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TOPIC 1 mark

2 mark

4 mark

5 mark theory

5 mark problem

TOTAL 10 14 4 8 4

Geometrical Optics

1 1 1 1 C

Physical Optics 3 3 1 PC 1

Electrostastics 1 2 1 T/P 1 C

Atomic Physics 3 4 1 T/P 4 1 C

Current Electricity

2 4 1 T/P 2 1 C

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n1 sin i = n2 sin r

Laws of refraction

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LATERAL SHIFT

When a ray of light is incident

obliquely on a parallel sided glass

slab the emergent ray shifts

laterally . The perpendicular

distance between the direction of

the incident ray and emergent ray

is called “lateral shift’’.

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N`

P

A

i

r i-r

B

D CN

T

S

R

t

Q

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ABCD=Principal section

of a glass slab

PQ = incident ray

QR = refracted ray

RS = emergent ray

NN`= normal

RT = lateral shift

BC = t = thickness of

the slab

i = angle of incidence

r = angle of refraction

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Consider ▲NQR,cos r =

From the ▲QRT,sin(i-r)=

QN

QR

QR=QN

cos r

RT

QR

RT cos r

QN

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N`

i

rN

It R

PQ

O

i

r

NORMAL SHIFT

The apparent shift in the position of an

object placed in one medium and viewed

along the normal, from the other medium.

S

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OP = incident ray

PQ = refracted ray produced

backwards meet ON at I

NN`= normal

OI = normal shift

RPO = i = angle of incidence = NPO

QPS = r = angle of refraction = NIP

t = thickness of the optical medium

From the ▲ NPO

From the ▲ NPI

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n1

sin i = n2

sin r

Here n1

= n n2

= 1

P is a point which very

close to N

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TOTAL INTERNAL REFLECTION

denser medium

i Ci>C

r 900

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A

i1 i1r1 r2

dR

S

Z

M

N

Q

P

B C

REFRACTION THROUGH A PRISM

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ABC = principal section

of the prism

BC = Base

A = Angle of the prism

PQ = Incident ray

QR = Refracted ray

RS = Emergent ray

ZMS = d = Angle of deviation

i1

= Angle of incidence

r1

= Angle of refraction

r2

= Angle of incidence at the

face AC

i2

= Angle of emergence

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Consider the Quadrilateral AQNR

In the ΔQNR

A + QNR = 1800

r1

+ r2

+ QNR = 1800

equating the above two equations, we have

A + QNR = r1

+ r2

+ QNR

A = r1

+ r2

---------------------(1)

d = (i1

– r1) + (i

2– r

2)

d = (i1

+ i2) – (r

1+ r

2) = i

1+ i

2- A (from 1)

d + A = i1

+ i2

---------------------(2)

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(1) A = 2r or r = A/2

(2) D + A = 2i or i = (A+D)/2

D

i

d

i

at minimum deviation

i1

= i2

= i

r1

= r2

= r

d = D

i1 i2

d

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REFRACTION THROUGH

CURVED SURFACES

Curved surface: It is a part of a sphere.

Principal axis: The horizontal line passing through the centre of the curved surface.

Pole(P): Centre of the curved surface.

Centre of curvature(C): It is a point on the principal axis which is the centre of the sphere in which curved surface is a part.

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Radius of curvature(R): It is the

distance between centre of curvature and the

pole

Principal focus(F): It is a point on the

principal axis where parallel rays from infinity

converge or appear to diverge from that point

Focal length (f): It is the distance

between the principal focus and the pole.

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Relation between n,u,v&R

C O I P

M

ϴ

i

rN

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OM = incident ray

MN = refracted ray

MI = refracted ray produced

backwards

CP = radius of curvature (R)

OP = object distance (u)

IP = image distance (v)

i = angle of incidence

r = angle of refraction

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Applying sine rule to the ▲s CMO & CMI

Dividing equation (2) by equation (1)

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= X

=

=

We know that n1

sin i = n2

sin r

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Dividing the above equation through out by uvR

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M is a point which is very

close to P, therefore

IM ≈ IP = -v (because image

is virtual)

OM ≈ OP = u

CO = CP – OP = R - u

CI = CP - IP = R + v

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Lens maker’s formula

u v I

A

C

B DP

O I`

V`

E F

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OE = incident ray

FI = refracted ray

O = point object

I` = virtual object for the surface ADC

OP = u =object distance

R1

,R2

=radii of curvatures of the

surfaces ABC and ADC

PI =v=image distance

P I` =v` =image distance for the first

surface

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For the refraction through surface ABC,

For the refraction through surface ADC,

Negative sign indicates that the

object is virtual

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Adding (1) and (2)

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Two thin lenses in contact

L1 L2

O I I`

u vV`

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Let us consider two thin lenses in

contact let f1

and f2

be their focal

lengths

let u=object distance

v =image distance

v`=virtual image distance

here I`is the image formed

For refraction through first lens,

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For refraction through second lens,

here I`serves as virtual object for the

second lens.

Adding equations (1) and (2)

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let F be the effective focal length

of the combination.

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Questions carrying ONE mark each

1. Define critical angle.

2. Mention an application of total internal

reflection.

3. What is dispersion of light?

4. Write the expression for the deviation produced

by a thin prism.

5. Define dispersive power of the material of a prism.

6. What is angular dispersion?

7. Write the condition for dispersion without deviation.

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14. A thin convex lens of focal length 0.1m & a thin

concave lens of focal length 0.1m are placed co-

axially in contact. What is the net power of the

combination?

8. Define power of a lens.

9. What is magnification?

10. Write the relation between power and focal

length of a lens.

11. Write the expression for the equivalent focal

length of two thin lenses separated by a distance.

12. What is a thin prism?

13. What is the lateral shift produced by a parallel

sided glass slab for grazing incidence of light?

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Questions carrying two marks

1.Mention the conditions for total internal

reflection to occur.

2.Arrive at the relation between critical angle and

refractive index.

•the light ray should travel from denser medium to rarer

medium

•the angle of incidence must be greater than critical

angle

c

Here n1=n, i=c, n

2=1 & r=90

0

n sinc=1 sin 900

n sinc=1

n=

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3.Write the ray diagram showing the

experimental arrangement for pure spectrum.

S

L1L2

R

V

P

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6.What is an impure spectrum? give an example.

5.What is a pure spectrum? give an

example.

If the constituent colours are seen distinctly in a

spectrum, it is called pure spectrum.

Example: Spectrum produced in the laboratory

A spectrum in which the constituent colours overlap

and are not seen distinctly is called impure spectrum.

Example: Rainbow

4.What is an optical fibre? on what principle

does it work?

An optical fibre is a thin transparent fibre which

transmits light along any desired path. It works on

the princple of TIR.

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8.Calculate the angle of deviation produced by a

thin prism of refracting angle5˚ and refractive

index 1.5.

7.Calculate the refractive index of a pair of media

given critical angle as 40˚.

given C=40˚ ,n=1

d=?,A=5˚,n=1.5 d=(n-1)A

=(1.5-1)X5˚

=0.5X5˚=2.5˚

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9.An object is kept at a distance of 0.12m

from a convex lens forms an image at a

distance of 0.18m.calculate the

magnification produced.

u=0.12m,v=0.18m, m=?

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10.Two thin convex lenses of focal

lengths 0.15m & 0.2m are separated by

a distance of 0.6m. Find the effective

focal length of the combination.

given:f1=0.15,f

2=0.2,d=0.06,F=?

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Question carrying FIVE mark each

1. What is lateral shift? Derive an expression for

lateral shift produced by a parallel sided glass

slab for oblique incidence.

2. What is normal shift? Derive the expression for

normal shift produced when an object in a

denser medium is viewed normally through air.

3. Derive the expression for refractive index in

terms of the angle of the prism and angle of

minimum deviation.

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4.Derive the relation connecting n,u,v & R for

refraction at a spherical surface concave

towards a point object in a denser

medium.

5.Derive lens maker’s formula.

6.Derive the expression for the effective

focal length of two thin lenses in

contact.

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•Read the given problem twice

•Write the given data

•Write a relevant formula/e connecting the given data

•Substitute and simplify

•Write the result with S.I unit

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List of formulae

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March 2010 : Find the real position of an air

bubble in a glass cube of side 0.3 m if the bubble

appears to be at a distance of 0.05 m from one face

and at 0.15 m from the opposite face.

Solution:

Here apparent distance from both the faces are given.

A.D1

= 0.05 m

A.D2

= 0.15 m

Let ‘x’ be the real distance from the first

face.

Real distance from the opposite face is

(0.3-x) m

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We know that

Therefore the real position of the air

bubble in the glass cube is 0.1 m from

the first face or 0.2 m from the

opposite face.

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March 2008: Focal length of convex lens is

0.1 m. A liquid lens is formed between a plane

surface and one face of this lens of radius of

curvature 0.12m. The converging combination

formed is found to have a focal length 0.18 m.

Calculate the refractive index of liquid.

Solution:

Given F=0.18 m, f1

= 0.1m, R1=0.12m, n=?

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Here radius of curvature of the plane surface

is infinity

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March 2011: A concave lens when placed in

air has a focal length of 0.4 m and has a refractive

index of 3/2. What will be its focal length when

immersed in water of refractive index 4/3.

Solution:

Given: f=0.4 m, n of glass=3/2, f of lens when

immersed in water=? R.I of water 4/3

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Simplify and arrive at

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March 2006: A small air bubble is found

0.04 m inside the glass sphere of refractive

index 1.5 When viewed normally from outside it

is seen at a depth of 0.03 m. Calculate the

radius of curvature of the glass sphere.

Solution:

Given

u=0.04m, n=1.5, v=0.03m, R?

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R=0.119m