - ~J ... (i) 310 = ( 3 ~ 2 - 1 )( ~1 - ~J ... (ii)

From Eqs. (i) and (ii), we get 9

11--- s· 32. For convex mirror (positive focal length) image is always

smaller in size. For concave mirror (negative focal length) image is smaller when object lies beyond 2f.

33. Total internal reflection takes place when ray of light travels from denser to rarer medium.

F h · 8 llz d . 8 113 urt er, Sill 12 = - an Sill 13 = - . Ill Ill

Since, 113 > 113

Ill Ill

812 > 813 Smaller the value of critical angle more the chance of total internal reflection.

34. Using, llz - 111 = llz -Ill , we get v u R

or

1.5 1.0 1.5 -1.0

v 20

v=±60cm

' 1 1 . 1 -35. As !l2':-->-->~2':J2=1.414.

sine sin45° 11 -v2

:. Possible value of J..l are 1.5 and 1.6. ,


36. In air

1 . ( 1 1 ) p = f = (1.5 -1) Rl - Rz = 2

_!_J= _3_ = 4 R2 0.5 ·

In water

P ' ( 1.5 1) 4 4.5-4 = 4 1 3

- X =-4-x4=0.5D.

A+B 1 37. !l=~ => ll""~

Therefore refrective index for A. 4000A will greater then 1.5.

38. Use _.!_ = (~ - 1 )[__.!__ - __.!__] f !lm Rl Rz

Since (IlL< !lm) because lens is of water and IlL= llw ·

39. !l=-1-= 1 _1 __ _± sine sin48 .6 0.75 3

40. Light cannot undergoes total internal reflection when it is travelling from air to water, ie, from rarer to denser medium. ·

41. sine=_.!_= -1

- = 3. = 0.6667 ll 3/2 3

e = sin-1(0.6667) = 41.8°

42 F 1 · el . rom ll = -.- ,s1n =-Sill e !l

As llv > !lr

ev < e,. 43. As shown in figure

OS= h When angle of incidence is slightly greater than e, light undergoes total internal reflection. :. Diameter of circle of light coming from water surfaee = 2 r = 2 (OB) = 2 OS tan e = 2h tan C.

----- ---------- ----------- -----·

~-=-=~-=-=-- h c -~~-=~-=-=· -------- ---------------- ----------- ------ --------· --------- ---------,------~---_-, ------- ~ --·s----------·

44. In air or water, a convex lens made of glass behaves as a convergent lens but when it is placed in carbon disulfide, it behaves as a divergent lens. Therefore, when a convergent lens is placed inside -a transparent medium of refractive index greater than that of material of lens, it behaves as a divergent lens. It simply concludes that property of a lens whether the ray is diverging or converging depends on the surrounding medium.

45. If a mirror is placed in a medium other than air its focal length does not change asf = R/2, but for ,the lens

_.!._ = Cang -1)(__.!__ _ __.!__)

1

fa Rl Rz ·

and __.!__ = C.vng -1)(__.!__ _ __.!__) fw Rl Rz

As w ng <0ng, hence focal length oflens in water increase.

The refractive index of water is 4/3 and that of air is 1. Hence, llw > lla·


46. The magnifying power of telescope in relaxed state is

m = fo . , ) ' h .'.s

So, for high magnification, the focal length of objective length should be larger than of eye-piece.

Resolving power of a telescope· = _d_: .. 1.22A

For high resolving power, diameter (d) of objective should be higher.

47. We know that power of lens is a reciprocal of its focal " 1 1 . length, hence P = - =

50 = 2 D

f-100

Since, lens is concave hence, its power will be 2D. If the objective is placed at infinity then u==,v=?,f=50cm

1 1 1 From the formula, --- = -v u f 1 1 1

v 00 -50

v=-SOcm

Thus, concave lens will form an image of the object at infinity at a distance of 50 em.

48. When glass surface is made rough, then light incident on it is scattered in different directions. Due to which its transparency decreases. There is no effect of roughness on absorption of light.

49. Refraction index of diamond w.r.t. liquid zlld = -_-1-, smC

or

.J6 1

.J3 =sine

" 1 " so smC=-=sm4 F2

c = 45°

50 Red glass transmits only red light and absorbs all the colours of white light. Thus, when green flower is seen through red glass it absorbs the green colour, so it appears to be dark.

51. Here, u = -mf, u = v,f = -f

52.

Using lens formula !_..!. = ..!., we have v u f

1 1 ( 1) -1 1 -=-- 1+- =-(m+1)=-(m+1) v f m mf u

or

v 1 or -=--

u m+1

. 'fi . v 1 Lmear magm catiOn =- = ---u (m+ 1)

The lens formula is 1 1 1 -=---! v u

... (i)

The graph between u and vwill be curve as shown in figure . Let the coordinates of P be (x, x), then u = -x and v = x

1 1 1 2 FromEq.(i), -=---=- orx = 2f

f X -X X

:. Coordinate of point Pare (2f, 2f) .

f

53. When lens is in air, then

..!_=(aJ.l. -1)(_!_ _J_J.· f g ' R R · ' l 2

1 ' (1 1J ~=(1.5-1) ---10 R

1 R

2 .

or

or ( ;1- ;J=~ When lens is in water, then

~ = r:Jlg -1J(_!_ _ _!_J f llw Rl Rz

=(~-1)x!=_..!._ 1.33 5 40

f'=40cm 54. For a telescope

I= fo a fe

-~- = _Q2_ => ~ = so 0.5° 0.03

55. !tis observed if Li=Le=~LA 4

and

Here,

and

. _A+om [=--- .

2 LA= 60°

L . 3 t=Le=-LA

4

o = 2x~x60°-60° = 30° m 4

56. Incident ray and finally reflected ray are parallel to each other means, o = 180°

From => =>

1 57. Jl oc-

A.

0=360°-20 180° = 360° -20

e = 90°

1-lrarer < 1-ldenser

Actenser < Ararer

ie, wavelength decreases.

58. allg =-.-1-=>sinC=-

1-.

smC allg

As J..l for violet colour is maximum, so sin C is minimum and hence critical angle C is minimum for violet colour.

59. As A.r > lvv :. ov > or The statement given is true neither, in prism spectrum nor in grating spectrum.

~ I

I I

60.

61.

_!_ = C~-t-1) (__!_ - __!_ J· F ·R1 R2

-1- · = (1.5 -l)(_!_+_!_J

0.06 R1 R2

(Taking R1 positive and R2 negative)

1 1 1 100 -+-=----R1 R2 0.06 X 0.5 3

R1 1 Now, -=- or R2 = 2R1

R2 2

3 100 2R1 3

9 R1 =

200 = 0.045m

R2 = 2R1 = 2 x 0.045 m = 0.09 m.

sinC = _!_ = -1

- = ~ = 0.75 11 4 / 3 4

c = 48° :. Angle with horizontal = 90°- C

= 90° - 48° = 42°

. ·.

62. AI> it is clear from figure distance travelled in. glass place

=DB= OC cosr

OB=-t_ . cosr

90° 63. AI> the ray turns through 90°, therefore, C =- = 45°

2

~-t=-1-=_1_= lin =J2=1.414. sin C sin 45° 1/ -v 2

64. p = _!_ = (!l-1)(_!_ _ _!_J F R1 R2

66.

= (1.4 -1)(_!_ + 100) = 40 = 2.66D 00 15 15

sin(A+om)/2 sin(60°+60°)/2 ~-t= =

sinA/2 sin60° /2

= sin60° = .J3 = 1.73 sin30° .2xl / 2

6 7. From the lens formula !.. -!.. = _!_ = constant v u f

u is always negative, vis positive.

68. The distance in the experiment are measured by a vernier scare provided on the microscope.

-j t i


' '

R1 . Rz ."',. ! I I

=(~-1)_(-~-~) 20 -20

1 1 1 =-+-=-

40 120 30

4 f = -x30 = 40cm 3

70. The angle subtended at the eye becomes 10 times larger. This happens only when the tree appears 10 times nearer.

l . 71. 8=-,8=1 mm.

r

So,

and

1 1t 1 8=-=-x - rad

60° 180 60 1=3m

l x=r=-

8 3

1t 1 = 10 krn. - x -180 60

72. For dispersion without deviation 81 - 82 = 0, ie , 81 = o2

Cl-11 -l)Al = C~-tz -l)A2

or (1.54 -1)4° = (1.72-1)A2

73. The image formed by a concave mirror is always real when the object is placed between focal point and infinity. But when the object is placed between focal point and pole, then the image is always virtual. But when the object is virtual ie, object is placed far away from the focal point, then the image is certainly real. Hence, the required answer is certainly real if the object is virtual.

74. Given that, the refractive index of the lens w.r.t. air, aflg = 1.60 and the refractive index of water w.r.t . air aflg = 1.33 The focal length of the lens in air, f = 20 em. We know that for a lens

_!_ = C~-t-1) (__!_- __!_ J f R1 R2

When the lens is in the air

__!_ = ( ~l -1)(_!_- __!_ 1 20 a g R

1 R

2)

or -=(1.60-1) ---1 ( 1 1 J 20 R1 R2

or __!_ = 0.6ox(_!_ _ _!_J 20 R1 __ R2

When the lens is in the wate~

... (i)

__!_=( 11 -l)(_!_ _ _!_J f' w g R1 R2

I


or ;.-(:~: 1)(~1 ~2 ) or -\ = ( a !-lg -a 1-lw )(J__J_)

f a!-lw R1 Rz

]__ = (1.60 -1.33 )(__!_ _ __!_) f' 1.33 R1 R2

·or ]__- 27 (__!_ _ __!_ ) · ... (ii) f' 133 R1 R2

On dividing Eq. (i),by Eq. (ii), we get

.[_ 0.60x133 20 ·. 27

or f' = 20x2.95cm ~ 60cm.

Hence, its focal length is three times longer than in air.

d 25 75. M=1+-=1+-=6

f 5 76. The limit of resolution of an optical instrument arises on , account of diffraction of light.

77. om= A,!J- = 1.5,A =? sin(A + om)12 sin(A + A)l2

1-1 = sinAI2 = sinA12

2 sin A I 2 cos A I 2

sinAI2 3 A -= 2cos -2 2 A 3

cos-=-2 4

~ = cos-1 (%}A= 2cos-

1 (%) 78. In displacement method,

size of object, 0 = ~11 12

. . 32 =9xi2 cm

J2 = 1

79. Shift =x(1-~J=6(1- 3 ~ 2 ) 1 =6 x -=2 em

3

80. Length of tube 10 em f a + fe = 10 em

Magnification m = fo = 4 fe

fo = 4fe Putting in Eq. (i), I

5fe-10 em or fe = 2 em and fa= 8 em

fa = 8 em, fe = 2 em Hence, L4 and L1 will be used.

,

81. Bi-convex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens. Focal length of bi-convex lens

]_ = (n -1)(2_- __!__) f R1 Rz

1 2 -=(n-1)-f R

f=-R-. 2(n -1)

... (i)

For plano-convex lens

]__ = (n -1)(]_- .!..) f1 R =

R !1 = (n-1) ·

.. . (ii)

Comparing Eqs. (i) and (ii), we see that focal length becomes double.

1 As power of lens P ex: -,--......,-,---:

focal length

Hence, power will become half.

New power = .± = 2D 2

82. In normal adjustment,

L = fo + fe = 200 + 4 = 204 em.

83. For o = (!J-- 1)A

As 1-lv = 1.525 and 1-lr = 1.520

ie 1-lv > 1-lr

ov >or or D2 > D1 or D1 > D2

84. As object is at the centre of the sphere, the image must be ~~~~00~ .

:. Distance of virtual image from centre of sphere = 6 em.

. 1 1 8 6 85. From smC=-=--=0.75,C=4. 0

1-1 1.33

86.

d = 2 x tan C = 2 x 8 tan 48.6°

= 16 x 1.13 = 18.08 em

1inm -4 A=--=5x10 mm

2000 = 5 x 10-7 m = 5000 A

A.'= 2:. = 5000

= 40ooA 1-1 1.25

87. When a ray of light moves from one medium to other, its velocity changes. This change depends on refractive index of the medium. Light travels from denser to rater medium, ie, from medium of higher refractive index to lower refractive index. So, in second (rarer) medium, its velocity in.creases. ~

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23.Ray Optics