Ray Optics

Level 1

1. If an object is not symmetrically placed between the two plane mirrors, inclined at an angle of 400, then the

total number of images formed as

A. infinite B. 7 C. 8 D. 9

2. A man of height 180 cm is standing in front of a plane mirror. His eyes are at a height of 170 cm from the floor.

What should be the minimum length of the plane mirror for the man to see his full length image?

A. 180 cm B. 170 cm C. 90 cm D. 85 cm

3. Two adjacent walls of a big room are mirror surfaced. A man standing at a large distance in front of one

moving horizontally with a speed of 3 m/s. Another man standing in front of the other mirror, at a large

distance, observes the image to be moving with a speed of 5 m/s. What is the actual speed of the insect?

A. 3 m/s B. 5 m/s C. s/m15 D. s/m34

4. A man moves towards a plane mirror with a velocity v in a direction making an angle θ with the normal to the

mirror. The amplitude of velocity of the image relative to man normal to the mirror will be

A. 2v B. θsinv2

C. 2v sin θ D. 2v cos θ

5. A virtual object situated between the pole and the principal focus of a convex mirror produces an image which

is

A. real, diminished and inverted B. virtual, diminished and upright

C. real, magnified ad upright D. virtual, diminished and inverted

6. A convex mirror has a focal length f. A real object is placed at a distance f in front of it from the pole, produces

an image at

A. infinity B. f C. 2f D. f/2

7. A concave mirror of focal length f produces an image p times the size of the object. If the image is real, then

the distance of the object from the mirror is

A. (p – 1)f B. (p + 1)f C. fp1p

− D. f

p1p

+

8. A person standing up in front of a mirror finds his upright image larger than himself. This implies that the

mirror is

A. convex and spherical B. convex and cylindrical with axis horizontal

C. convex and cylindrical with axis vertical D. concave

9. Two plane mirrors are placed perpendicular to each other. A ray of light incident on one mirror at an angle of

incidence θ such that the ray finally reflects from the second mirror. Then the finally reflected ray will be

A. inclined at an angle 2θ with that incident on the first mirror

B. inclined at an angle (90 - θ) with the incident on the first mirror

C. parallel to that incident on the first mirror

D. perpendicular to that incident on the first mirror

10. How far should an object be held from a concave mirror of focal length 40 cm so as to obtain a virtual image

twice the size of the object?

A. 10 cm B. 20 cm C. 50 cm D. 60 cm

11. A ray of light is incident on a prism of refracting angle 600 and refractive index of the material of prism 2 . At

what angle the ray must be incident on the prism so that it suffers a minimum deviation

A. 300 B. 450 C. 600 D.

−

311sin

12. The refractive index of glass is 1.6. A light beam of wavelength 6400 Å in air has a wavelength in glass equal

to

A. 4000 Å B. 4233 Å C. Å51

616400.

.× D. 10240 Å

13. A crown glass prism of refractive index 1.6 is immersed in water (refractive index of water = 1.2). A light beam

incident normally (figure) on the face BC is totally reflected to reach the face AC if

A.

≤ −

32

sinθ 1 B.

≥ −

43

cosθ 1

C.

≥ −

43sinθ 1 D.

<<

−−

43sinθ

32sin 11

14. A glass slab of thickness 6 cm contains the same number of waves as 10 cm thick water layer when the same

monochromatic beam of light is allowed to incident on them. If the absolute value of refractive index of water

is 4/3, what will be absolute value of refractive index of glass?

A. 5/3 B. 20/9 C. 4/3 D. 1.5

15. When light enters from a denser medium to a rarer medium at an angle of incidence α, the reflected and

refracted lights goes in mutually perpendicular directions. The critical angle is

A. sin-1 (cos α) B. sin-1 (tan α) C. sin-1 (cot α) D. sin-1 (1)

16. A ray of light traveling horizontally is incident on a transparent sphere )3µ( = . If it emerges from the other

end of the parallel diameter, the angle of incidence will be

A. 300 B. 600 C.

−

31sin 1 D.

−

31cos 1

A

C B θ

Water

17. A thin biconvex lens has focal length 30 cm. The lens is immersed in water of refractive index 4/3. The

refractive index for the material of lens is 1.5. The focal length of the lens in water is

A. 30 cm B. 7.5 cm C. 15 cm D. 120 cm

18. A convex lens shown in the figure is made up of two types of transparent materials. A point source of light is

placed on its principal axis. If reflections from the boundaries between layers are ignored, the lens will form

A. only one image B. two images

C. infinite images D. no image at all

19. A simple microscope uses

A. a convex lens of large focal length B. a concave lens of large focal length

C. a convex lens of small focal length D. a concave lens of small focal length

20. A person cannot see objects clearly beyond 50 cm. The power of the lens to correct the vision is

A. + 2D B. – 2D C. 0.5 D D. – 0.5 D

21. The aperature of a telescope objective is increased to produce

A. large image B. larger resolving power

C. smaller chromatic aberration D. smaller spherical aberration

22. If a Galilean telescope has objective and eyepiece of focal lengths 200 cm and 4 cm respectively, the

magnifying power of the telescope for the person of normal vision is

A. 42 B. 50 C. 84 D. 100

23. In Q. 22, the length of the telescope for the person of normal vision is

A. 195.26 cm B. 196 cm C. 204 cm D. 200 cm

24. How many images are formed by the lens shown in the figure, if an object is placed on its axis?

A. 1

B. 2

C. 3

D. 4

25. Two plane mirrors M1 and M2 each of length 1 m are placed parallel to each other with their mirror surfaces

facing, as shown in the figure. The separation between the mirrors is 1 cm. If a ray of light incidents on one end

of mirror M2 at an angle of incidence 450, how many reflections the ray will make before going out from the

other end?

A. 50

B. 51

C. 100

D. 101

µ2 µ1

1 m

1 cm 450

26. In a lamp and scale arrangement, a light spot reflects from a plane mirror and falls on a scale 1.5 m from the

mirror. When the mirror is rotated to a small angle, the spot moves a distance 4 cm on the horizontal scale. The

angle through which the mirror is rotated is

A. 1.530 B. 2.670 C. 3.140 D. 0.390

27. A plane mirror is placed in a yz-plane facing towards x-axis. The mirror is moving parallel to y-axis with a

speed of 5 cm/s. A point object is moving in front of the mirror with a velocity k5j4i3v +++=�

. The

velocity of the image with respect to the plane mirror is

A. s/cmk5j4i3v −−−=�

B. s/cmk5j4i3v −+−=�

C. s/cmk5j4i3v ++−=�

D. s/cmk5j4i3v +−=�

28. A short linear object of length b lies along the axis o a concave mirror of focal length f at a distance u from the

pole of the mirror. The size of the image is numerically equal to

A. 2/1

ffub

− B.

2/1

fufb

− C.

−ffu

b D. 2

fufb

−

29. A mirror produces on a screen an image of the sun 2 cm in diameter. If the sun’s disc subtends an angle 0.1

radian on the surface of the earth, then the radius of curvature of the mirror will be

A. 20 cm B. 40 cm C. 200 cm D. 400 cm

30. A small strip of plane mirror A is set with its plane normal to the principal axis of a convex mirror B and placed

10 cm n front of B which it partly covers. An object is placed 20 cm from A and the two virtual images formed

by reflection in A and B coincide without parallax. Find the radius of curvature of B

A. 20 cm B. 22.5 cm C. 27.5 cm D. 30 cm

31. A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive

index of water is 4/3 and the fish is 12 cm below the surface of water, what is the radius of the circle?

A. 5312× B. 5312 ×× C.

7312× D.

7312×

32. A U shaped wire is placed in front of a concave mirror of radius of curvature 20 cm as shown in the figure. The

total length of the image of the wire ABCD is nearly

A. 2.5 cm

B. 6 cm

C. 12.5 cm

D. 15 cm

33. A diver of a motor car A sees the image of another car B in the mirror of focal length 20 cm. The car B is 2 m

broad and 1 m high and is actually 4 m behind the car A. The size of the image of car B as seen in the mirror of

car A is

A. breadth = ;m112

height = m111

B. breadth = ;m212

height = m211

5cm A D

B C

30cm 10cm

C. breadth = ;m114

height = m112

D. breadth = ;m214

height = m214

34. In Q. 33, if the car B is overtaking the car A at the relative speed of

20 m/s, what will be the speed of the image?

A. s/m)11(

202

B. s/m)21(

202

C. s/m)11(

102

D. s/m)21(

102

35. A fish ‘F’ in the pond is at a depth of a 0.8 m from the surface of water )34µ( water = and is moving vertically

upwards with velocity 2 m/s. At the same instant a bid ‘B’ is at a height of 6 m from the water surface ad is

moving vertically downwards with velocity 3 m/s. If at this instant both are on the same vertical line as shown

in the figure, then this instant, the height of B, observed by F (from itself)

A. 6.8 m

B. 6.6 m

C. 5.3 m

D. 8.8 m

36. In Q. 35, the velocity of B, observed by F (relative to itself) is

A. 4.5 m/s B. 5.0 m/s C. 5.5 m/s D. 6.0 m/s

37. In Q./ 35, the velocity of F, observed by B (relative to itself) is

A. 4.5 m/s B. 5.0 m/s C. 5.5 m/s D. 6.0 m/s

38. A rectangular glass block of thickness 10 cm and refractive index 1.5 is placed over a small coin. A beaker ifs

filled with water of refractive index 4/3 to a height of 10 cm and is placed over the glass block. When the coin

is viewed at normal incidence, the apparent position of the coin is

A. 18.33 cm B. 20.0 cm C. 14.17 cm D. 23.33 cm

39. A crown glass prism of refractive indices for red and violet rays 1.514 and 1.523 respectively has a prism angle

100. The angular dispersion produced by this prism will be

A. 0.030 B. 0.060 C. 0.090 D. 0.120

40. An object is placed at 8 cm from the upper face of a glass slab of thickness 6 cm. The lower face of the slab is

silvered as shown in the figure. If the refractive index of the material of slab is 1.5, find the position of the

image

A. 6 cm behind the silvered face

B. 8 cm behind the silvered face

C. 10 cm behind the silvered face

D. 4 cm behind the silvered face

41. For a thin convex lens which one of the following best represents object distance u versus real image distance

v-graph?

A. B. C. D.

Water 0.8 m

0.6 m 3 m/s

Air

2 m/s F

×××× O

µ = 1.5 6 cm

8 cm

v v v v

42. If u and v represents the object and real image distances respectively for a thin convex lens, then the sloe tan θ,

in the adjoining graph represents

A. the focal length f of the lens

B. the reciprocal of the focal length of lens

C. f2

D. 1/f2

43. An achromatic prism is made by combining two prisms P1 (µF = 1.532 and µC = 1.515) and P2(µF = 1.666 and

refractive indices of the flint and crown glasses of the prisms respectively. If the angle of prism P1 is 100, the

angle of prism P2 will be

A. 200 B. 10.60 C. 8.40 . 50

Answers

1. D 2. C 3. D 4. D 5. C

6. D 7. D 8. D 9. C 10. B

11. B 12. A 13. C 14. B 15.

16. B 17. D 18. B 19. C 20. B

21. B 22. B 23. B 24. A 25. D

26. A 27. C 28. D 29. B 30. D

31. D 32. B 33. B 34. B 35. D

36. D 37. A 38. C 39. C 40. C

41. B 42. B 43. D

Solutions Level 1

1. Here, n940360360 ===

θ. This is an odd number and the object is placed on the bisector of the angle between

the mirrors. Hence, the number of images is n – 1 = 4.

3. Let the insect is at the position A and is moving with speed v in a direction θ from x-axis. Mirror M1 measures

the velocity vx = v cos θ = 3 m/s and mirror M2 measures the velocity of insect vy =

v sin θ = 5 m/s

Then, s/m34)5()3(v 222 =+=

θ O

P

u

u/v

4. Velocity of the image of man relative to the man in the direction normal to the mirror is

v cos θ- (- v cos θ) = 2v cos θ

5. Objects (real or virtual) placed in front of a convex mirror between its pole and focus are magnified.

6. For the convex mirror, Given u = - f

Using mirror’s formula, f1

u1

v1 =+

f1

f1

v1 =

−+

2f

v = (virtual image behind the mirror)

7. Given, magnification puv

OI

m === or v = up

Using mirror, formula f1

u1

v1 =+

Here the image of the real object is also real therefore v, u and f all are on the negative side. Hence,

f1

u1

up1 =

−+

−

f11

p1

u1 =

+ f

p1p

u

+=⇒

8. The image formed by a concave mirror for an objected placed between its center of curvature and pole is

always magnified.

10. For the virtual image, 2uv

m +=−

+= u2v −=⇒

Using mirror’s formula

f1

u1

v1 =+

f1

u1

u21

−=+

− or 2u = - f = - 40 u = - 20 cm

11. In the minimum deviation position of the prism, the refractive index of its material is given by

δ+

==µ

2A

sin

2A

sin

rsinisin

m

00

30sin22

60sin22Asinisin =

=

µ=∴

or 21isin = i = 450

12. Given, aµg = 1.6 and λa = 6400 Å g

a

a

gga λ

λ=

µ

µ=µ Å4000

6.1Å6400

ga

ag ==

µλ

=λ∴

13. Given, µg = 1.6, µw = 1.2 Then, 2.16.1

w

ggw =

µ

µ=µ

For the total internal reflection, the angle of incidence i must be greater than the critical angle C.

i.e., ∠i = ∠θ > ∠C But gw

1Csin

µ=

=

µµ

= −−

1612sinsinCor 1

g

w1

= −

43

sinC 1 Hence,

>θ −

43sin 1

14. Given, Number of waves in 6 cm thick glass = Number of waves in 10 cm thick layer of water.

wg

86λ

=λ

or 35

610

g

w ==λλ

But 35

w

g

g

w =µ

µ=

λλ

920

34

35

35

wg =×=µ×=µ∴

15. α + θ = 900 (given)

θα==

µµ

=µsinsin

rsinisin

2

112

θα=

α−α=

sinsin

)90sin(sin

0 = tan α

α

=µ∴tan1

21 and critical angle )(tansin1

sinC 1

21

1 α=

µ= −−

16. Since, AC = CB ∠r = ∠α also ∠i = ∠(r + α) = ∠2 r

Now, 3rsinr2sin

rsinisin ===µ 3

rsinrcosrsin2 =

or 030cos23

rcos == r = 300 and i = 600

17. Given, 89

3/45.1

wa

gaga ==

µ

µ=µ

Focal length of the lens is given by

−−µ=

21 R1

R1)1(

f1

When the lens is in air )i(....R1

R1

)15.1(301

21

−−=

When lens is inside water )ii(...R1

R11

89

'f1

21

−

−=

Dividing equations (i) by (ii),

48/12/1

189

15.130'f ==

−

−= f′ = 120 cm

18. Since the lens is made up of two kinds of transparent material, it has two refractive indices for the incident

beam of light. Hence, there will be two focal lengths of the lens and therefore two images will be observed.

20. The person is suffering with myopia (or near sightedness) in which the person can see near objects but not the

far objects. For him the distance objects (u = - ∞) must be focussed at the distance to which he can see the

objects distinctly. In the given problem v = - 50 cm. This is called the far point for the person having myopic

defect.

Using lens formula, u1

v1

f1 ==

∞−−

−= 1

501

f1

This gives f = - 50 cm and the power of the lens

D2cm50cm100

p −=−

=

21. Large aperture of telescope objective covers large number of rays coming from distinct object. This will

increases the resolving power d/D of the objective. Where d is the diameter of the objective and D is the

distance between object and the objective of the telescope.

22. 504

200ff

Me

0 ===

23. Length of the telescope for the final image at infinity,

l = f0 – fe = 200 – 4 = 196 cm

25. The horizontal distance covered by the ray in one reflection between the mirrors = 1 cm.

∴ Number of reflections for a journey of 100 cm horizontal distance = 100.

Total number of reflection including the first one = 100 + 1 = 101.

26. Dd

2tan =θ

When θ is small Dd

22tan =θ≈θ D2d=θ

1005.14×

=

radian1067.2 2−×= reesdeg1067.2180 2−××

π=

14.31067.2180 2−××= = 1.530

27. Mirror is placed in yz-plane and moving along +y-direction. As shown in the figure, only

x-component of the velocity will be reversed y-and z-components remains unaffected. Hence, the correct

choice is C.

28. The object is short in size, therefore the magnification produced by the concave mirror is given by

dudv

objecttheofSizeimagetheofSize

M ==

where du and dv are the small axial width of the image and object repectively.

Using mirror’s formula

)i(....f1

u1

v1 =+

0u1

dudv

v1

22=−− ∵( focal length of the mirror is fixed)

)ii(....uv

dudvM

2

2−==∴

And also from equation (i) fu

1vu =+ or

ffu1

fu

vu −=−=

∴ Magnification 2

2

2

fuf

uvM

−−=−=

By definition, 2

fuf

bI

objecttheofSizeimagetheotSize

M

−−===

29. A real image on the screen is produced. Therefore the mirror used is concave in nature. Further the sun rays

reaching the surface are parallel rays, therefore its image is formed at the focus of the mirror.

radiusarc=θ rad1.0

f2 ==

cm20rad1.0cm2

f == and R = 2f = 40 cm

31. If C is the critical angle, then hr

Ctan = or Ctan

hr =

But µ

= 1Csin

211

1

CcosCsinCtan

µ−

µ== 1

12 −µ

=

So, 1

hr

2 −µ= cm

7312

134

122

×=

−

=

32. Focal length of the concave mirror cm10220

2R

f −=−==

For the left arm (AB) of the U-tube, u = - 40 cm

and from mirrors formula, u1

v1

f1 +=

401

v1

101

1 −+=

− or cm

340

v1−=

For the right arm (CD) of the U-tube u2 = - 30 cm

So, 301

v1

101

2 −+=− v2 = - 15 cm

Now magnification in AB is A′B′

31

403/40

uv

AB'B'A

1

1 === 3AB

'B'A = cm35=

Also magnification in CD is C′D′

cm21

3015

uv

CD'D'C

2

2 === cm

25

2CD

'D'C ==

Thus, total length of the image of the wire is

'D'C'C'B'B'A ++= 25

34015

35 +

−+= cm635

25

35

35 ≈=++=

33. Mirror fitted in the car is convex f = 20 cm; u = - 400 cm,

u1

v1

f1 += or 1cm

40021

4001

201

u1

f1

v1 −=

−−=−= cm

21400v =

Magnification M 211

40021/400

uv

objectofSizeimageofsize

==−

=

Breadth of car B )Acarofbreadth(211 ×= m

212

2211 =×=

Height of car B m2111

211 =×=

34. Using mirror’s formula, u1

v1

f1 +=

Differentiating with respect to time t, taking of constant,

0dtdu

u1

dtdv.

v1

22 =−−

∴ Speed of image = 2

2

uv= (speed of object)

20)m4(

m214

2

2

×

= Speed of image = s/m)21(

202

−

Minus sign indicates that the image of car B is approaching towards the drive of car A.

35. If an object is at a real depth h in the denser (water) medium and is viewed normally from the rare (air)

medium, then it appears at a distance µh

where µ is the refractive index of denser medium relative to the rare

medium.

When the object is situated in the rare medium and is viewed normally from the denser medium, it appears at a

distance ‘µh’.

Here, the height of B (rare medium) as observed by F (in denser medium) is

H = 0.8 + µh = m8.86348.0 =×+

38. 3/4

105.1

10hhdepthApparent

2

2

1

1 +=µ

+µ

= cm17.14430

320 =+=

39. A white beam of light inside the prism is dispersed into different colours. The angular dipersion for the extreme

colours, red and violet, is given by

ω = δV - δR = (µV – 1)Å – (µR – 1)Å = (µV - µR)Å = (1.523 – 1.514) 100 = 0.090

40. For refractive at the first surface ab of the glass slab, v = - 8 × 1.5 = - 12 cm

The image I1 (v = - 12 cm) will serve as an object for the plane mirror cd and its virtual image is formed at I2 at

a distance u2 [u2 = - 12 + 6) = - 18 cm] behind the mirror.

This image I2 will serve as an object for the refracting surface ab for which v = (- 18 + 6) = - 24 cm. The final

image I is formed at a distance v from the face ab.

R

11

u1

v

1 −µ=−µ

R

1uv

1or

µ−=µ−

∞

µ−=−

µ− 118v

1or cm16

5.12418

vor −=−=µ

−=

Thus, the final image I is formed at a distance 16 cm from the refracting face ab which means at a distance 16 –

6 = 10 cm behind the silvered face cd.

41. For a thin convex lens,

f1

u1

v1 =−

uffu

u1

f1

v1 +=+=

fuuf

v+

=

For the real image u is negative and v is positive. Therefore

fu

uffu

ufv−

=+−

−= or

uf

1

fv−

=

As u increases, the value of v decreases but not linearly.

42. For an achromatic combination of two prisms P1 and P2, the angular dispersions θ1 and θ2 are numerically equal

to θ1 = θ2

(µF - µC) . A1 = (µF - µC) . A2 (1.523 – 1.515) 100 = (1.666 – 1.650)A2

02 5

016.010008.0

A =×=

Level 2

1. White light is passed through a prism of angle 50. If the refractive indices for red and blue colours are 1.641 and

1.659 respectively, calculate the angle of dispersion between them

A. 0.030 B. 0.090 C. 0.120 D. 0.180

2. A ray of light is incident at an angle of 600 on one face of a prism which has an angle of 300. The ray emerging

out of the prism makes an angle 300 with the incident ray. Then the

A. emergent ray is parallel to the face through which it emerges

B. emergent ray is perpendicular to the face through which it emerges

C. emergent ray makes an angle 450 to the face through which it emerges

D. emergent ray makes an angle 600 to the face through which it emerges

3. A beam of light consisting of red, green and blue colours is incident on a right angled prism ABC as shown in

the figure. The refractive index of the material of the prism for red, green and blue wavelengths, respectively

are 1.39, 1.44 and 1.47. The prism will

A. separate part of red colour from green and blue colours

B. separate part of blue colour from red and green colours

C. separate part of the three colours from one another

D. not separate even partially any colour from the other two colours

4. Spherical aberration in a thin lens can be reduced by

A. using a monochromatic light B. using a double combination

C. using a circular annular mask over the lens D. increasing the size of the lens

5. An eye specialist prescribes spectacles having a combination of a convex lens of focal length 40 cm in contact

with a concave lens of focal length 25 cm. The power of this lens combination is

A. + 1.5 D B. – 1.5 D C. + 6.67 D D. – 6.67 D

6. A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index

1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the

glass. The line PQ cuts the surface at point O and PO = OQ. The distance PO is equal to

A. 5 R B. 3 R C. 2 R D. 1.5 R

7. A concave lens of glass, refractive index 1.5 has both surfaces of the same radius of curvature R. On immersion

in a medium of refractive index 1.75, it will behave as a

A. convergent lens of focal length 3.5 R B. convergent lens of focal length 3.0 R

C. divergent lens of focal length 3.5 R D. divergent lens of focal length 3.0 R

8. A diverging beam of light from a point source S having divergence angle α, falls symmetrically on a glass slab

as shown in figure. The angles of incidence for the two extreme rays are equal. If the thickness of the slab is t

and refractive index is ‘µ’, then the divergence angle of the emergent beam is

A. zero

900

450

A

B C

α

i i

S

B. α

C.

−

µ1

sin 1

D.

−

µ1

sin2 1

9. A hollow double concave lens is made of a very thin transparent material. It can be filled with air or either of

two liquids L1 and L2 and having refractive indices µ1 and µ2 respectively

(µ2 > µ1 > 1). The lens will diverge a parallel beam of light if it is filled with

A. air and placed in air B. air and immersed in L1

C. L1 and immersed in L2 D. L2 and immersed in L1

10. A ray of light passes through four transparent media with refractive indices µ1, µ2, µ3 and µ4 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

A. µ1 = µ2 B. µ2 = µ3

C. µ3 = µ4 D. µ4 = µ1

11. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of

identical shape and of same material as P are now added as shown in the figure. The ray will now suffer

A. greater deviation B. no deviation

C. same deviation as before D. total internal reflection

12. An observer can see through a pin hole the top end of a thin rod of height h, placed as shown in the figure. The

beaker height is 3h and its radius is h. When the beaker is filled with a liquid upto a height 2h, he can see the

lower end of the rod. Then the refractive index of the liquid is

A. 25

B.

25

C.

23

D. 23

13. The size of the image of an object, which is at infinity as formed by a convex lens of focal length 30 cm is 1.6

cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance 26

cm from the convex lens, the size of the final image will be

A. 0.8 cm B. 1.2 cm C. 2.0 cm D. 2.4 cm

14. The image formed by an objective of a compound microscope is

A

B C

D µ1 µ2 µ3 µ4

P R

Q

2h 3h

h

2h

A. virtual and diminished B. real and diminished

C. real and enlarged D. virtual and enlarged

15. To get three images of a single object, one should have two plane mirrors at an angle of

A. 600 B. 900 C. 1200 D. 300

16. A planoconvex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now

this lens has been used to form the image of an object. At what distance from this lens, an object be placed in

order to have a real image of the size of the object?

A. 20 cm B. 30 cm C. 60 cm D. 80 cm

17. A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive

index of water is 4/3 and the fish is 12 cm below the water surface, the radius of this circle in cm is

A. 736 B. 7

36 C. 536 D. 54

18. A thin glass (refractive index 1.5) lens has optical power of – 5D in air. Its optical power in a liquid medium

with refractive index 1.6 will be

A. 1 D B. – 1D C. 25 D D. – 25D

19. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be the angles of

minimum deviation for red and blue light respectively in a prism of this glass, then

A. D1 < D2

B. D1 = D2

C. D1 can be less tan or greater than D2 depending upon the angle of prism

D. D1 > D2

20. An isosceles prism of prism angle 1200 has a refractive index of 1.44. Two parallel monochromatic rays enter

the prism parallel to each other in air as shown in figure. The rays emerging from the opposite faces

A. are parallel to each other B. are diverging

C. make an angle 2[sin-1 (0.72)] with each other

D. make an angle 2[sin-1(0.72) – 300] with each other

21. A ray of light is incidence at 600 on a prism of refracting angle 300. The emerging ray is at an angle 30 with the

incident ray. The value of refractive index of the prism is

A. 2/3 B. 4/3 C. 3 D. 32

22. One of the refracting surfaces of a prism of refractive index 2 is silvered. The angle of the prism is equal to

the critical angle of a medium of refractive index 2. A ray of light incident on the unsilvered surface passes

through the prism and retraces its path after reflection at the silvered face. Then the angle of incidence on the

unsilvered surface is

A. 00 B. 300 C. 450 D. 600

1200

23. Focal lengths of objects and eyepiece of telescope are 200 cm and 4 cm respectively. What is the length of

telescope for normal adjustment?

A. 196 cm B. 204 cm C. 225 cm D. 250 cm

Answers

1. B 2. B 3. A 4. C 5. B

6. A 7. A 8. B 9. D 10. D

11. C 12. B 13. C 14. C 15. B

16. A 17. B 18. A 19. A 20. D

21. C 22. C 23. B

Solutions Level 2

1. The angle of dispersion θ = δb - δr where δ = (µ - 1)Å (for small angled prism)

∴ θ = (µb - µr) Å = (1.659 – 1.641) × 500 = 0.090

2. δ = i + i′ - A ∴ i′ = A + δ - i = 30 + 30 – 60 = 0

This means that the angle of emergence is 00 i.e., the emergent ray is perpendicular to the face through which it

emerges.

3. For the normal incidence, the beam refracts through the face AB undeviated and falls on the face AC at an

angle of incidence 450. Now the critical angles for the red, green and blue colours are respectively,

01

r

1r 46

39.11sin1sinC =

=

µ= −− 01

g

1g 44

44.11

sin1

sinC =

=

µ= −−

01

b

1b 43

47.11sin1sinC =

=

µ= −−

The critical angle for total internal reflection at face AC is 450 = C

Here C < Cr but C > Cb and also C > Cg

Hence, the blue and green colours get total internally reflected from the face AC while red colour transmits

from the face AC.

4. Spherical aberration is produced due to large focal length of para-axial zones and small focal length of marginal

zones of the lens. This can be reduced b using a circular annular mask (or aperture) AB so that either the

marginal zones or the para-axial zones of the lens are blocked. Further the monochromatic light eliminates the

chromatic aberration. A doublet combination satisfying the condition 2

1

2

1

ff

−=ωω

also minimize chromatic

aberration. Increasing the size of the lens increases its resolving power.

5. 21

21 f100

f100

PPP +=+= where focal length f is in centimetre.

So, D5.145.225

10040

100P −=−=

−+

+=

6. For the refraction at a spherical surface

Ruv

1221 µ−µ=

µ−

µ

Let PO = x, then u = OP = - x and v = PQ = + x

R

0.15.1x5.1

x0.1 −=

−− or R5x

R5.0

x5.2 =⇒=

7. Refractive index of glass with respect to air is

5.10.15.1

a

gga ==

µ

µ=µ

Refractive index of glass with respect to the given medium is

857.075.150.1

m

ggm ==

µ

µ=µ

Further, the focal length of the concave lens in air is given by

−−µ=

21ga

air R1

R1)1(

f1

R1

R1

R1

)15.1( −=

+−

−−= and that in the medium is

−−µ=

21gm

medium R1

R1)1(

f1

R286.0

R1

R1)1857.0( +=

+−

−−=

R5.3286.0R

for medium +=+=

8. The light rays passing through the transparent slab with parallel faces does not suffer any deviation but only

displaced parallel to itself. Hence, divergence angle of the emergent beam will remain unchanged i.e., same as

that of incident beam.

9. The focal length of the lens is given by

−−µ=

2121 R

1R1)1(

f1

Here 21µ is the refractive index of the lens material 2 relative to the medium 1 in which lens is placed.

Here, RRandRR, 211

221 +=−=

µµ

=µ

So,

+−

−

µµ−µ

=R1

R1

f1

1

12 [For concave lens]

R2

1

12

µµ−µ

−= or )(2

Rf

12

1

µ−µµ

−=

f will remain negative if µ2 > µ1. Where µ2 is the refractive index of the medium filled in the lens and µ1 that of

medium outside the lens. Since, the refractive index of liquid L2 is greater than that of liquid L1, hence choice D

is correct, where lens is filled with liquid L2 and immersed in L1.

10. If a ray coming through a medium A, passing through different media and finally emerges through the same

medium as that of A, then the final emergent ray appears to be parallel to the incident ray.

11. If a ray incidents on a prism such that it undergoes inside the prism, parallel to the base of the prism, then the

ray suffers minimum deviation.

Since the additional prisms Q and R are identical in shape and of same material as that of P, the refracted ray in

prism P suffers no deviation at the interface between P and Q and also between Q and R. Hence, the ray

emerges out of R suffer the same deviation as that prodcued by the prism P.

13. The convex lens focusses the image AB at 30 cm. Concave lens shift this image to A’B’. For the concave lens f

= - 20 cm; u = + 4 cm

Using lens formula,

ui

v1

f1 −= cm5v

41

v1

201 =⇒−=

−

Now, 45

uv

AB'B'A ==

45

uv

AB'B'A ==∴

cm0.26.145

)AB(45

'B'A =×==∴

15. 31360

n =−θ

= 0904

360 ==θ

16. When the curved surface of the planoconvex lens is silvered, it behaves as a concave mirror of focal length F

where

2R

2F1 µ= cm10

5.1230

2R

F 2 =×

=µ

=

Further given that the size of the image is equal to the size of the object i.e., magnification

M = 1

But 1OI

uvM ==−= or v = - u

Using lens formula u1

v1

F1 −=

u2

u1

u1

101 −=−

−=

u = - 20 cm (This is the real image of the object)

17. 431Csin =

µ= But

43

)12(r

rCsin22

=+

=

169

144rr

2

2=

+ cm

736

7312ror =×=

18. In air, the focal length of the lens is

−−=

−−µ=

2121ga

a R1

R1)15.1(

R1

R1)1(

f1

)i(R1

R15.0

21

−=

In liquid medium, its focal length is

−−µ=

21gl

l R1

R1)1(

f1

where 6.15.1

la

gagl =

µ

µ=µ

−

−=∴21l R1

R1

5.16.15.1

f1

)ii(R1

R1

6.11.0

21

−−=

Dividing equation (i) and (ii), we get

81.0

6.15.0ff

a

l −=×−= al f8f −=

But optical power of lens in air is given to be – 5D.

Therefore, D5f1Pa

a −== or cm20cm5

100fa −=−=

m6.1orcm160)20(8fl =−×−= and the optical power of the lens in the liquid medium is

D16.16.1

fP

ll +==µ=

19. The angle of minimum deviation for a thin prism is given b D = (µ - 1) Å

Since µblue > µred ∴ Dblue > Dred or D2 > D1

20. Deviation produced b the prism is given by δ = (i1 + i2) – A

Here i1 = 0 and A = 300 and also

44.1rsinisins

2

2 =µ= since, r2 = 300

sin i2 = 1.44 sin 300 = 0.72 or i2 = sin-1 (0.72) ∴ δ = sin-1 (0.72) - 300

If the rays the emerging from the opposite faces at an angle θ, then

θ = 2δ = 2[sin-1 (0.72) – 300]

21. δ = i + i’ – A

Here, i = 600, A = 300 and δ = 300 i’ = δ - i + A = 300 – 600 + 300 = 00

Thus the ra, emerge normally so that r’ = 00

Since, r + r’ = A = 300 ∴ r = 300

and 32/12/3

30sin60sin

rsinisin

0

0====µ

22. Critical angle,

=

µ= −−

21sin1sinC 11 = 300 = angle of prism (A)

Since, the ray retraces its path, it incidents on the silvered face normally,

∴ r’ = 0 since, A = r + r’ ∴ r = A – r’ = 300

and now, rsinisin

=µ or sin i = µ sin r 2130sin2 0 == ∴ i = 450

23. Length of the telescope = f0 + fe = 200 + 4 = 204 cm.