ray opticsteaching notes(30!09!09)

8/14/2019 RAY OPTICSTeaching Notes(30!09!09)

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Mohammed

Asif

Name :

Roll No. :

Topic : Optics

Ph : 9391326657, 64606657

Teaching Notes (Optic)

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 108

m/siii) It is transverse in nature

In the spectrum of e.m.w. it lies between u.v. and infra-red region and has wavelength between4000 to 7000

O

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.

Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. 1

Ph: 040 64606657, 9391326657. www.asifiitphysics.vriti.com
http://www.asifiitphysics.vriti.com/http://www.asifiitphysics.vriti.com/


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

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

real object.




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Virtual Object: In this case incident ray are converging and point of convergence is the positionof virtual object. Virtual object cannot be seen by human eye be cause for an object can imageto 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, Realimage 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 raysare diverging)

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.




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The direction of incident energy is called incident ray and the direction in which energy is

thrown back 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 = .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 surfaceremains uncharged, while component perpendicular to reflecting surface (i.e. along normal)

reverse in direction.

^^^

1kzjyixr ++=

,

^^^

2kzjyixr +=




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3) Vector form of laws of Reflection:

^^^^^

.2 NNIIR

=R Unit vector along the reflected ray

^

IUnit vector along the Incident ray

^

NUnit vector along the normal ray

Image formed by a plane mirror: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




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( )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

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: If 060= then 516160

360 ===

(b) If360 is odd integer, then number of images formed is

360=

Ex: If 040= (which is not the complete part of 1800) then 940

360 ==

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

For reflection i2=




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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.

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 =0

60=i2

1=

== 120302180 0 A.C.W.00

2 1203021802 === i( ) =+= 120240

12ornet

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




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

successive reflections?

Solution: F,B.D

i2

1=

== 100502180 0

== 002

140202180

== 002

160102180

( ) =++=0

260100160140100 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 = 22

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

21 +=Net Now in OBC

22 += 0180=++ COBBCOOBC




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( ) += 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/mDifferentiating both sides w.r.t. time (t)

( ) ( )mImO xdt

dx

dt

d// =

x

mI

x

mO VV

=

//

mxIxmxOx VVVV

=

OxmxIx VVV

= 2Similarly yI/m = yO/m

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

y

mO

y

mIVV

=

//

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

=/




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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

Angle between 21 II VandVIs 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 image 2/1 to mirror

02

= VVV mI




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( ) ( ) smVI /106222/1 == For component of velocity of image parallel to the mirror

( ) smVI

/811 = Velocity of time ( ) ( )22

1 nIIIVVV +=

m16464100 =+=

= 5

4tan

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

3c3s13c o1 03 7c o3 7s i n1 0 03 7s i n1 0 02

jijiVI M

scji /4 82 8^^

+=

1212 , IMIMIMIM VVV

=




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s/4 81 2 8

^^

c mji

+=

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

smjVi /56 0s i n1 0

^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.




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0VVVV

mmi=

( ) ( ) 511 =iVsmsmVi /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:

( )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 the r1 to the planeof 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

=




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++

^0

^0

^0

^0

6s6c o6 0s i n6 0c o s 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 theangle made by the reflected ray with the horizontal?

Solution: AO = Incident ray

OB = Reflected ray

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

horizontal is zero.




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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.

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

shown.




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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 by2222 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 by222222 cbazyx ++=++ , where a, b and c are co-ordinates of object.

Minimum size of Mirror to see Full-Image:

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.

So 1111 NEMNAM =As 1111' NEMandNAMs are similar.

( )AEENSayxEM2

1111 === ..(1)

Similarly the light rays BM2 and M2E are incident and reflection rays respectivelySo 2222 NEMNBM =

2222 NEMandNBMsS are similar

( ) ( )BEENSayyEM2

1212

=== (2)

Adding equation (1) and (2) yield

=+yx 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 theparticle 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.

Solution:




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(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. line r1 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 mirrorP Q = Inversed shadow of PQ on the screen

Let = angle of incidenceThen PS = h tan and QS = h secFrom the properly of image P Q = ( ) hh 2cossec2 =




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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.

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

point P. Then co-ordinates of point P are

Solution: From figure

2

9,

2

9 == yx

=2

,2

qqP




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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:

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-Diagrams1) A ray of light parallel to principal axis passes (or) appears to pass through four after

reflection.




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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.

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, v andfof a mirror is known as mirror formula

+=

vuf111

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




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fvu

111 =+ .(1)

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

0.1

.1

22=

dt

dv

vdt

du

u

0.1

.1

22==

dt

dv

udt

dv

v

dt

dy

u

v

dt

dv.

2

2

= (2)

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 ABABpS and pBA '' are similar

PAPA

ABBA

'''= (1)

Applying sign conversionuPAAB =+= 0

vPABA == '1''Equation (1) can be rewritten as




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u

v

O

I

=+

u

v

O

I=

=

u

vm same for convex-mirror also.

Magnification in terms of u, v and fa) As we know that

fvu

111=+

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

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 ==




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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=

v

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

BA

BA

uu

VVm

=2

For tip A( )Lxu +=

2

R

f =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.

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




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

Solution: smiiiV /1 129

^^^

0=

+=

^

2 iVm

=

m/s

( )2

3020

20=

=

=uf

fm

( )11

/2

11/

=

MOMI VmV

= -(-2)2 11^

i

= -44^

im/s.

( )I

mI

n

mImI VVV +=

///

( ) smjx /24122 ==

I

mI

n

mI

n

mI VVV

+

=

///




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smji /2 44 4^^

=

^^

/ 22 44 4 ijiVVV mmII =

==

=

=

ji 24 6^

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

fu

v =

==




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( )

( ) 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 =

=

=

( )

( ) 4

3

23

7

24

7

=

=

=f

f

ff

uu

VVM

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

uf

fbI

uf

f

b

I

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

4114

2

2

2 =

==

R

R

u

Vm

11

32 =m

So radius of second image113

.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?

Solution: cmu 60=




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u

Vm =

602

1

=V

(or) cmV 30+=

60

1

30

1

60

1111=+=+=

vuf

cmf 60+=

In second case

3100

3

1 uV

u

Vm ===

Asfvu

111 =+

60

131=

uucmu 120=

Q: 4) Two objects A and B when placed one after another in front of a concave mirror of focal-length

10cm, 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

==

As1

114hh = and 122 hh = , cmf 10=

cmu 501

=

5010

10

4

1 2

+

=

u

cmu 202

=

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 ==




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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+==

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 isn

1times the object. The distance of

the object which isn

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

Solution:u

V=+=

1

uV =

uVf

111

+=

111+

=

uf




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( ) 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

( )

( )

( )

( ) ( )402/25

2/25

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

Solution:uf

fm

=

u++=

+

30

30

4

1

cmu 90=

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

=

VcmV 60=

0

2

.Vu

VV

i

=

( )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




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of 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

01 Luf

fL

=

( ) 32055 0

0

LL =

=

For part QR

0

2

2L

uf

fL

=

( ) 92055 0

0

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.

Let OF = x and IF = y

From triangle OMC

( ) sinsinsinOMOMOC == (1)

And from triangle ICM

sinsin

IMIC = .(2)




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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)y

x1

(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 andvice-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.

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

Since2fxy = 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 xu = and yV = we have( ) ( ) 2ffyfx =This is the equation of a rectangular hyperbola with its origin shifted to ( )ff, and x being




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

negative, v can be positive (or) negative, f is negative

ffandyvxu === ,

We have, form

( ) ( ) 2ffvfu =( ) ( ) 2ffyfu =++

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 OFv

1VERSUS

u

1

From mirror formula

fuv

111=+

Putting xu=

1and y

v=1 , we have

yyx

1=+

It is the equation of a straight line having a slope +1 (or) -1 according as u and v bear 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:




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u is always ve

v can be positive (or) negative andfis ve.

For a convex mirror u is always negative v is always positive andfis always positive.

CONCEPT OF CRITICLA ANGLEWhen 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 900 .

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

r

ia

b

sin

sin=




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090sin

sin Ca

b i=

Ca

b isin=

=

C

ab

isin

1

C depends on colour and and temp

CRed > Cviolet CRed < Cviolet

If temp CQ: 1) The sum (diameter d) subtends an angle radius at the pole of a concave mirror of focal

length f. 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

u

Or fv =

It means image is formed at focus

Taking '' f as radius and using

franddwhenr

===

fdorfd ==

REFRACTION AT SPHERICAL SURFACE:

From sle OBC and IBC

We have +=i




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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=+ 21As i is small, and so rand, are also small. Thus

( ) tantan +=+

R

h

u

h

++

=

And ( )V

h

R

hr =

=

++

v

h

R

h

R

h

u

h21

After simplifying we get

Ruv1212 =

Rv

11

1

2

1

2 =

=Ruv

112

1

2

1 This formula is derived for convex surface and for real

Image

From denser to rarer medium

Ruv2121 =

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




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=sin constant( ) yyf sinsin1 = ..(1)

Slope of curve at A

( )ydx

dy= 90tan

dx

dyy = cot

From equation (1)

( ){ }

sin

sin22

=yf

dx

dy

Case-I: ( ) ( )..eiyf= Refractive index varies as function ofx.According to Snells Law

=sin constantFor initial refraction at the air medium interface

( )00

90sinsin1 =( )

0090sinsin =

00cossin =

Here 0angle of refracting ray at point A with OX

So0

0

sincos

=




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And 20

2

0

sin1sin

=

Now Snells Law at M gives( ) 00 sinsin =xxf

Or( ) 2

0

2

0sin

1sin

=

xfx

( )xfx

22

0 sinsin

=

Now slope of tangent at M is given as

xdx

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 =+= andyyf2/1

ydx

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=gSolution: For refraction at first face

Ruv

1212

=

cm25=1

1=




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2/32=

R = 140cmcmV 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

UsingRuv

2121

=

=

+ 5.11

140

5.11

v

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

PROBLEMS ON REFRACTION1) 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 thelateral 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=

2cos

=

12

tan

2

=

r

12

2

=

tD

24=t

D




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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

6/2sin

4/sin

sin

sin

=== r

rr

i

( )

6cos

64sin

cos

sin

=

=t

r

ritD

=

6tan.

4cos

4sin

t

=

3

1

2

1

2

1t

( )136

=t

D ( )

6

623 =

t

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

m

m

a

nn

==

1

2

12nn m =

2660200033.12

==n in water3000200050.12 ==n in glass

Q: 4) Light from a sodium-lamp ( )nm58= traverses a distance of 60m in a chloroform ( )45.1= in a certain time. Determine the time difference occurring when light happens to traverse the

same length in ethyl ether ( )35.1=

Solution:mc

dt= ( )

sm

mtchloroform /103

45.160

81 =

For ethyl ether

( )sm

mt re

/103

35.1608. 2

=

21ttt =




43/68

( )30.145.11030

608

= m

sec100.28=

Q: 5) A rectangular glass slab of thickness 12cm is placed over a small coin kept on a table. A thin

transparent beaker filled with wager to a height 6cm & placed over the block. Find the

apparent shift of the position of the coin, when viewed from a point directly above it?

Solution:

=

1

11

11

tS

=

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

=

== wCd

t

ns911 10104 =

ns04.0=




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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=

21

3

2

3

60sinsin

0

===

r

030=r

PC = QC 030=== rPQCCPQ

Angle between reflected ray QR and refracted ray QS at the other face0

60180 = r000

906030180 ==

REFRACTION AT SPERICAL SURFACESQ: 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 +==

Rv

1212

2

=




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uRV1122

+

=

( ) 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 =Vcmf 12=

+==

ufVfuV

111111

1

1912

7

19

1

12

11

1

==V

7

19121

=V

When the particle is at left extreme position

cmfVcmu 12?,21 2 === ( )

2112

9

21

1

12

1111111

22

==+==ufV

orfuV

Amplitude of oscillation of image2

21VV

=




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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 imageSolution: Differentiating the

a) Equationuvf

111= w.r.t. time

dt

du

udt

dv

V .

1

.

1

0 22 +=

dt

du

u

V

dt

dv.

2

2

=

cmVV

12040

11

30

1=

=

smdt

ducmu /01.0,40 ==

smdt

dv/01.0

4040

120120 =

sm /09.0=

b)

2

2

2

1

===fV

uV

dudvM

=

f

V

dt

d

f

V

dt

dm112

dt

dv

ff

V.

112

=

109.0

30

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?




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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

Now 10

1

10

11

2 =V

010

1

10

11

2

=+=V

=2

V

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




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cmR

f 405.0

20

15.1

20

11

2

1 =

=

=

=

For second plano concave lens

cmR

f 605.0

30

15.1

30

11

2

2 ==

=

=

The focal length of the liquid lens is given by

( )

+=

21

2

3

1111RRf

R1 = 20cm, R 2 = 30cm 2/52 =cmf 83 =

cmffff 12

1

8

1

60

1

40

11111

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 are

Above 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

x

C

dx

V

dxdt 1

1

0

Total time of travel is

+=

a

a

x

C

dxt

0

01

C

a0

2

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?




50/68

Solution: )fbfb VVV =

fb VV +=16 V

Vb=

416 += bV V

12

3

4 =

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 theshadow 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

3

sin =

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 abird 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:




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DR

DA

.

.=

y

y '=

yy = '

'yxh +=

yxh +=Differentiating

dtdy

dtdx

dtdh +=

dt

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=

v




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mv 6.0=

Q: 13) A pole 5m high is situated on a horizontal surface. Sun rays are incident at an angle 300 with the

vertical. The size of shadow in horizontal surface is______

Solution:

530tan

0 BC=

mBC3

530tan5

0 ==

Q: 14) The Sun subtends an angle 05.0= at the pole of a concave mirror. The radius is curvature ofconcave 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==

u

R

D

D

S

i

2=




53/68


54/68


55/68

Ar=2

And from equation (4)

( ) Aiim +=Aim = 2

( )2

2 mm AioriA +==+

from Snells Law

( )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

111sin.sin ri =

1

0sin.90sin1 r=

=

1sin

1r

( )Crorr

=

=

1

1

1

1sin

We know that Arr =+ 21 ( )12 rAr = ( )CAr =2

For surface CD




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22sinsin ir =

( )CAi = sinsin 2( )[ ] = Ai sinsin 12

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

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

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

CB

A




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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 - drAs ( ) Ad VV 1=

( ) Ad VS 1= ( ) ( ) AAdd rVrV 11 =

( ) A

rV11

+=

( ) 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 0Necessary condition

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

In this Situation




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Net deviation = ( ) ( ) ( ) ( )

+=+=+ +''1

11'1'1'RV

RVAA

( ) ( )( )

( )AA

RV

RV

''1'1

=

( )( )

( )

( )

( )

=

''

1'

11

RV

RVA

1

1 1

Usually


59/68

Solution: ( ) ( )00

2415.11 === Aprism

mirrorprismtotal +=

( ) ( )iA 21801 +=

( ) ( )00 22180415.1 +=1781762

00 =+=

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 =

2222200

12015

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




60/68

using Snells Law

0

22

max

max 90sin1203

4=

+rVr

22

max

2

max209916 += rr

22

max2097 =r

mcmr2268.020

7

9

max ==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

2190==+ Arr

1290 rr = ..(2)

190 rC =

( )1

90sinsin rC =

1cossin rC =

1cos

1r=

and 21

2

1

11cos1sin

== rr

Snells law AR

10 sin45sin1 r

2

0 1145sin

=

12

1 2 =

12

1 2 =

5.12

32 ==

5.1=

AXIAL CHROMATIC ABERRATION IN LENSES:The inability of a lens to focus all the colours of components of white light at one point is called

axial longitudinal chromatic aberration. When white light focus on a lens, it suffers dispersion.

The constituent colours are focused at different positions on the principal axis by the lens.




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The red colour deviates at least and the violet deviates the most. So red colour is focused away

form the violet colour. As a result of this, the image produced by the lens suffering from

chromatic aberration is coloured and blurred.

Cause of Chromatic Aberration :

For a lens ( )

+=

21

1111RRf

(1)

The value of R.I of the material depends upon the wavelength of the incident light

(i.e.) 2

BA += (Cauhys Relation)

It means the value of is different for different colours ( )rv > . So form equation (1) the value of f is also different for different colours. For instance fv < fr.violet colour is focused near the lens as compared to the red colour. All other colours are focused in between the violet and red colored. Thus we get a coloured image of an object.

Boundary of such an image is not definite so image is blurred EXPRESSION FOR AXIAL (or) LONGITUDINAL CHAROMATIC ABERRATION:

L.C.A vr ff ..(2)

We know that ( )

=

21

111

1

RRf ..(3)

For red colour

( )

=

21

111

1

RRfr

r

(4)

For violet colour

( )

=21

1111RRf

v

r

.(5)

Subtracting (4) form (5) we get

( ) ( )

=

2121

111

111

11

RRRRffrv

rv




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[ ]1111

21

+

= rv

RR

( )

=

21

11

RRff

ffrv

vr

vr ..(6)

from equation (3), we can write

( ) fRR 1

111

21 =

equation (6) become

( )( ) fff

ff rv

vr

vr

1

=

But vr ff so( )

( )

=

=1

.2 rv

vr andfff

fffr =Thus L.C.A = Dispersive power x focal length of the lens

Note: Since vew += , but f can be +ve for convex lens (or) f can be -ve for convex lens

L.C.A = +ve foe convex lens= -ve for canvcave lens

LENSES WITH DIFFERENT MEDIAN ON EITHER SIDE:Consider a lens made of a material with refractive index with a liquid a on the left and a

light b on the right. What is the governing equation for this system. Let an object placed im

medium a . A ray of light form the object undergoes f=refraction at surface 1 and then

refraction at surface 2 before emerging in medium b . Let the radius of curvature of the

surface be R1 and that of the second surface be R2.

At surface 1

== 21 a and R = R1

11Ruv

aa = .(1)

At surface 2

b == 21 and R = R2 , 1v=

21 Rvv

bb = ..(2)

Adding (1) and (2) we get




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Example: A biconvex lens of focal length 20cm made of glass of refractive index 1.5, has water33.1= on one side and sir on the other. An object is placed 15cm form the lens on the

side with water. Where is the image formed?

Solution:

21 RRuv

baab +

=.(1)

Here 33.1,5.1 == a1=b

Since R is not given ux can get it from

( )

=Rf

21

1 RRR == 21

( )

=R

215.1

20

1

R

2

2

1

20

1

=

R

1

20

1=

cmR 201

+= , and cmR 202 =Substitute in equation we get

cmv 3.10=

PRISM: Students normally think that a prism should have only three edges as shown. However ahexagon shaped element can also be a prism.

A prism is defined as an element with two plane refracting surfaces at an angle to each other.




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The angle of the prism is the angle between the plane surfaces. Now it is important that a light ray be incident on one surface and emerge from the another

plane surface.

Example:-

Example-(1): The path of a ray of light passing through on equilateral glass prism as shown. The ray

of light striking at AC is incident at critical angle. Find the refractive index of glass.

Solution: Total deviation of the ray is 1080

0108=++ BCACAB ..(1)

0

160==+ Ar C

And0

260==+ Ar C

This implies that 21 rr = . (2)(i.e.) i1 = i2 = ..(3)

Now 11 riAB =

CAC 21800 = and

22riBC =

Hence ( ) ( ) 0220

1110820180 =++ riri C

( ) ( ) 021210 1082180 =+++ Crrii

( ) ( )21210

11

010822180 rrandiiri C ===++

( ) 001

01086022180 =+ i

0

1 24=i (4)




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Now ( )Cr

i

==

0

0

1

1

60sin

24sin

sin

sin

( ) 024sin60sin = C

=

C

sin

1

Hence( ) 024sinsin

60sin=

C

C

( )0

0

0

0

60sin

24sin

60sin.sin

60sin=

C

C

0

0

0

00

60sin

24sin

60sin.sin

sin.60coscos60sin =

C

CC

469.0866.0

407.060cotcot

0 ==C

047.1cot =C

45.1cot1sin

1 2 =+== CC

Example-(2): Figure shows an irregular block of material of refractive index 2 . A ray of light

strikes the face AB as shown in figure. After refraction it is incident on a spherical surface CD

of radius of curvature 0.4m and enter a medium of refractive index 1.514 to meet PQ at .Find the distance o upto two places of decimal?

Solution: By Snells Law

r

i

sin

sin

1

2 =

2

1

22

1sin

sin

45sin

1

20

=== rr

030=r

It shows that the refractive ray thus be comes parallel to AD inside the block. So parallel ray is

incident on spherical surface CD.

514.1,2,4.0, 21 ==== mRu

FromRuv

1212

= we have




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4.0

25.125.1 =uv

mv 606=( ) mOei 06.6.. =

Example-(3): A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point P is

kept at a distance of mR form it. Find the value of m for which a ray from P will energy

parallel to the table as shown.Solution: Refraction at plane surface

5.1,1,,211 ==== RmRu

Ruv

1212

=

=

15.115.1

1 mRv

Rmv 5.11 =For refraction at concaved surface:

( ) ( ) RmRRu 15.15.12 +=+=1,5.1,,

2122==== vRR

2

12

2

1

2

2

Ruv

=

( ) RRm =

+

5.11

15.1

5.11

315.1 =+m

3

4=m

Example-(4): A ball is kept at a height 0y above the surface of a transparent sphere of radius R,

made of material of refractive index . At t = 0, the ball is dropped to fall normally on

the sphere. Find the speed of the image formed as a function of time forg

yt 0

2< ,

consider the image by a single refraction?

Solution: If '' y is the distance falls by the ball

In time t, then2

2

1gty =

The distance of the ball form the point P of the sphere




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= 2

02

1gtyu

And velocity of ball tgv =0By refraction formula

Ruv

11 =

..(1)

R

gty

v +

=

1

2

1

1

2

0

( )

=2

0

2

0

2

1

2

11

gtyR

Rgty

V

Differentiating equation (1) we have

01122

=

dt

du

udt

dv

v

02

2

2

2

.1

.1

vu

vV

dt

du

u

v

dt

dv

i ==After substituting the values of u, v and v0 we get

( )2

22

0

2

2

11

=

Rtgy

tgRVi




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ray opticsteaching notes(30!09!09)

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