notes on optics for aieee

7/29/2019 Notes on Optics for AIEEE

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(Optics)

The study of light and vision is called optics.

Light is a form of energy which is propagated as Electromagnetic waves which produces the

sensation of sight in us.

Geometrical optics treats propagation of light in terms of rays and is valid only if wavelength of

light much lesses than the size of obstacles.

i) Light does not require a medium for its propagation

ii) Its speed in free space (vaccum) is 3 x 108m/s

iii) It is transverse in nature

In the spectrum of e.m.w. it lies between u.v. and infra-red region and has wavelength between

4000 to 7000O

A . i.e ( )mtom 7.04.0

Indigo is not distensible from blue.

BASIC - DEFINATIONS

Source:

A body which emits light is called source. Source can be a point one (or) extended one.

(a) Self-luminous-source: The source which possess light of it own.

Ex:- Sun, Electric arc, Candle, etc.


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(b) Non-luminous-Source: It is a source of light which does not possesses light of its own but acts as

source of light by reflecting the light received by it.

Ex: Moon, object around us, Book.etc.

Isotropic Source: It gives out light uniformly in all directions.

Non-isotropic Source: It do not give out light uniformly in all direction.

Medium: Substance through which light propagates is called medium

Ray: The straight line path along with the light travels in a homogeneous medium is called a ray.

A single ray cannot be propagated form a source of light.

Beam: A bundle can bunch of rays is called beam it is called beam it is of following 3 types

Convergent-beam: In this case diameter of beam decreases in the direction of ray

Divergent Beam: It is a beam is with all the rays meet at a point when produced backward and the

diameter of beam goes on increasing as the rays proceed forward.

Parallel Beam: It is beam in which all the rays constituting the beam move parallel to each other

and diameter of beam remains same

Object: An optical object is decided by incident rays only. It is if two kinds


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Real Object: In this case incident rays are diverging and point of divergence is the position of real

object.

Virtual Object: In this case incident ray are converging and point of convergence is the position

of virtual object. Virtual object cannot be seen by human eye be cause for an object can image to

be seen by eyes, ray received by eyes must be diverging.

Image: An optical image is decided by reflected (or) refracted rays only. It is of two types.

(a) Real Image: This is formed due to real intersection of reflected (or) refracted rays, Real image

can be obtained on screen.

Virtual-Image: This is formed due to apparent intersection of reflected (or) refracted light rays.

Virtual image cant be obtained on screen.

(Note: Human ray cant distinguish between real and virtual image because in both case rays are

diverging)


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REFLECTION:

The phenomenon by virtue of which incident light energy is partly or completely sent back into

the same medium from which it is coming after being obstructed by a surface is called reflection.

The direction of incident energy is called incident ray and the direction in which energy is thrownback is called reflected ray. It is of two types.

LAWS OF REFLECTION:1) First Law: The incident ray, the reflected ray and the normal to the reflecting surface at the

point of incidence, all lie in one plane which isr

' to the reflecting surface.

2) The angle of incidence is equal to the angle of reflection ri = .


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Note:

1) The laws of reflection are valid for any smooth reflecting surface irrespective of geometry.

2) Whenever reflection takes place, the component of incident ray parallel to reflecting surface

remains uncharged, while component perpendicular to reflecting surface (i.e. along normal)

reverse in direction.

^^^

1 kzjyixr ++=

,^^^

2 kzjyixr +=

3) Vector form of laws of Reflection:

^^^^^

.2 NNIIR

=

R Unit vector along the reflected ray

^

I Unit vector along the Incident ray

^

N Unit vector along the normal ray

Image formed by a plane mirror:


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a) Point Source: For construction of image of a point source it is sufficient to consider any two

rays falling on mirror. The point of intersection of corresponding reflected rays give the position

of image as shown in figure.

OA = AI

( )ABIABD

Image I lies as much behind the mirror as the object is in front of it.

b) Extended source:

Characteristics of the image formed by a plane mirror:

1) The image formed by a plane mirror is Virtual

2) The image formed by a plane mirror is Erect

3) The image formed by a plane mirror is of same size as object.

4) The image formed by a plane mirror is at the same distance behind the mirror is the object is

infront of it.

5) The image is laterally inverted (i.e.) right appear as left and vice-versa.

6) Note: If two plane mirror faring each other are inclined at an angle with each other, then


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number of images are formed due to multiple reflection. This principle is used in the toy

kaleidoscope.

(a) If

360is even integer, then number of images formed is 1

360=

Ex: If0

60= then 516160

360 ===

(b) If

360is odd integer, then number of images formed is

360=

Ex: If040= (which is not the complete part of 1800) then 9

40

360==

Deviation ( ): The angle between incident and reflected (or) refracted ray is termed as deviation.

For reflection i2=

Cases: When i = 0 (Normal incidence)

=max

When2

=i (Grazing incidence)

0min =

Multiple Reflection:

= inet i = deviation due to single reflection.Note while summing up, sense of rotation is taken into account.


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Q: 1) Two plane mirror are inclined to each other such that a ray of light incident on the first mirror and

parallels to the second is reflected from the second mirror parallel to the first mirror. Determine

the angle between the two mirror. Also determine the total deviation produced in the incident ray

due to the two reflections.

Solution:

From figure 1803 =

060=

i21

=

== 120302180 0 A.C.W.

00

2 1203021802 === i

( ) =+= 12024012 ornet

Or From fig. 60180180 +=+= ( ) = 1202400 or

Q: 2) Calculate deviation suffered by incident ray in situation as shown in figure, after three successive

reflections?

Solution: F,B.D


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i21 =

== 100502180 0

== 002 140202180

== 002 160102180

( ) =++= 0260100160140100 ornet

Q: 3) Two plane mirrors are inclined to each other at an angle . A ray of light is reflected first at one

mirror and then at the other. Find the total deviation of the ray?

Solution: Let = Angle of incidence for M1

= Angle of incidence for M2

=1 Deviation due to M1

=2 Deviation due to M2

From figure

21 =


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22 =

Also ray is rotated in same secure (i.e.) anticlockwise

21 +=Net Now in OBC

22 += 0180=++ COBBCOOBC

( ) += 22Net ( ) ( )000 1809090 =++

( ) 22 = Net =+

22 =Net

Velocity of Image:Let

xO/m = x-co-ordinate of object w.r.t. mirror

xI/m = x-co-ordinate of image w.r.t. mirror

yO/m = y-co-ordinate of object w.r.t. mirror

yI/m = y-co-ordinate of image w.r.t. mirror

For plane mirror

xO/m = -xI/m

Differentiating both sides w.r.t. time (t)

( ) ( )mImO xd

dx

d

d// =

x

mI

x

mO VV

=

//

mxIxmxOx VVVV

=

OxmxIx VVV

= 2

Similarly yI/m = yO/m

Differentiating both sides w.r.t. time we get


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y

mO

y

mI VV

=

//

In nutshell, for solving numerical problems involving calculation of velocity of image of object

with respect to any observer, always calculate velocity of image first with respect to mirror using

following points.

11

/

11

/

=

mOmI VV

1

/

1

/

=

mOmI VV

1

/

11

//

+

= mImImI VVV

Velocity of image with respect to required observer is then calculated using basic equation for

relative motion.

BABA VVV

=/

Note: If the velocity of the object (w.r.t mirror) is not in a direction normal to the mirror, then the

velocity of the object can be resolved into two components one normal to the mirror (vn) and the

other along the mirror (vp). The image has velocities Vn and VP, normal to and along the mirror.

Q: 1) Point object is moving with a speed V before an arrangement of two mirrors as shown in figure.

Find the velocity of image in mirror M1 w.r.t. image in mirror M2?

Solution: 212/1

= VVV = sin2V

F.B.D


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Angle between21 II

VandV

Is 2 their magnitude is V.

Q: 2) Find the velocity of image of a moving particle in situation as shown in figure.

Solution: Analysis:

For component of velocity of image2/1 to mirror

02

= VVV mI

( ) ( ) smVI /106222/1 ==

For component of velocity of image parallel to the mirror

( ) smVI /811 =

Velocity of time ( ) ( )221 nIII VVV +=


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m16464100 =+=

=

54tan 1

Q: 3) Two plane mirror are placed as shown in the figure below:

A point object is approaching the intersection point of mirror with a speed of 100cm/s. The

velocity of the image of object formed by M2 w.r.t. velocity of image of object formed by M1 is:

Solution: The components of various velocities are as shown in the figure below


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2IMV

is given by the vector sum of components of velocity of image w.r.t. M2 along the normal

and r1 to the normal.

++

+=

^00

^02

^00

^02

37cos37sin10037cos10037cos37sin10037sin1002 jijiV IM

scmji /4828^^

+=

1212 , IMIMIMIM VVV

=

sec/48128^^

cmji

+=

Q: 4) In the situation show in figure, find the velocity of image?

Solution: Along x direction, applying

( )mmi VVVV == 0

( ) ( )000 30cos560cos1030cos5 =iV

( ) smiVi /315^

+=

Along y-direction V0 = Vi


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smjVi /560sin10^

0 ==

Velocity of the image ( ) smji /5315^^

++=

Q: 5) An object moves with 5m/s towards right while the mirror moves with 1m/s towards the left as

shown. Find the velocity of image.

Solution: Take as +ve direction.

0VVVV mmi =

( ) ( ) 511 =iV

smsmVi /7/7 = and direction towards left.

Q: 6) Find the region on y-axis in which reflected rays are present object is at A(2, 0) and MN is a

plane mirror, as shown

Solution:


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( )0,6'=A

( )6,0' =M ( )9,0'=N

Q: 7) An object moves towards a plane mirror with a speed v at an angle 600

to ther1 to the plane of

the mirror. What is the relative velocity between the object and the emage?

a) V b) V2

3c)

2

Vd)

2

V

Solution: IOOI VVV

=

++

^0

^0

^0

^0 60sin60cos60sin60cos jViVjViV

Q: 8) A ray of light making angle 200 with the horizontal is incident on a plane mirror with itself

inclined to the horizontal at angle 100, with normal away from the incident ray. What is the angle

made by the reflected ray with the horizontal?

Solution: AO = Incident ray

OB = Reflected ray


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The reflected ray goes along the horizontal. Hence angle made by the reflected ray with the

horizontal is zero.

Q: 9) A ray of light making angle 100 with the horizontal is incident on a plane mirror making angle

with the horizontal. What should be the value of , so that the reflected ray goes vertically

upwards?

a) 300 b) 400 c) 500 d) 600

Solution:

Number of Images Formed by two Inclined-Plane Mirrors:a) When mirror are parallel: In this case, infinite images are formed due to multiple reflections.


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b) When mirror are perpendicular: In this case, three images are formed. The ray diagram is

shown.

Note that the third image is formed due to rays undergoing two successive reflection. Also, object

and its images lie on a circle whose equation is given by 2222 bayx +=+ .

When an object is placed in front of arrangement of three mutually perpendicular mirror, then

total seven images are formed.

Further, object and its image lie on a sphere whose equation is given by

222222 cbazyx ++=++ , where a, b and c are co-ordinates of object.

Minimum size of Mirror to see Full-Image:


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AB is the person with E as his eyes,

M1 M2 = plane mirror infront of him.

For the length of the mirror to be minimum, the rays coming from the extreme top and bottom

portions of his body. (i.e.) A and B, Should after reflection, be able to just enter his eyes.

The light ray AM, is incident ray and M1E the reflected ray.

So1111 NEMNAM =

As1111' NEMandNAMs are similar.

( )AEENSayxEM2

1111 === ..(1)

Similarly the light rays BM2 and M2E are incident and reflection rays respectively

So2222 NEMNBM =

2222NEMandNBMsS are similar

( ) ( )BEENSayyEM2

1212 === (2)

Adding equation (1) and (2) yield

=+ length of mirror ( ) ( )2

1

2

1

2

1==+= ABBCAE (Height of person)

Note:- Minimum size is independent of distance between man and mirror.

Q: 1) A plane mirror is inclined at an angle with the horizontal surface. A particle is projected from

point P (see fig.) at t = 0 with a velocity v at angle with the horizontal. The image of the

particle is observed from the frame of the particle projected. Assuming the particle does not

collide the mirror, find the (a) time when the image will come momentarily at rest w.r.t. the

particle (b) path of the image as seen by the particle.


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Solution:

(a) The image will appear to be at rest w.r.t. the particle at the instant, the velocity of the

particle is parallel to the mirror.

tan=x

y

V

V

tan

cos

sin =V

gtV

( )g

Vt

tantancos =

(b) St. liner1 to mirror

Q: 2) An a oblong object PQ of height h stands erect on a flat horizontal mirror. Sun rays fall on the

object at a certain angle. Find the length of the shadow on screen placed beyond the shadow on

the mirror.

Solution: PS = Shadow on the mirror

P Q = Inversed shadow of PQ on the screen


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Let = angle of incidence

Then PS = h tan and QS = h sec

From the properly of image P Q = ( ) hh 2cossec2 =

Q: 3) A plane mirror is placed at parallel of y-axis, facing the positive x-axis. An object starts form

(2m, 0, 0) with a velocity of smji /22^^

+ . The relative velocity of image with respect to

object is along

Solution: ( ) ( )220 22 +==

IVV

smV /220 =

Relative velocity of image with respect to object is in negative x-direction as shown in figure.


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Q: 4) A reflection surface is represented by the equation222 ayx =+ . A ray traveling in negative x-

direction is directed towards positive y-direction after reflection from the surface at some pointP. Then co-ordinates of point P are

Solution: From figure

2

9,

2

9 == yx

=2

,2

qqP

Q: 5) A ray is traveling along x-axis in negative x-direction. A plane mirror is placed at origin facing

the ray. What should be the angle of plane mirror with the x-axis so that the ray of light offer

reflecting from the plane mirror passes through point (1m, m3 )?

Solution:

Q: 6) Two plane mirror A and B are aligned parallel to each other as shown in the figure. A light ray is

incident at an angle 300 at a point just inside one end of A. The plane of incidence coincides with

the plane of the figure. The maximum number of times the ray undergoes reflection

(including the first one) before it emerges out is____

Solution:


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3

2.030tan2.0 0 ==d

303/2.0

32.. == reflectionofNoMax

REFLECTIN FROM CURVED SURFACE: (Spherical Surface only)

A curved mirror is a smooth reflecting part of any geometry. The nomenclature of curved mirror

depends on the geometry of reflecting surface. There are different types of curved mirror like

paraboloidal, ellipsoidal, cylindrical, spherical .etc.


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Sign-Conversion:

Rules for Ray-Diagrams

1) A ray of light parallel to principal axis passes (or) appears to pass through four after

reflection.

2) A ray of light passing through focus (or) appears to pass through focus becomes parallel to

principal-axis after reflection.

3) A ray of light passing through (or) appears to pass through centre of curvature is reflected

back.


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4) A ray of light hitting pole is reflected making equal angle with principal oxis

Note: 1) Focal length and radius of curvature of plane mirror =

2) Concave mirror = Convergent mirror

Convex mirror = Divergent mirror

Relation between focal-length and radius of curvature:

=

2

Rf Both for concave and convex mirror.

Mirror formula (or) Mirror Equation:The relation between u, vand fof a mirror is known as mirror formula

+=

vuf

111

Relation between the speeds of object and image formed by a spherical mirrorWe know that, mirror formula is given by

fvu

111

=+ .(1)

Differentiating both sides w.r.t. time (t), we get

0.1

.1

22=

d

dv

d

du

u

0.1

.1

22==

dt

dv

udt

dv

dt

dy

u

v

dt

dv.

2

2

= (2)


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Since timeofspeedvdt

dvi ==

objectofspeedvdt

du== 0

0

2

.vu

vvi

= .(3)

From equation (1),

( )fx

f

u

vor

fu

fuv

=

=

Hence equation (2) become

0.vfu

fvi

=

Linear magnification: It is defined as the ratio of the size (or) height of the image to the size (or)

height of the object.

imageofheight

imageofheight

imageofsize

imageofsizem ==

=o

IM

Magnification produced by concave mirror:

='' BA image of object AB

ABpS and pBA '' are similar

PA

PA

AB

BA ''' = (1)

Applying sign conversion

uPAAB =+= 0

vPABA == '1''


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Equation (1) can be rewritten as

uv

OI =+

v

O

I=

=

u

vm same for convex-mirror also.

Magnification in terms of u, v and f

a) As we know thatfvu111 =+

Multiplying both sides by u we get

f

u

v

u

u

u=+

f

u

v

u=+1

f

fu

f

u

v

u == 1

fu

f

m

v

=

Sinceu

vm =

( )

=

=uf

fmor

fu

fm

b) As we know thatfvu

111 =+

Multiplying both sides by V, we get

f

v

v

v

u

v=+

f

v

u

v=+1


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f

fv

f

v

u

v == 1

Sinceu

vm =

( )f

vfmor

f

fv =

=

Note: a) +ve magnification mean both object and image are upright

b) ve magnification means, object and image have different orientation (i.e.) if object is

upright, then image is inverted.

LATERAL-MAGNIFICATION (mL)

0L

L

objectoflength

imageoflengthm iL ==

For extended objects the lateral magnification can be obtained by independently imaging the two

end points and calculating the length of the image. There is no direct formula to obtain the

magnification.

However, if the length of the object is small, them the lateral magnification can be directly

obtained from equation

fvu

111=+

Differentiating both sides, we get

022

=dv

u

du

== Lm

u

v

du

dv2

2

Q: 1) What do we do if the size of the object is large as compared to the distance u?

Analysis:

For extended object


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BA

BA

uu

VVm

=2

For tip A

( )Lxu +=

2

Rf =

BVV =

fuv

111=+

RlxvB

211 =

+ from which VB can be obtained

Subtracting VB from VA, we can calculate the length of the image.

Combinations of mirrors:

What do we do if we have a combination of mirror? If an object is placed between the mirrors,

how do we find the final position of he image?

Analysis: In such situations, we need to simply solve for the reflection at each of the mirror

keeping in mind that the image formed by the first mirror is the object of the second mirror and so

on.

Case must be taken to correctly apply the sign conversion at each of the mirror.

Q: 1) Find the velocity of image in situation as shown in figure?

Solution: smiiiV /1129^^^

0 =

+=


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^

2 iVm =

m/s

( ) 23020

20

=

== uff

m

( )11

/2

11/

=

MOMI VmV

= -(-2)2 11^

i

= -44^

i m/s.

( )I

mI

n

mImI VVV

+

=

///

( ) smjx /24122 ==

I

mI

n

mI

n

mI VVV

+

=

///

smji /2444^^

=

^^

/ 22444 ijiVVV mmII =

==

=


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=

ji 2446^

Q: 2) A thin rod of length3

fis placed along the principal axis of a concave mirror of focal-length f

such that its image just touches the rod, calculate magnification?

Solution: Since image touches the rod, the rod must be placed with one end at centre of curvature.

Case I Case II

3

5

32

fffu

=

=

ff =

( )

( ) 25

3

5

3

5

f

ff

ff

fu

fuv =

=

=

( )

( ) 23

23

5

23

5

f

ff

ff

uu

VVm

CA

CA =

=

=

3

7

32 fffx =

+=

ff =

( )

( ) 47

3

7

3

7

f

ff

ff

fu

fuV =

=

=


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( )

( ) 4

3

23

7

24

7

=

=

= ff

ff

uu

VV

M CA

CA

CONCEPTUAL POINTS

It a hole is formed at the center of mirror, the image position and size will not change.

The intensity will reduce depending on the size of the hole.

For all object positions a convex-mirror forms a virtual and erect image

PROBLEMS OF MIRRORS

Q: 1) 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 mirror. The size of image approximately is

Solution:

22

=

=

uf

f

u

VMaxial

2

=uf

f

O

I

=

=

22

uffbI

uff

bI

Q: 2) Two spherical mirrors M1 and M2 one convex and other concave having same radius of curvature

R are arranged coaxially at a distance 2R (consider their pole separation to be 2R). A bead of

radius a is placed at the pole of the convex mirror as shown. The ratio of the sizes of the first

three images of the bead is

Solution: The first image is formed due to the reflection from concave mirror M2


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( ) RRV =

+

2

2

11

1

RRV 2

4

2

11

1

=

RV 2

31

1

=

3

1

2

3

2

3

21/

1

=

=

R

R

mR

V.

object distance3

4

3

22

RRR ==

=

+

2

2

3

4

11

2RRV

RRV 4

221

2

+=

11

42

RV =

11

3

3

411

4

2

22 =

==

R

R

u

Vm

11

32 = m

So radius of second image

113

.

11

32

aaa ==

Similarly radius of third image is41

3

aa =

41

1:

11

1:

3

1 Answer

Q: 3) When an object is placed at a distance of 60cm from a convex spherical mirror, the magnification

produced is 1/2. where should the object be placed to get a magnification of 1/3?


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Solution: cmu 60=

u

Vm =

602

1

=

V(or) cmV 30+=

60

1

30

1

60

1111=+=+=

vuf

cmf 60+=

In second case

3100

31 uV

uVm ===

Asfvu

111 =+

60

131 =uu

cmu 120=

Q: 4) Two objects A and B when placed one after another in front of a concave mirror of focal-length10cm, form images if same size. Size of object A is 4 times that of B. If object A is placed at a

distance of 50cm from the mirror, what should be the distance of B from the mirror?

Solution: For object A For object B

11

2

uf

f

h

hm

==

11

2

'

''

uf

f

h

hm

==

1

2

2

2

1

1

1

2

' uf

uf

h

h

h

h

m

m

==

As 111 4hh = and

1

22hh = , cmf 10=

cmu 501

=

5010

10

4

1 2

+

=u

cmu 202 =


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Q: 5) A concave mirror of focal length 10cm is placed at a distance of 35cm form a wal. How far from

the wall should an object be placed to get in image on the wall?

Solution: cmVcmf 35,10 ==

Vfu

111 =

14

1

35

1

10

1 =+=

cmu 14=

Distance of the object form wall

= 35 14 = 21 cm

Q: 6) An object is placed at a distance of 36cm form a convex mirror. A plane mirror is placed in

between so that the two virtual images so formed coincide. If the plane mirror is at a distance if

24cm from the object, find the radius of curvature of the convex mirror.

Solution: cmuOP 36==

cmPIV 12+==


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18

1

36

31

12

1

36

1111

=+

=+=+= Vuf

cmf 18=

cmfR 361822 ===

Q: 7) A convex mirror of focal length f forms an image which is1

times the object. The distance of

the object which is

n

1times the object. The distance of the object from the mirror is

Solution:u

V=+=

1

u

V =

uVf

111 +=

111 +

=uf

( ) fu 1=

Q: 8) An object of size 7.5cm is placed in front of a convex mirror of radius of curvature 25cm at a

distance of 40cm. The size of the image should be

Solution:uf

f

O

I

== 40=u

( )

( )

( )

( ) ( )402/252/25

2/

2/

5.7 =

=

uR

RI

cmI 78.1+=

Q: 9) The image formed by a convex mirror of focal length 30cm is a quarter of the size of the object.

The distance of the object from the mirror is


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Solution:uf

fm

=

u++=

+

30

30

4

1

cmu 90=

Q: 10) A concave mirror of focal length f(in air) is immersed in water ( )3/4= . The focal length of

the mirror in water will be

a) f b) f3

4c) f

4

3d) f

3

7

Solution: On immersing a mirror in water, focal length of the mirror remains uncharged.

Q: 11) An object is 20cm away form a concave mirror with focal-length 15cm. If the object moves with

a speed of 5m/s along the axis, then the speed of the image will be

Solution:15

1

20

11

=

V

cmV 60=

0

2

.Vu

VVi

=

( )5.20

602

=

sm/45=

Q: 12) A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical

when Sun-light falls normally in the mirror, it is focused at distance of 32cm form the mirror. If

the tank filled with water

=

3

4 upto a height of 20cm, then the Sunlight will now get

focused at

Ans: 9cm above water level

Q: 13) A small piece of wire bent into an L shape with upright and horizontal portions of equal-

lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of


38/73

curvature is 10cm. If the bend is 20cm from the pole of the mirror, then the ratio of the lengths of

the images of the upright and horizontal portions of the wire is

Solution: cmR

f 52

10

2===

For part PQ

01L

uf

fL

=

( ) 32055 0

0

LL =

=

For part QR

0

2

2 Luf

f

L

=

( ) 9205

5 00

2

LL =

=

1

3

2

1 =L

L

CONCEPT OF NEWTONS FORMULA (FOR A MIRROR)

In this formula, the object and image distance are expressed w.r.t. focus. Consider an object O

kept beyond C of a concave mirror, and whose image is formed at I with in C.


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Let OF = x and IF = y

From triangle OMC

( ) sinsinsinOMOMOC

=

= (1)

And from triangle ICM

sinsinIMIC

= .(2)

Dividing equation (1) and (2) yields

IP

OP

IM

OM

IC

OC== (since M is close to P)

(or)yf

fx

yf

fx

++

=

yffyxfxyfyxffx +=+ 22

(or) 2fyx =

yxf=

1) As2

fyx = (or) yx1

(i.e.) The distance of object and image form the focus are inversely proportional to each other. In

other words, the more the object distance (from the focus), the less will be the image distance

(from the focus) and vice versa

2) If yx ;0 and if 0; yx . If the object is at focus the image is a far off distance

and vice-versa.

3) From fxiffxy == ;2 , then fy= .

Thus, if the object be at C, then image will also be at C (for a concave mirror) and if the object is

at P, then the image will also be at P (for a convex mirror)

4) Since2f is necessarily +ve for both types of mirror, so x and y bear the same sign, which

implies that both the object and the image always lie an the same side of focus.


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A) GRAPH OF |x| Versus |y| :

Since 2

fxy=represents a rectangular hyperbola, existing in the first and third quadrant

( 2f being positive).

The graph of |y| vs |x| will be a rectangular hyperbola existing only in the first quadrant.

B) GRAPH OF U Versus V :

Since2fxy=

( )( ) 2ffvfu =

For a convex mirror, u is always negative and V is always positive. Further f is also positive.

Putting u = and yV = we have

( ) ( ) 2ffyfx =

This is the equation of a rectangular hyperbola with its origin shifted to ( )ff, and x being

always negative while y lies between O and f. (see figure) for a concave mirror, u is always

negative, vcan be positive (or) negative, f is negative

ffandyvxu === ,

We have, form

( )( ) 2ffvfu =

( )( ) 2ffyfu =++


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Or ( )[ ] ( )[ ] 2ffyfx =

Evidently it is again an equation of a rectangular hyperbola with origin of coordinates shifted to

the point (-f, -f) (see figure )

3) GRAPH OF1

VERSUSu

1

From mirror formula

fuv

111 =+

Putting x=1

and y=1

, we have

yyx

1=+

It is the equation of a straight line having a slope +1 (or) -1 according as uand vbear the same

(or) opposite signs. The intercepts on x and y axis are each ( )f

orf

11 + according as the

object and image are to the right (or) left of the mirror.

For a concave mirror:

u is always ve

vcan be positive (or) negative and fis ve.


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For a convex mirroru is always negative v is always positive and fis always positive.

CONCEPT OF CRITICLA ANGLE

When a ray of light is traveling form denser medium to rarer medium, it get refracted and the ray

derivates away form the normal.

If we keep increasing angle of incidence then at an angle, the angle of refraction becomes 90 0 .

This is known as Critical Angle (c).

When angle of incidence is increased, further the ray gets reflected back in the same medium.

This phenomena is known as T.I.R.

According to Snells Law


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r

ia

b

sin

sin=

090sin

sin Ca

b i=

Cab isin=

=

C

ab

isin

1

C depends on colour and and temp

CRed > Cviolet CRed < Cviolet If temp C

Q: 1) The sum (diameter d) subtends an angle radius at the pole of a concave mirror of focal lengthf. Find te diameter of the image of sun formed by mirror?

Solution:fuv

111=+ we get

fv =

11(u is very large so 0

1

Or fv = It means image is formed at focus

Taking '' f as radius and using

franddwhenr

=== ll

fdorf

d ==


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REFRACTION AT SPHERICAL SURFACE:

From sle OBC and IBC

We have +=i

And ( ) rrorrr =+=

From Snells law

r

i

sin

sin

1

2 =

ri sinsin 21 = For small angle of incidence I, we can write

rrandii sinsin

ri 21 =

[ ] [ ]r=+ 21

As i is small, and so rand , are also small. Thus

( ) tantan +=+

hh

++

=

And ( )V

hhr =

=

++

v

h

R

h

R

h

u

h21

After simplifying we get


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Ru1212 =

Rv

11

1

2

1

2 =

=u

112

1

2

1 This formula is derived for convex surface and for

real

Image

From denser to rarer medium

Ru2121

=

Q: How can we derive a mathematical expression for the equation of a ray in the medium? The

medium is of variable refractive index. Ray of light is incident at an angle at air mediuminterface?

Analysis: Here two cases a rise. Refractive index is varying either as function of y (or) function

of x.

Case-I: ( ) ( )..eiyf= Refractive index varies with y

At some height h angle of incidence is y and refractive index is ( )yf from Snells Law

= sin constant

( ) yyf sinsin1 = ..(1)

Slope of curve at A


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( )yd

dy= 90tan

d

dyy= cot

From equation (1)

( ){ }

sin

sin 22

=yf

d

dy

Case-I: ( ) ( )..eiyf= Refractive index varies as function ofx.

According to Snells Law

= sin constant

For initial refraction at the air medium interface

( )00 90sinsin1 =

( )00 90sinsin =

00 cossin =

Here 0 angle of refracting ray at point A with OX

So0

0

sincos

=

And 20

2

0

sin1sin

=

Now Snells Law at M gives


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( ) 00 sinsin =xxf

Or( ) 20

20 sin1sin

=

xfx

( )xfx

22

0 sinsin

=

Now slope of tangent at M is given as

xd

dytan=

( ){ }

22

0

2

22

0

sin

sin

+

=xfdx

dy

Q: 1) If y+= 1 and ray of light is incident at grazing incidence at origin, then find equation of

path of refracted ray.

Solution: We can use result derived above for which

( ) 0901 =+= andyyf

2/1

yd

dy=

So4

2xy=

Q: 2) An object is at a distance 25cm form the curved surface of a glass hemisphere of radius 10cm.

Find the position of the image and draw the ray diagram 5.1=g

Solution: For refraction at first face


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Ru1212 =

cm25=

11 =

2/32 =

R = 140cm

cmV 150=

The rays are converging beyond of at 140cm form Q. Again refraction takes place atthe plane surface.

For refraction at second face

15.1, 12 === R

cm140+=

?=v

UsingRu

2121 =

=

+ 5.11

140

5.11

3.93=v The ray meet axis at 93.3cm form point Q.

PROBLEMS ON REFRACTION


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1) A light ray is incident at an angle of incidence double that of refraction on one face of a parallel

sided transparent slab of refractive index ''

and thickness t. Find the lateral

displacement of the ray?

Solution: ri 2=

( ) ( )rt

r

rt

r

rrt

r

ritD tan

cos

sin

cos

2sin

cos

sin==

=

=

Asr

i

sin

sin=

r

rr

r

r

sin

cos.sin2

sin

2sin==

rcos2=

2

cos

=

12

tan

2

=

r

12

2

=

tD

24 =

tD

Q: 2) A light ray is incident on a transparent slab of R. I. 2= , at an angle of incidence 4/ .

Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it

could have suffered?

Solution: 2,4

==

i


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6/2sin

4/sin

sin

sin

=== r

rr

i

( )

6cos

64sin

cos

sin

=

=t

r

ritD

=

6tan.

4cos

4sin

t

= 31

2

1

2

1

t

( )136

=t

D ( )

6

623 =

D

Q: 3) Light of a certain colour has 2000 waves in one millimeter of air. Find the number of waves of

that light in one millimeter length of water and glass respectively?

Solution: =a wavelength in air

=m wavelength in medium

The number of waves of that light in a length of d will be

a

dn

=1

Andm

dn

=2

mm

a

n

n

== 12

12 nn m=

2660200033.12 ==n in water

3000200050.12 ==n in glass


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52/73

=

3

2112

cm4=

=

2

22

11

tS

=

4

316

= 1.5 cm

cmSSS 5.55.1421 =+=+=

Q: 6) The time taken by light to cover a distance of 9mm in water is____

Solution: smCw /104

9

3/4

103 88

=

=

sec104109

4109 118

3

=

==

wC

dt

ns911 10104 =

ns04.0=

Q: 7) A ray incident at an angle of incident 600 enters a glass sphere of R.I 3= . The ray is

reflected and reflected and refracted at the farther surface of the sphere. The angle between

reflected and refracted rays at this surface is_____

Solution:r

i

sin

sin=

2

1

3

2

3

60sinsin

0

===

r

030= r


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PC = QC

030=== rPQCCPQ

Angle between reflected ray QR and refracted ray QS at the other face

0

60180 = r 000 906030180 ==

REFRACTION AT SPERICAL SURFACES

Q: 1) Sunshine recorder globe of 30cm diameter is made of glass of refractive index 5.1= . A ray

enters the globe parallel to the axis. Find the position from the centre of the sphere where the ray

crosses the principal axis?

Solution:

For first refraction (Rares to deuces)

5.12 == u

cmR 1511 +==

R1212

2

=

uRV1122 +

=


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( ) 3011

15

15.15.1=

+

=V

cmV 45=

For second refraction (douses to rarer)

( ) cmucmR 153045',15 ===

UsingRuV

2121

''

=

( ) cmVorV

5.74

30'

30

1

15

5.11

15

5.1

'

1===

=

Distance of image from centre of globe is (15 + 7.5) = 22.5cm

Q: 2) A particle executes a simple harmonic motion of amplitude 1.0cm along the principal axis of a

convex lens of focal length 12cm. The mean position of oscillation is at 20cm from the lens. Find

the amplitude of oscillation of the image of the particle?

Solution: When the particle is

At R

cmu 19=

?1 =V

cmf 12=

+==

ufVfuV

111111

1

1912

7

19

1

12

11

1 ==

V


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7

19121

=V

When the particle is at left extreme position

cmfVcmu 12?,21 2 ===

( )2112

9

21

1

12

1111111

22 ==+==

ufVor

fuV

Amplitude of oscillation of image2

21VV =

cm

=9

212

7

1912

2

1

cm2857.2=

Q: 3) A point object is moving with velocity 0.01m/s on principal axis towards a convex lens of local-

length 0.3m when object is a distance of 0.4m form the lens, find

a) Rate of charge of position of the image and

b) Rate of charge of lateral magnification of image

Solution: Differentiating the

a) Equationuvf

111= w.r.t. time

d

du

ud

dv

V

.1

.1

022

+=

dt

du

u

V

dt

dv.

2

2

=

cmVV

12040

11

30

1 =

=

smd

ducmu /01.0,40 ==


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smd

dv/01.0

4040

120120 =

sm/09.0=

b)

2

2

2

1

===

f

V

u

V

du

dvM

=

f

V

dt

d

f

V

dt

dm112

dt

dv

ff

V.

112

=

109.030

1201

30

21

2

=

= s

dt

dv

f

V

f

sec/018.0=

Q: 4) Find the position of the image formed by the lens combination given in figure?

Solution: Image formed by first lens

( )10

1

30

11111

1111

=

=V

orfuV

( ) cmVorV

1510

1

30

111

1

==+

The image formed by the first lens server as the object for the second.

This is at a distance of (15 5)cm i.e. 10cm to the right of the second lens. It is a virtual object

Now10

1

10

11

2 =

V


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010

1

10

11

2

=+=V

=2V

The virtual image is formed at an infinite distance to the left of the second lens. This acts as an

object for the third lens.

30

111

3

=

V

30

11

3

=V

cmV 303 =

The final image is formed 30cm to the fight of the third lens.

Q: 5) Two Plano-concave lenses of glass of refractive index 1.5 have radii of curvature 20cm and 30cm

respectively. They are placed in contact with the curved surfaces towards each other and the

space between them is filled with a liquid of refractive index 5/2. Find the focal length of the

combination.

Solution: For first plano concave lens

cmR

f 405.0

20

15.1

20

11

21 =

=

=

=

For second plano concave lens

cmR

f 605.0

30

15.1

30

11

22 =

=

=

=

The focal length of the liquid lens is given by

( )

+=

21

2

3

111

1

RRf

R1 = 20cm, R2 = 30cm 2/52 =


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cmf 83 =

cmffff 12

181

601

4011111

321

=+=++=

cmf 12=

Q: 6) Given the object image and principal axis find the positions and nature of the lens

Solution: First join the object and image

If the one point is above the optical axis and the other below it, then the lens is always a convex

lens.

If object and image points areAbove the principal axis and image point is higher, then the lens is convex and is present between

the image and object points.

Other wise the lens is concave.

Q: 7) For the given positions of the objects and the image in figure determine the location and the

nature of the lens used?

Solution:


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Q: 8) A ray of light passes through a medium whose refractive index varies with distance as

+=

a

x10 . If the ray enters the medium parallel to the x-axis, what will be the trajectory of

the ray and what will be the time taken for the ray to travel a distance a?

Solution: The ray enters normally and proceeds along a straight line. At a distance x in the

medium consider a slab of thickness dx. Velocity of the light ray at this point is

Time taken to cross the distance dx is

+=

+

==a

x

C

dx

a

xCdx

V

dxdt 1

10

Total time of travel is

+=

a

a

x

C

dxt

0

0 1


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C

a02

3 =

Q: 9) A fish is rising up vertically inside a pond with velocity 4cm/s and notices a bird which is diving

vertically downward and in velocity appears to be 16cm/s (to the fish). What is the velocity of the

diving bird, if R. I of water is 4/3?

Solution: )fbfb VVV =

fb VV +=16 V

Vb=

416 += bV V12

34 =

12=bV scmV /9124

3==

Q: 10) Solar rays are incident at 450

on the surface of water ( )3/4= . What is the length of the

shadow of a pole of length 1.2m erected at the bottom of the pond, if the pole is vertical assuming

that 0.2m of the pole is above the water surface?

Solution: Applying Snells law at point c

sin3

445sin1

0 =

24

3sin =


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Here mCDAE 2.0==

BEBE

CE

BC=== 1tan

( )21sin

BE

BE

+=

( )2124

3

BE

BE

+=

mBE 625.0=

The length of shadow = AB = AE + EB = 0.2+0.625 = 0.825

Q: 11) In a lake, a fish rising vertically to the surface of water uniformly at the rate of 3m/s, observes a

bird diving vertically towards the water at a rate of 9m/s vertically above it. The actual velocity of

the dive of the bird is_______( )3/4=

Solution:

DR

DA

.

.=

y

y'=

yy = '

'yxh +=

yxh +=

Differentiating

dt

dy

d

dx

dt

dh+=


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d

dy+= 39

( )sm

td

yd/5.4

3/4

6==

Q: 12) A convex lens of focal length 0.2m is cut into two halves each of which is displaced by 0.0005m

and a point object is placed at a distance of 0.3m form the lens, as shown in figure. The position

of the image is ________

Solution:

uvf

111 =

ufv

111+=

mu 3.0= 2.0=f

3.0

1

2.0

11=

mv 6.0=

Q: 13) A pole 5m high is situated on a horizontal surface. Sun rays are incident at an angle 30

0

with thevertical. The size of shadow in horizontal surface is______

Solution:


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530tan 0

BC=

mBC3

530tan5 0

==

Q: 14) The Sun subtends an angle05.0= at the pole of a concave mirror. The radius is curvature of

concave mirror is R = 1.5m. The size of image formed by the concave mirror is_____

Solution: As Sun is at infinity image is formed at the focus of mirror

1) 1085.02

1 0 =

105180

5.02

1 =

= 0.654 cm

u

DPOQ S== Or

=S

i

D

Dmagnification

u

f

u

v==


64/73

u

R

D

D

S

i

2=

Si Du

RD =

2

MAGNIFICAIOTN IN CASE OF CURVED SURFACE:

Consider an extended object QO placedr1 to the principal axis at the point O. The image of

the point O is formed at I. Image of extended object is IQ

From 'PIQandPOQles we can say that

PO

OQ=1tan and

IP

QI 'tan 2 =

But since21 and are very small, we can approximate

Also2211 sinsin = Sign conversion

2211 = OQh +=0

PI

IQ

PO

OQ '.. 21 = 'IQhi =

( )( )

( )vih

uu

h =

.2

01 POu =

u

V

h

ih

2

1

0

= PIV +=

LATERAL MAGNIFICATION

The lateral magnification is


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d

dvmL =

Which can be obtained by differentiating equation

( )u

1212 =

Thus 0.2

1

2

2 =+udu

dv

REFRACTION AT PRISM

A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at

a suitable angle.

a) Calculation of angle of Derivation:

In passing through the prism, ray KL suffers two refractions and has tweed through an

=LQPN . (with is angle of direction)

In PLM ,

PMLPLM +=

( ) ( )2211 riri +=

( ) ( )2121 rrii ++= (1)

In OLM

0

21 180=++ rrO (2)

In Quadrilateral ALOM,

As 0180=+ ML ( Sum of 4 angles of a Quad = 360)0180=+ OA


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using equation (2) we can write

OArrO

+++ 21

Arr =+ 21 .(3)

Putting (3) in (1)

( ) Aii += 21 (4) (for bigger refracting angles)

If '' is the refractive index of the material of the prism, then according to Snells Law

1

1

1

1

sin

sin

r

i

r

i== (when angles are small)

11

ri similarly22

ri =

Putting the above two in equation (4) we get

( ) Aruru += 21

( ) Arr += 21

( )A1= (when refracting angle are small)

This is the angle through which a ray derivate on passing through a thin prism of small refracting

angle A.

Prism formula:

In minimum derivation position

i1 = i2 = i and r1 = r2 = r

from equation (3)

Arr =+

Ar =2

And from equation (4)

( ) Aiim +=

Aim = 2

( )2

2 mmA

ioriA

+

==+

from Snells Law


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( )2/sin2

sin

sin

sin

A

A

r

im

+

==

Note : The deviation through a prism is maximum when i1 = 900.

Thus Aim += 290

MAXIMUM DEVIATION:

For0

1

90=i

?2 = i

At surface BD

111 sin.sin ri =

1

0 sin.90sin1 r=

=

1

sin 1r

( )Crorr =

= 111 1sin

We know that Arr =+ 21

( )12 rAr =

( )CAr =2

For surface CD

22 sinsin ir =


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( )CAi = sinsin 2

( )[ ] =

Ai sinsin1

2

CONCEPT: Sometimes a part of a prism is given and the keep on thinking whether how should

we proceed? To solve such problems first complete the prism then solve as the problems of prism

are solved

Defects in Image formed by leuses:

The defect in the lens on account of which it does not form a white point image of white point

object is defined as Aberration.

Axial Chromatic Aberration: The variation of the image distance form the lens with the colour

measures axial chromatic Aberration.

Lateral Chromatic Aberration: The variation in the size of the image with colour measures the

lateral transverse chromatic aberration.

Concept of Dispersion of Light:Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent

colours on passing through a prism. (accuse due to wavelength).

The band of seven colours so obtained is called the visible spectrum.

The order of colours from the lower end of spectrum is VIBGYOR.

Violet colour deviates through maxi. Value and red colour deviates through the minimum

angle.

Causes of Dispersion: Each colour has in own wavelength according to Cauchys formula


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R.I. of a material depends on wavelength ( )

..........42 +++=

CBA

For a prism of small refracting angle of deviation is

Angular dispersion: It is the angle in which all colous of white light are contained

dv = deviation of violet colour

dr= deviation of red colour

( ) Angular dispersion = dv - dr

As ( ) Ad VV 1=

( ) Ad VS 1=

( ) ( )AAdd rVrV 11 =

( ) ArV 11 +=

( ) Add rVrV =

Dispersive Power (w): Ratio of angular deviation to the mean position produced by the prism.

( )

( ) ( )11 =

=

=

d

d

dd rvrv

Note: Single prism produces both deviation and dispersion simultaneously. It cannot give deviation

without dispersion (or) dispersion without deviation. However a suitable combination of two

prisms can do so.

Dispersion of light occurs because velocity of light in a material depends upon its colour

There is no dispersion of light refracted through a rectangular glass slab.

Combinations of prim:

I) Deviation without dispersion:

Net dispersion = 0, Net deviation 0

Necessary condition


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( ) ( ) 011 =+ RVRV

( ) ( ) 0''' =+ AA RVRV

In this Situation

Net deviation = ( ) ( ) ( ) ( )

+=+=+ +

''111'1'1'

RV

RVAA

( ) ( )( )

( )AA

RV

RV

''1'1

=

( )( )

( )( )

( )

=''

1'

11

RV

RVA

1

1 1

Usually


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In this situation

Net dispersion ( ) ( ) ''' AA rvrv +=

( ) ( )( )( )AA rvrv

1'1''

+=

( )( )( )

AA rvrv1'

1''

+=

( )( ) ( )

=1'1

1''

rvrvA

[ ]' =

If >' , the resultant dispersion is negative. (i.e.) opposite to that produced by the first prism.

This prism which produces dispersion without derivation is called direct vision prism.

Q: 1) A prism having an apex angle 40 and refractive index 1.5 is located in front of a vertical plane

mirror. Through what total angle is the ray deviated after reflected form the mirror?

Solution: ( ) ( ) 00 2415.11 === Aprism

mirrorprismtotal +=

( ) ( )iA 21801 +=

( ) ( )00 22180415.1 +=


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178176200 =+=

Q: 2) A container contains water upto a height of 20cm and there is a point source at the centre of the

bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of

the from is 2.0m above the wager surface.

(a) Find the radius of the shadow of the ring formed on the ceiling if r = 15cm.

(b) Find the maximum value of r for which the shadow of the ring is formed on the ceiling

( )3/4=w ?

Solution: a) Using Snells Law

ri sinsin =

2222 2001

2015

15

3

4

+=

+

x

x

cmx3

800=

Radius of shadow =3

80015 +

cm3

845=

m81.2=

b) For shadow to be formed angle of incidence should be less than critical angle

using Snells Law

0

22

max

max 90sin1203

4 =+rV

r


73/73

22

max

2

max 209916 += rr

22max 2097 =r

mcmr 2268.0207

9max ==

Q: 3) Monochromatic light falls on a right angled prism at an angle of incidence 450. The emergent

light is found to slide along the face AC. Find the refractive index of material of prism?

Solution:

Cr =2 ..(1)

0

21 90==+ Arr

12 90 rr = ..(2)

190 rC = ( )

190sinsin rC =

1cossin rC =

1cos1

r=

and21

2

1

11cos1sin

== rr

Snells law AR

1

0 sin45sin1 r

2

0 1145sin

=

11 2 =

notes on optics for aieee

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