Optical Instruments

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OPTICAL INSTRUMENTS

Preface

This chapter deals with the application of whatever we learned in ray optics. Solving the problemsof this chapter shows the skill and level of clearity of concept of ray optics. In this chapter youwill be able to know the working principal of various instruments which aids our vision.

This book consists of theoretical & practical explanations of all the concepts involved in thechapter. Each article is followed by a ladder of illustration. At the end of the theory part, thereare miscellaneous solved examples which involve the application of multiple concepts of thischapter.

Students are advised to go through all these solved examples in order to develop betterunderstanding of the chapter and to have better grasping level in the class.

Total No.of questions in Optical Instruments are -

In Chapter Examples.............................................................. 06

Solved Examples ................................................................... 04

Total No. of questions .......................................................... 12


1. INTRODUCTION

1.1 Definition

Optical instruments are used primarily to assist

the eye in viewing an object.

1.2 Types of Instruments

Depending upon the use , optical instrumentscan be categorised in the following way :

OPTICAL INSTRUMENT

ProjectorCamera

Pinhole

Astronomical

Refracting Reflecting

SimpleLens

Terrestrial

Telescope Microscope

Gallalian Binocular

2. FILM OR SLIDE PROJECTOR

It projects real, inverted and magnified image ofan object, when the object is placed between F& 2F and screen between 2F and .

2.1 Magnification :If the sides of slide are a and b andmagnification of projection lens is m , each sidewill become m times so that the area of imageformed.Ai = ma x mb = m2 ab = m2A0

2.2 Intensity :As area of image becomes m2 times that ofobject, the intensity of image will become 1/m2

times that of object, asIA0 = Ii Ai = Ii m

2 A0 Ii = ( I/ m2)

Examplesbased on Projector

Ex.1 A slide projector lens has a focal length10 cm. It throws an image of a 2cm x 2 cmslide on a screen 5 m from the lens. Find (a)the size of the picture on the screen and (b)the ratio of illumination of the slide and thepicture on the screen.

Sol. (a) As here f = 10 cm and v = 5m = 500 cm

So from lens formula v1

– u1

= f1

, we have

5001

– u1

= 101

i.e., u = –

49500

cm

So that m = uv

=

49/500500

= – 49

Here negative sign means the image isinverted with respect to object. Now as hereobject is (2 cm × 2cm ) so the size of pictureon the screenAi = ( 2 × 49 cm) × ( 2 × 49cm)

= ( 98 × 98) cm2

(b) As light energy passing per sec through slideis equal to that in the picture on the screen.IA = Ii Ai

SoiII

= AA i =

ba

mbma

= m2

i.e.iII

=(–49)2 = ( 49 × 49)

i.e. intensity in picture on the screen will be[1/(49x49)] times lesser than that on the slide.This is why in case of projector for observingimage on a screen the source of light mustbe very powerful and the room dark.

v

Projection lens Image/ Screen

>>

>

u

Film/ SlideCondensing lens

Carbon Arc

Mirror

f

> >> >

> >

> >

>

>

>

> >

<>

>>>

Mirror Condensing lens Film/Slide Projection lens Image/Screen

Carbon Arc


3. CAMERA

Camera is of two types viz. pin hole camera andlens camera.

3.1 Pinhole Camera :It is based on recti linear propagation of light andforms real, inverted image on the screen.

= uh0 =

vhi

0

i

hh

= vu

Note :The image formed on the screen is neither ashadow ( as it is not dark) nor a true image( as the rays do not intersect each other andcan not be seen as an aerial image inabsence of screen.

3.2 Lens Camera :

. . . . >>

> >

2F

2F

O I

F

In it a converging lens whose aperture anddistance from the film can be adjusted isused. Usually object is real and between and 2F;so the image is real, inverted,diminished and between F and 2F.(A) Exposure :

For a particular film, a definite amount oflight energy must i.e. incident on the filmfor proper exposure. i.e.,I x S x t = constant where, I = intensityof light, S = area of aperture of lens andt is the exposure time.i.e. ID2 t = constant where D = aperture

(B) f- number:It is the ratio of focal length to theaperture of lens. i.e. f - no. = f/D

Examplesbased on Camera

Ex.2 A pinhole camera with 75 mm high film is tobe used to take a picture of a 10 m hightree. The film is 150 mm from the pinhole.How far should the camera be from the treeto include the full height of the tree?

Sol. In case of pinhole camera

= vI

= uO

i.e., u = v x IO

Here I = 75 mm, O = 10 m and v = 150 mm.,

So u = (10 m) x mm75

mm150 = 20 m

4. MICROSCOPE

It is an optical instrument used to increase thevisual angle of near objects which are too smallto be seen by naked eye.Microscopes are of two types v iz. simplemicroscope and compound microscope.

4.1 Simple Microscope :It is also known as magnifying glass ormagnifier and consists of a convergent lens withobject between its focus and optical centre andeye close to it. The image formed by it is erect,virtual, enlarged and on same side of lensbetween object and infinity.

.....

>

2F F(

Fu

v

0

OI

0 (

D

O

u v

O I


HereMagnifying power

= unaidedeyeforanglevisualMaximuminstrumentwithanglevisual

=0

Now, = vhi =

uh0 with, 0 = 0h /D

M.P. = 0

= uh0 x

0hD

= uD

Now, two possibilities are there :(A) Image is at infinity (far point)If v = , u = f (from lens formula)

So, M.P. = uD

= fD

Note :Here parallel beam of light enters the eye i.e.,eye is least strained.(B) Image is at D (Near point)

In this situation v = – D, so that,D1

–u

1

=f1

or,uD

= 1 + fD

So, M.P. = 1 + fD

Note :Here final image is closest to eye i.e. , eye isunder maximum strain.

4.2 Compound Microscope

v

• • •

ue

uf0

feL

O

Fe

F0

ve

F0

It consists of two convergent lenses of shortfocal lengths and apertures arrangedco-axially.Lens 0f is the objective or field lens and efis the eye-piece or ocular. Objective has smalleraperture and focal length than eye-piece. Theseparation between objective and eye-piece canbe varied.

Magnifying Power = 0

= e

i

uh

×0h

D =

0

i

hh

×eu

D

But for objective, m = uv

i.e.,0

i

hh

= – uv

so, M.P. = – uv

euD

where, eu + u = L

Now two possibilities are there :(A) Final image is at infinity ( far point)

eu = ef M.P. = – uv

efD

where L = u + ef

Note :A microscope is usually considered tooperate in this mode unless statedotherwise.

(B) Final image is at D ( near point)

For eye - piece . ev = D ,

D1

– eu

1

= ef1

eu1

= D1

efD1

M.P.= – uv

efD1 with L = v +

DfDf

e

e

Note :In case, u f0 and, L = v + ue v so that

| M.P.| 0fL

x efD

IMPORTANT POINTS1. As magnifying power is negative, the image

seen in a microscope is always truly inverted,i.e., left is turned right with upside downsimultaneously.

2. Resolving Power : The minimum distancebetween two lines at which they are justdistinct is called limit of resolution andreciprocal of limit of resolution is calledresolving power.

R.P. = x1

1

= 2 sin /


Examplesbased on Microscope

Ex.3 A man with normal near point (25 cm) readsa book with small print using a magnifyingglass, a thin convex lens of focal length5 cm. (a) What is the closest and farthestdistance at which he can read the book whenviewing through the magnifying glass ? Whatis the maximum and minimum MP possibleusing the above simple microscope ?

Sol.(a) As for normal eye far and near points are and 25 cm respectively so for magnifiervmax = – and vmin = – 25 cm. However, fora lens as

v1

– u1

= f1

i.e. u = 1)v/f(f

So u will be minimum when v = min=– 25 cm

i.e. (u)min. = 125/55

= – 6

25= – 4.17 cm.

And u will be maximum when v = max =

i.e. (u)max. = 1/55

= – 5 cm.

So the closest and farthest distances of thebook from the magnifier ( or eye) for clearviewing are 4.17 cm and 5 cm respectively.

(b) As in case of simple magnif ierMP = (D/u). So MP will be minimum whenu = max = 5 cm.

i.e. (MP)min = 525

= 5

fD

And MP will be maximum when u = min= (25/6) cm

i.e. (MP) max. = )6/25(25

= 6

fD1

Ex.4 If the focal length of a magnifier is 5 cmcalculate (a) the power of the lens( b) themagnifying power of the lens for relaxed andstrained eye.

Sol.(a) As power of a lens is reciprocal of focal lengthin m,

P = m10x51

2 = 05.01

diopter = 20 D

(b) For relaxed eye, MP is minimum and will be

MP = fD

= 525

= 5

While for strained eye, MP is maximum andwill be

MP = 1 + fD

= 1 + 5 = 6

5. TELESCOPE

It is an optical instrument used to increase thevisual angle of distant large objects.

Telescopes mainly are of two types v iz.astronomical and terrestrial.

5.1 Astronomical Telescope :

It consists of two converging lenses placedcoaxially with objective having large aperture anda large focal length while the eye- piece havingsmaller aperture and focal length. The separationbetween eye- piece and objective can be varied.

Astronomical Telescope

L

IMF0

U=0 0

Fe

v=f0

f0

F0

feue

) ((

Eye lens

Field lens

VV V

>>

V V V Vy∙ ∙


Magnifying power

= eyeunaidedforanglevisual

instrumentwithanglevisual =

0

,

0 = h/ 0f & = h/(– ue)

M.P. = –

e

0

uf

with L = f0 + ue

Now two possibilities are there(A) Final image is at infinity ( far point)

Here, v = ue = feSo, M.P. = – (f0/ fe) with, L = f0 + fe

Note :Usually , a telescope operates in this modeunless stated otherwise.

(B) Final image is at D ( near point)

Here, v = D D1

– eu

1

= ef1

eu1 =

ef1

Df

1 e

So, M.P. = – e

0

ff

Df1 e

with L = f0 + Df

Df

e

e

Note :1. The above discussion is that of the refracting

telescope.2. Reflecting Telescope :

If the field lens of refracting telescope isreplaced by a converging mirror, then thetelescope becomes a reflecting one, whereM.P. = f0 /fe

3. Infrared Telescope :It is used to see distant objects in dark.

v v> > > > > > >

> >> > > > >ELECTRON BEAM....... .... >

>Eye piece

Light

Focusing electrodesVacuum- tube

Infrared

Radiation

Rock-saltfield lens

Cesium- oxidePhoto- cathode Fluorescent

Screen

f0

fe

IR

5.2 Terrestrial Telescope :

>

IF

IM

f0 ff0

IMF0

f0 2f 2ff0

Erecting lensFieldlens

u =

If a lens of short focal length f is placed at 2ffrom the intermediate image at a distance2f on the other side of it and this image will actas an object for eye- lens which will produceerect image with respect to the object; this lensis cal led erecting lens and as for i tm = – 1, the MP and length of telescope forrelaxed eye will be

M.P. = – e

0

ff

(–1) = e

0

ff

, L = 0f + ef + 4 f

(A) Galilean Telescope

f0f0

IM

L

Here the convergent eye- piece ofastronomical telescope is replaced by adivergent lens. Here M.P. = f0 / fe with ,L = f0 – fe

Note :In this telescope as the intermediate imageis outside the tube, the telescope cannot beused for measurements. This was not thecase for all previous telescope.


(B) Binocular

When two telescopes are mounted parallelto each other so that an object can be seenby both the eyes simultaneously, thearrangement is called a binocular. Here, thelength of each tube is reduced by using aset of totally reflecting prisms which provideintense, erect image f ree from lateralinversion. Binocular gives proper 3-D image.

Examplesbased on Telescope

Ex.5 An astronomical telescope has an angularmagnification of magnitude 5 for distantobjects. The separation between the objectiveand eye- piece is 36 cm and the final imageis formed at infinity. Determine the focallength of objective and eye-piece.

Sol. In case of astronomical telescope if objectand final image both are at infinity.MP = – (f0/fe) and L = f0 + feSo here –(f0/fe) = – 5 and f0 + fe = 36Solving these for f0 and fe, we getf0 = 30 cm and fe = 6 cm

Ex.6 A telescope has an objective of focal length50 cm and an eye- piece of focal length 5 cm.The least distance of distinct vision is 25 cm.The telescope is focused for distinct vision ona scale 2 m away from the objective. Calculate(a) magnification produced and (b) separationbetween objective and eye- piece.

Sol. As objective has focal length 50 cm and objectis 2 m from it , it will form the image ofobject at a distance v such that

v1

– 2001

= 50

1i.e., v = 3

200 cm.

with m0 = uv

= 200)3/200(

= – 31

and as focal length of eye- piece is 5 cm andit forms as image 25 cm in front of it, thedistance of object (image formed by objective)from it will be

251

–

eu1

= 51

i.e., ue = – 625

cm

with me = e

e

uv

= – 6/2525

= 6

and hence(a) Magnification m=m0xme =(– 1/3) x 6 = – 2,

i.e., final image is inverted, virtual, double ofobject and is at a distance of 25 cm in frontof eye lens.

(b) As distance of intermediate image (which isbetween the two lenses), from objective is(200/3) cm while from the eye- lens is(25/6) cm., so separation between theobjective and eye-piece,

L = 3200

+ 6

25 = 6

425 = 70.83 cm.


SOLVED EXAMPLESEx.1 A 35 mm film is to be projected on a 20 m

wide screen situated at a distance of 40 mfrom the film- projector. Calculate the distanceof the film from the projection lens and focallength of projection lens.

Sol. As in case of projector,

m = OI

= uv

So –

cm5.3cm100x20

= u100x40

i.e., u = – 7 cmi.e., film is at a distance of 7 cm in front ofprojection lens.

And from lens formula v1

– u1

= f1

, here we

have

40001

– 7

1

= f1

or f 7 cm = 70 mm[as ( 1/4000) << (1/7) ]i.e., focal length of projection lens is 70 mm.

Ex.2 A compound microscope has an objective offocal length 2 cm and an eye- piece of focallength 5 cm. If an object is placed at adistance of 2.4 cm in front of the field lens,find the magnifying power of the instrumentand length of the tube if (a) final image is atinfinity (b) final image is at least distance ofdistinct vision ( = 25 cm).

Sol. As object is at a distance of 2.4 cm. in frontof field lens of focal length 2 cm. field lenswill form its image at distance v such that

v1

– 4.2

1

= 21

i.e., v = 12 cm.

so that m = uv

= 4.2

12

= – 5

(a) If final image is at infinite ( far point)

For eye- piece,1

– eu1

= 51

i.e. , ue = – 5 cm.

and m = (D/Fe) = (25/ 5) = 5

So, MP = m x m = – 5 x 5 = – 25

and L = v + ue = 12 + 5 = 17 cm.In this situation eye is said to be relaxed andfor a given microscope MP is minimum whilelength of tube maximum.(b) If final image is at D ( Near point)

For eye- piece, 251

–

eu1

= 51

i.e., ue = 625

= – 4.17 cm.

and, m = [ 1 + (D/fe) ] = [1+ (25/5)] = 6

So, MP = m x m = – 5 x 6 = – 30

and, L = v + ue = 12 + 4.17 = 16.17 cm.In this situation eye is said to be strainedand for a given microscope MP is maximumwhile length of tube is minimum.

Note : In case (b) me = eu

D= 6/25

25

= 6 = m

So here MP = m × m = m × me = linearmagnification

Ex.3 In a compound microscope the objective andthe eye- piece have focal lengths of 0.95 cmand 5 cm respectively, and are kept at adistance of 20 cm. The last image is formedat a distance of 25 cm from the eye- piece.Calculate the position of object and the totalmagnification.

Sol. As final image is at 25 cm in front of eyepiece

251

–

eu1

= 51

i.e., ue = – 625

And so, me = e

e

uv

= 6/2525

= 6 ...(1)

Now for objective,v = L – ue = 20 – (25/6) = (95/6)So if object is at a distance u from theobjective,

956

– u1

= 95.01

i.e., u = – 9495

cm


i.e. object is at a distance (95/94) cm infront of field lens.

Also, m = uv

= )94/95()6/95(

= –

6

94 ...(2)

So total magnification,

M = m xme = –

694

x (6) = – 94

i.e., final image is inverted, virtual and 94times that of object.

Ex.4 A Galilean telescope consists of an objectiveof focal length 12 cm and eye- piece of focallength 4 cm. What should be the separationof the two lenses when the virtual image ofa distant object is formed at a distance of 24cm from the eye- piece? What is themagnifying power of telescope under thiscondition?

Sol. As object is distant, i.e., u = – , so

v1

– 1

= 0f1

i.e. v = f0 = 12 cm

i.e. objective will form the image IM at itsfocus which is at a distance of 12 cm fromO. Now as eye- piece of focal length – 4 cmforms image I at a distance of 24 cm from it,

241

–

eu1

= 4

1

ue = 524

= 4.8 cm.

i.e, the distance of IM from eye lens EA is4.8 cm. So the length of tubeL = OA – EA = 12 – 4.8 = 7.2 cm.Now by definition :

MP = 0

0tan

tan

= OA/AB

EA/AB=

EAOA

So, MP = e

0

uf

= 8.412

= 4

10= 2.5

i.e. , the image is erect, virtual, and is at adistance of 24 – 7.2 = 16.8 cm in front ofobjective.

>4.8 cmue

AIMB

L= 7.2 cm

24 cm

f0

I

O

0

0

>

>>>>

>>>

fe

f0=12 cm

E))

((

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Optical Instruments