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INTRODUCTION

The branch of physics which deals with the nature of light, its sources, properties, effects and vision is called optics. Light is that form of energy which makes objects visible to our eye.

USES

The science of optics is by far an important part of our life and our economy. It is due to optics that we have discovered our universe from the microscopic virus to the largest galaxy. It is due to optics that we see the colours of a rainbow, sparkling of diamond, twinkling of stars, shining of air bubble in water etc.

NATURE, PROPERTIES AND BEHAVIOR

1. Huygen gave wave theory of light. Maxwell showed that light is an electromagnetic wave, which consists of oscillation of electric field and magnetic field perpendicular to each other and perpendicular to direction of propagation in space.

2. The wavelength of light typically varies form 400nm to 700 nm. The limits of the visible spectrum are not well defined because the eye sensitivity curve approaches the axis asymptotically at both long and short wavelengths. This theory explained beautifully reflection, refraction, interference, diffraction and polarization.

Photoelectric effect, Compton effect and several other phenomena associated with emission and absorption of light could not be explained by wave theory. Therefore, quantum theory of light was developed, mainly by Planck and Einstein. At present it is believed that light has dual nature i.e., it propagates as a wave but interacts with matter as a particle.

CLASSIFICATION INTO GEOMETRICAL AND WAVE OPTICS, AND ITS NECESSITY To make the study of optics easy, it is divided into two parts.

(a) Geometrical or Ray Optics: The phenomena of rectilinear propagation of light, reflection and refraction, are studied in this section.

(b) Wave Optics: The wave behavior of light (like diffraction and interference) is studied in this section. Geometrical optics can be treated as the limiting case of wave optics when size of obstacle or opening is very much large than the wavelength of light. Under such conditions, wave nature of light can be ignored and light can be assumed to be travelling in straight line (rectilinear

propagation). But when size of obstacle or opening is comparable to wavelength of light, rectilinear propagation is no longer valid and resulting phenomena are explained by using wave nature of light.

REFLECTION AND ITS LAWS

When light falls on the surface of a material it is either re-emitted without change in frequency or is absorbed in the material and turned into heat. When the re-emitted light is returned into the same medium from which it comes, it is called reflected light and the process is known reflection.

LAW OF REFLECTION:

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(I) (I) (N) (N) (R) (R) (P) i r i r Reflecting Surface (P) T T

Note: To apply laws of reflection draw tangent (T) and Normal (N) at the point of reflection (P) 1st Law : Incident ray (I), normal (N) and reflected ray (R) are in same plane.

2nd Law : Angle of incidence = Angle of reflection ⇒ ∠ = ∠i r

Mirrors:

1. Plane Mirror 2. Concave Mirror

3. Convex Mirror

REFLECTION AT PLANE SURFACE WITH DIFFERENT EXAMPLES Image Formation of Point Object by Plane Mirror:

1. Point of intersection of incident light ray is known as object.

2. The point of intersection of reflected rays or refracted ray is known as image. Note: The object and image may be real or virtual

3. For real object the image formed by plane mirror is virtual.

I M

MO=MI O

4. For virtual object image formed by plane mirror is real.

M I R

MO=MI O

Field of view: Region in which diverging rays from object or image are actually present is known as field of view.

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Field of view

Field of view

O I

I

Case 2: Field of view for virtual object

Field of view

O I

Image formation of extended object by plane mirror :

Case 1: Case 2: A B A’ B’ A B A’ B’

NGE OF VISION, DEVIATION PRODUCED BY MIRROR

Angle of Deviation: The angle between the direction of incident and reflected (refracted) light ray is known as angle of deviation. i r δ i r δ angle of deviation δ = π − +(i r) Assume ∠ = ∠ = αi r , then δ = π − α(2 ) 180º 90º (angle of incidence) (angle of deviation) Normal incidence: (N) (R) (I) (N) (R) (I) i r 0, ∠ = ∠ = δ = π

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Grazing Incidence: (N) (R) (I) (N) (R) (I) i 90 r ∠ = ° = ∠ and δ =0 PROPERTIES OF IMAGE FORMED BY PLANE MIRROR

Properties of Image Formed by a Plane Mirror of an Extended Object (i) size of the object = size of the image

(ii) image is laterally inverted.

Illustration 1: A man is standing at distance x from plane mirror in front of him. He wants to see the entire wall in mirror which is at distance y behind the man. Find the size of mirror ?

O A B M1 M2 A’ B’ O’ h y x x

Solution: From ∆O' M M1 2 and DO' AB

1 2 M M x h = 2x y+ size of mirror, 1 2 hx M M (2x y) = + Note : If 1 2 h x y, M M 3 = =

Rotation of Plane Mirror:

1. If a plane mirror is rotated through an angle q about an axis in the plane of mirror then reflected ray, image and spot are rotated through an angle 2q in the same sense.

2. If plane mirror is rotated about an axis perpendicular to plane of mirror then reflected ray image, spot do not rotate.

Velocity of image formed by a plane mirror :

(x ,o)IM (x ,o)oM (o, o)

OM

x →x co-ordinate of object relative to mirror

IM

x →x co-ordinate of image relative to mirror

Here, IM OM x = −x differentiating, OM IM dx dx dt = − dt

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IM OM

v v

⇒ = −

velocity of image relative to mirror = – velocity of object relative to mirror.

Illustration 2: Find velocity of image when object and mirror both are moving towards each other with velocity 2 m/s and 3 ms-1. How they are moving ?

Solution: Here vOM = −vIM 1 1 0 M I M I v v = −(v v ) ( 2ms ) ( 3ms )+− −= − + −v ( 3) 1 I v 8ms− ⇒ = −

Combination of Plane Mirrors

To find net deviation produced by combination of plane mirror and deviation produced by each mirror. While adding the deviation ensure that they must be in same direction either clockwise or anticlockwise.

Illustration 3: Two plane mirrors are inclined at 30º as shown in figure. A light ray is incident at angle 45º. Find total deviation produced by combination of mirror after two successive reflection.

30º 45º

M1

Solution: deviation at mirror M ,1 δ =1 180° − ×2 45° =90° ↑(clockwise)

deviation at mirror M ,δ =2 2 180º - 2 × 15º = 150º ¯ (anti clockwise)

total deviation δ = δ − δ =2 1 150º - 90º = 60º (anti clockwise)

IMAGES FORMED BY TWO MIRRORS

CASE 1: When the mirrors are parallel to each other

M1 M2 I1’ I2’ y I1 I2 (2x+y) O x (2y+x) y x

Above figure shows image formed by object placed at distance y, from M1 and at distance x from M2. Number of image formed by parallel plane mirrors is infinity.

CASE 2: (when the mirrors are inclined at angle q)

1. All the images formed by two mirrors lie on circle having centre C. Here if angle between mirror is q then image will formed on circle at angle (2p - q). If angle q is less number of

image formed will be more. θ

C 2 -π θ

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(a) 360 n= −1 θ If i 360 ( ) is even. 360

(ii) is odd and object is kept symmetrically   θ    θ  (b) If 360 n=   θ

  = fractional number then only integral part is taken.

(c)

360 n=  

θ

  For all other conditions .

Reflection at a spherical surface, paraxial and marginal rays Spherical mirrors are part of polished spherical surface

Convex mirror Concave mirror R C P P

1. Center of curvature (C): Center of circle of which mirror is a part. 2. Radius of curvature (R): Radius of circle of which mirror is a part . 3. Pole (P): Centre of mirror reflecting portion

4. Principal axis: Line which join pole to the centre of curvature 5. Diameter of mirror: Shortest distance between two ends of mirror. Concept of focus: According to figure, SC cosi CQ = Here SC=R2 So,

( )

R R 2 CQ CQ cosi 2cosi = ⇒ = Also, h sini R = C Q P i h S i 2i T

When i→o, CQ=R2 and h→o and TC PC→

Paraxial Ray: Rays whose angle of incidence are small are known as paraxial rays. Marginal Ray: Rays whose angle of incidence are not small are known as marginal rays.

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FOCUS, FOCAL PLANE, CENTRE OF CURVATURE, RADIUS OF CURVATURE Focus: F f P f F P F = Focus f = Focal length

If paraxial rays are parallel to the axis of mirror they will meet or appear to meet at a point on the axis, the point is known as focus and the distance between pole and focus is called focal length

Focal plane:

Plane perpendicular to principal axis and passing through focus is known as focal plane

F Primary focus

F Secondary focus’

(i) Rays coming from primary focus will become parallel after reflection. (ii) Parallel rays of light after reflection will meet at secondary focus. Centre of curvature:

It is the geometrical centre of the mirror. If light ray passes or appear to pass through centre of curvature then, it retraces its path.

P F C R P F C R R = Radius of curvature

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RULES FOR IMAGE FORMATION BY SPHERICAL MIRROR AND RAY DIAGRAM 1. (i) Rays parallel to principal axis passes through focus.

(ii) Rays passing through focus goes parallel to principal axis.

C F F

2. Ray passing through center of curvature returns in same path.

C

3. In case of convex mirror for real object, image is erect diminished virtual and between pole and focus

Position, nature and magnification of image and mirror formula SIGN CONVENTION:

1. Pole is taken as origin and principal axis is taken as x-axis 2. Direction of incident light is taken as direction of +ve x-axis 3. Object, focus, image are referred by their co-ordinates. 4. Height above principal axis is taken as positive.

MIRROR FORMULA:

1 1 1 v u f+ =

1. Mirror formula holds only for paraxial rays. 2. Proof : (for point object and concave mirror)

In ∆CMOβ = + αi …(i) In ∆CMI γ = + βi …(ii) From eqn. (i) and (ii),

( ) γ = β − α + β ⇒ γ = β − α2

⇒ α + γ = β2 …(iii)

∴ α β, and γ are very small MP MP tan , tan PO PI α = α = γ = γ = and MP tan PC β = β =

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Substitute the value in equation (iii), we will get M P I i β γ i C α O 1 1 2 PO PI PC+ = using, PO= −u, PI= −v and PC= −R 1 1 2 ( u) ( v) ( R) ⇒ + = − − − 1 1 2 v u R ⇒ + = 1 1 2 1 1 1 v u 2f v u f ⇒ + = ⇒ + = Magnification (m): height of image I m height of object O = =

Proof: ∆PAB and ∆PA 'B' are similar

A 'B' AB A 'B' PA ' PA ' =PA ⇒ AB = PA ( I) ( v) m I v ( O) ( u) O u − − ⇒ = ⇒ = = − + −

1.

For Concave mirrors

a.

If

the object is real, the image formed is real and inverted

b. If the object is real, the image formed is virtual and erect. c. If the object is virtual, the image formed is real and erect.

2.

For Convex mirrors

a.

If the object is real, the image formed is virtual and erect. b. If the object is virtual, the image formed is real and erect. c. If the object is virtual, the image formed is virtual and inverted.

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Here f, should be substituted with sign Illustration:

Lex

REFRACTION AND ITS LAWS, REFRACTIVE INDEX, SPEED OF LIGHT IN DIFFERENT MEDIA 1. Change in the speed of light as it passes from one medium to other is called refraction 2. Light is deviated due to medium when it is not along the normal

3. Velocity of light in medium

r r r

C

v = µ →

µ ε relative permeability of medium C → velocity of light in vacuum

ε →r relative permitivity of medium

4. Absolute refractive index of medium is defined by

m

C V µ =

V →velocity of light in medium. SNELL’S LAW:

(N) ® Normal (I) ® Incident Ray (R) ® Refracted Ray (I) (N) Medium(1) Medium(2) r i µ1 µ2 (R)

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1. 1 1 2 2 1 2 c V sini c sinr V µ µ = = = µ µ ⇒ µ1sini= µ2sinr

where, V1 and V2 are velocity of light in first and second medium. m1 and m2 are refractive index of first and second medium.

2. µ > µ ⇒2 1 i r> µ < µ ⇒2 1 i r< r i Denser medium Rarer medium µ1 µ2 µ1 µ2 r i Rarer medium Denser medium

SNELL’S LAW IN GENERAL FORM: 1. For different medium of RI

µ µ1, 2 and µ3……..

µ1sinθ = µ1 2sinθ2……. = constant µsinθ = constant

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REFRACTION AT PLANE SURFACE

Let us consider the situation as shown in figure. A point object O in medium (1) is viewed from me-dium (2). Medium (1) Medium (2) Interface A i r I M O u v i r

From Snell’s law, we have

µ1sini= µ2sinr …(i)

Rays are assumed to be paraxial (i.e., near normal viewing). So i and r are very small ∴ sin i tan i i≈ ≈

sin r tan r r≈ ≈

Hence, equation (i) can be rewritten as µ1tan i= µ2tan r or 1 2 AM AM MO MI µ = µ or 1 2 u v µ = µ − − or u1 v2 µ =µ In general, for refraction at plane surface following relation is used :

i r u v µ µ = …(ii)

where, µi = refractive index of the medium in which incident rays are present µr = refractive index of the medium in which refracted rays are present.

Particular Case (I):

When object is in denser medium and viewed from rarer medium. Using equation (ii), we get

Interface

H H’ I

O

Denser Medium (R.I. = )µD Rarer Medium (R.I. = )µR

0 R H v µ =µ − or R D H v= − µ ∴ R D H H′ = µ

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also, shift (S) = Real depth (H) – Apparent depth (H’)

or

Particular Case (II):

When object is in rarer and viewed from denser medium Using equation (ii), we get R D H v µ =µ − or v= − µHR D ∴ H' H= R Dµ Interface H H’ I O

Denser Medium (R.I. = )µD Rarer Medium (R.I. = )µR

Also, shift (S) = Apparent height (H’) – Real height (H) or S H(= R Dµ −1).

2. If refractive index of a medium is function of x,µ =f(x) then normal should be along x axis. 3. If refractive index of a medium is function of y,µ =f(y) then normal should along y- axis. Critical angle and T.I.R

i µR µD i<θC i=θC i µR µD µR µD i>θC i D

µ → Refractive index of denser medium ; µ →R Refractive index of rarer medium.

1. When a ray propagates from denser medium to rarer medium for 90º angle of refraction corresponding angle of incidence is known as critical angle.

2. At critical angle refraction takes place

R

D C R C

D

sin sin90 sin µ

µ θ = µ ° ⇒ θ =

µ

3. Critical angle does not depend on angle of incidence. REFRACTION THROUGH SLAB AND PRISM, DEVIATION Refraction through slab:

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d x S O

I1 I2

Air Slab Air

Consider refraction at air–slab interface

1 x v µ = − or v1= −µx

Now, I1 will act as object for refraction at slab–air interface

∴ 2 1 (d x) v µ = − + µ ⋅ or 2 d v = −x+  µ   ∴ shift (S) d 1 d x x  d 1  = + − + = µ µ    .

1. For real or virtval object the shift will be in the direction of incident wave

2. For multiple slabs, the net shift ‘S’ is given by

Dispersion by Prism:

Prism is combination of two plane refracting surface A → angle of prism

i → angle of incidence e → angle of emergence δ → angle of deviation

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1 2 A r r ⇒ = + …(i) deviation δ = δ + δ = −1 2 (i r ) (e r )1 + − 2 1 2 (i e) (r r ) ⇒ δ = + − +

Using equation (i), δ = +(i e) A− …(ii) d - i graph

(angle of incidence) Minimum Deviation:

For minimum deviation i e= , So, min min 2i A i 2 A δ + δ = − ⇒ = …(i) Also, 1 2 A r r r 2r A r 2 = = ⇒ = ⇒ = …(ii) Using Snell’s law,

1sini= µsinr min A A sin sin 2 2 δ + ⇒ = µ min ( A) sin 2 A sin 2 δ + ⇒ µ = Maximum Deviation: At maximum deviation e 90= ° and δmax =imin+90° −A To find imin in different condition r = θ2 C and r1= − θA C

A

imin r r2 e

1

Using Snell’s law, sinimin = µsin(A− θC)

[

]

1 min C i sin− sin(A ) ⇒ = µ − θ 1. If A 2= θC then imin =90°

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2. If A 2> θC then imin >0

3. If A < θC then

4. For imin< <i 90° light ray comes out. 5. For i i< min T.I.R. occurs.

Thin Prism:

For thin prism A 10≤ °

sini i,= sinr r ,1= 1 sinr2 =r2, sine e= and, 1 1 2 2 1sini sinr i r sinr isine e r = µ ⇒ = µ   µ = ⇒ = µ  …(i) deviation δ = +(i e) A− ⇒ δ = µ +(r r ) A1 2 −

using equation (i) and r r1+ =2 A, δ =A(µ −1)

Illustration 4: The cross-section of the glass prism has the form of an isosceles triangle. One of the equal faces is silvered. A ray of light incident normally on the other equal face and after being reflected twice, emerges through the base of prism along the normal. Find the angle of the prism.

Solution:

From the figure,

o 90 A α = − o i 90= − α =A …… (i) Also, β =90o − =2i 90o −2A And γ =90o − β =2A Thus, γ = =r 2A From geometry, o A+ γ + γ =180 or A=1805 =36o

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Dispersion of Light:

1. The angular splitting of a ray of white light into a number of components and spreading in different direction is called dispersion of light.

2. Angle of Dispersion : A θ Violet Mean ray (yellow) Red δr δyδv

Angle between the rays of the extreme colours in the refracted (dispersed) light is called angle of dispersion. θ = δ − δv r

3. Dispersive power ( w) of the medium of the material of prism is given by

y

angular dispersion deviation of mean ray (yellow)

θ

ω = =

δ

4. For small angled prism A 10≤ °

δ =v (nv −1)A δ =R (nR−1)A δ =y (n 1)At −

here n ,nv R and ny refractive index of material for violet, red and yellow colours respectively. y R v R y y (n n ) (n 1) δ − δ − ω = = δ −

If ny is not given in the problem then take

v R y n n n 2 + = 5. Cause of dispersion :

Dependence of n on l according to cauchy’s formula 2 b n( ) aλ = +

λ Combination of Two Prism:

(i) Achromatic Combination :

It is used for deviation without dispersion. For this combination θcomb=0 (nv1 −n )AR1 1=(nv2 −n )AR2 2 A1 A2 1 2

Net mean deviation δ =net [ny1 −1]A [n1− y2 −1]A2

1 1 2 2 v R v R net 1 2 n n n n 1 A 1 A 2 2 + −     δ =    

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Also 1 net 1 2 1  ω  δ = δ ω

  and ω δ + ω δ =1 1 2 2 0 where ω1 and ω2 are depressive powers for the two prisms and δ1 and δ2 are the mean deviation produced by them.

(ii) Direct vision combination

It is used for producing dispersion without deviation, condition for direct vision combination is

A1 A2 White

light Mean ray

(yellow)

Deviation of mean ray (dy) = 0

1 2 y 1 y 2 (n 1)A (n 1)A ⇒ − = − 1 1 2 2 v R v R 1 2 n n n n 1 A 1 A 2 2 + +     − = −        

Net angle of dispersion

1 1 2 2

net (nv n )A (nR 1 v n )AR 2

⇒ θ = − − −

or θ = δ ω − ωnet y1( 1 2)

REFRACTION AT SPHERICAL SURFACE (PARAXIAL RAY) CHROMATIC ABBERATION

In ∆ OCM, i = α + β …(i)

In ∆ CMI, β = + γr …(ii) According to Snell’s law, µ1sini= µ2sinr

For paraxial ray, µ = µ1i 2r

From eq. (i) and (ii), µ α + β = µ β − γ1( ) 2( )

M P I O α β Cγ r u v i µ1 µ2 1 2 ( 2 1) ⇒ µ α + µ γ = µ − µ β ⇒ µ1POPM+ µ2PMPI = µ − µ( 2 1)PMPC 1 2 ( 2 1) u v R µ µ µ − µ ⇒ + = − 2 1 2 1 ( ) v u R µ µ µ − µ ⇒ − =

Note: 1. The formula is applicable only for paraxial rays. 2. When surface is plane, R = ∞

2 1 V u µ µ = Magnification by Curved Surface :

From Snell’s law, µ1sini= µ2sinr For paraxial rays, µ = µ1i 2r

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1 2 1 2 ( O) ( I) OO' II' ( ) PO PI ( u) v µ + µ −     ⇒ µ = µ ⇒ = − +     So, magnification, 1 2 v I m O u µ = = µ O O’ P C i r I I’ u v

Illustration 5: A linear object of length 4 cm is placed at 30 cm from the plane surface of hemispherical glass of radius 10 cm. The hemispherical glass is surrounded by water. Find the final position and size of the image

Solution: For 1st surface 1 2

4, 3,u 20 3 2 µ = µ = = − cm, and R= +10cm, using 2 1 ( 2 1) v u R µ µ µ − µ − =

( ) ( ) (

3 4 3 4

)

2 3 2 3 v ( 20) 10 − ⇒ − = − v 30 ⇒ = − cm Using

( )

( )

1 2 4 ( 30) v A 'B' A 'B' 3 3 AB u (4cm) ( 20) 2 − µ = ⇒ = µ 4cm A’ B’ B A B” A’ 1 surfacest u v u’ v’ 5.3cm 5.3cm

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A 'B' 5.3

⇒ = cm.

A 'B' behaves as the object for plane surface

' ' 1 32, 2 43 µ = µ = and R= ∞, u'= −40

( ) ( )

' ' 2 1 43 32 v ' u' v ' ( 40) µ µ ⇒ = ⇒ = − Solving it we will get, v '= −25.4cm

Now using,

( )

( )

' 1 ' 2 v A "B" A 'B' u µ = µ

( )

(

)

( )

(

)

3 35.4 A "B" 2 A "B" 5.3 4 (5.3) 40 3 − ⇒ = ⇒ = − cm The final images in all the above cases are shown in figure.

30cm 20cm Right µ = 4/3 C A µ = 1 P Left B D O 15cm

Illustration 6: The slab of a material of refractive index 2 shown in figure has a curved surface APB of radius of curvature 10 cm and a plane sur-face CD. On the left of APB is air & on the right of CD is water with refractive indices as given in the figure. An object O is placed at a distance of 15 cm from the pole P as shown. What is the distance of final image of O from P as shown. What is the distance of final image of O from P as viewed from the left?

Solution: In case of refraction from a curved surface, we have

Here ; R = - 10 cm and u = - 15 cm So 1 2 1 2 , i.e., 30cm ( 15) 10 − − = ν = − ν − −

i.e., the curved surface will form virtual image I at a distance of 30 cm from P. Since, the image is virtual there will be no refraction at the plane surface CD (as the rays are not actually passing through the bound-ary), the distance of final image I from P will remain 30 cm.

THIN LENS AND ITS PROPERTIES

1. A lens is a homogenous transparent medium (such as glass) bounded by two curved surfaces or one curved and a one plane surface.

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Types of Lens:

Assume R and R1 2 are Radius of curvature of first and second surface.

(i) Biconvex Lens (ii) Plano convex lens

(iii)

Biconcave Lens

(iv)

Plano concave lens

(v) Concavo convex lens

2. Centre most point (P) of lens is known as optical centre

3. Line joining both the center of curvature and optical center is called principal axis.

(I) Real focus: (2nd focus, Main focus)

converging point of parallel beam of light is known as real focus (u= ∞, v f)=

F2 f2

f2 F2

(II) 1st Focus: If ray coming from a point or converging to a point after refraction become parallel then the point is known as 1st focus.

f1

F1

f1

F1

Note: (1) For convex lens f2 = +ve,f1= −ve (2) For concave lens f2 = −ve,f1= +ve REFRACTION THROUGH THIN LENS

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F C

f

(a) Refraction through a convex lens Focal Plane F C f (b) Refraction through a concave lens Focal Plane

LENS MAKER FORMULA, POWER OF LENS Lens Maker’s Formula:

s

µ →

refractive index of surrounding.

l

µ →

refractive index of lens For surface of radius R1:

µ µ − µ µ − s = l s I 1 ( ) v ' u R …(i)

and for surface of radius R2 :

s l s l 2 ( ) v v ' R µ µ = µ − µ …(ii) adding equation (i) and (ii)

l s s 1 2 ( ) 1 1 1 1 v u R R   µ − µ   ⇒ = µ   l s 1 2 1 1 1 1 1 v u R R µ    − =  µ   …(iii) If u = ∞ then v f=    − −        l s 1 2 ì 1= 1 1 1 R R (Lens Maker’s Formula) …(iv)

From equation (iii) and (iv)

1 = -1 1

f v u ® Lens Formula

Note: (Condition for application of Len’s makers formula)

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1. Medium on both sides of lens should be same 2. Lens should be thin

3. Light rays should be paraxial POWER OF LENS

Power of a lens is defined as the ratio of the refractive index of the surrounding medium and the focal length of the lens in that medium f. The unit of power is dioptre.

Mathematically it can be written as

ì P =

f . [f in metre]

RULES FOR IMAGE FORMATION AND MAGNIFICATION ETC. RAY DIAGRAM Position and Nature of Image formed for a given Position of Object.

The following rules are used for image formation in case of thin lenses :

(i) A ray passing through optical centre proceeds undeviated through the lens as shown in figure.

C

(a)

C

(b)

(ii) A ray passing parallel to the principal axis after refraction through the lens passes or appears to

pass through second focal point as shown in figure.

(a) f F2 C C (b) f F2

(iii) A ray passing through first focus or directed towards first focus, after refraction from the lens

becomes parallel to the principal axis as shown in figure.

(a) f F1 C C (b) f F1 Magnification:

This is defined as the ratio of size of image to the size of object.

Size of image (I) Magnification (m) =

Size of object (O) (a) Transverse or Lateral magnification (mT)

It is defined as the ratio of any transverse dimension of the image to the corresponding transverse dimension of the object. Let the size of the object and image be denoted by O and I respectively.

T I m O = .

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P Q L P Q v v m u u − = −

For small object, L

dv m

du =

Differentiating lens equation, we get

2 2 dv du 0 v v   − − − =   or dv du 0v2 − u2 = and therefore, 2 2 2 L 2 dv v v m m du u u   = = =  =  

Therefore, for small object longitudinal magnification is square of transverse magnification.

(a) For Convergent or Convex Lens

S. No. Position of Object Ray-Diagram Details of Image

1. At infinity Real, inverted, Diminished (m << -1) at F

F1

F2

2. Between infinity and 2F Real, inverted, Diminished (m < -1) between F and 2F

I O

3. At 2F Real, inverted, Equal m = -1 at 2F

I O

4. Between 2F and F Real, inverted, Enlarged (m >-1) between 2F and infinity

I O

2F F F 2F

5. At F Real, inverted, Enlarged (m >>-1) at infinity.

O

2F F F

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6. Between F and O Virtual Erect Enlarged (m>>-1)

2F F

F 2F

(b) For Divergent or Concave Lens

S. No. Position of Object Ray-Diagram Details of Image

1. At infinity Virtual, inverted, very small (m << -1), at F

F

2. Between F and optical centre Near Lens Virtual, Erect Diminished (m<+1)

F I O

Sign Convention:

Focal length of converging lens is taken as positive and that of the diverging lens is taken as negative, if it is kept in vacuum or in a medium whose refractive index is less than the lens.

1. For convex lens

a) If object is real, image formed is real & inverted b) If object is real, image formed is virtual & erect c) If object is virtual, image formed is real & erect 2. For concave lens

a) If object is real, image formed is virtual of erect b) If object is virtual, image formed is real & erect c) If object is virtual, image formed is virtual & inverted COMBINATION OF LENSES

Equivalent Focal Length of Combination of Lenses:

If number of lenses of focal lengths are placed coaxially in contact then the equivalent focal length of the combination is given by

1 2 n

1 1 1 ... 1

F f= +f + + f ...(iii)

In terms of power P p p= 1+ 2 +... p+ n SILVERING OF LENSES

When a lens is silvered, it behaves like a spherical mirror whose power (P )eq is given by Peq= ΣPi

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where Pi is the power of lens or mirror to be taken as many times as the number of refraction or reflection. For example, let us consider silvered equiconvex lens then

Peq=2P PL + M

as final image is formed after three optical events involving two refraction through lens and one reflection at plane mirror. The focal length of equivalent spherical mirror is given by

eq eq

1 f

P = −

If feq is –ve then mirror is concave and, if feq is +ve, then mirror is convex. FORMATION OF IMAGE BY DISPLACEMENT METHOD

In this method, an object and a screen are placed at a distance a apart. A convex lens of focal length f(f < a/4) can be placed in two position between the object and the screen such that a real image of the same object is formed on the screen in both the positions.

A B Screen A' B' B" u v V a d u

Let a be the distance between the object and the screen and let d be the distance between the two position of the lens. From the figure :

u + v = a v – u = d Solving, we get a d a d v , u 2 2 + − = =

Using the lens formula and simplifying, we get

2 2 a d f 4a − = ...(i) Let I1=A 'B'; I2 =A 'B ; O AB′′ = 1 1 I a d m O a d + = =

i.e. the one image is larger than the object

2 2 I a d m O a d − = =

+ i.e. the other image is shorter than the object

\ O= I I1 2 ...(ii)

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Illustration 7: For two positions of a converging lens between an object and a screen which are 96 cm apart,

two real images are formed. The ratio of the lengths of the two images is 4.84. Calculate the focal length of the lens.

Solution: Here 1 2 m 4.84 m = ⇒ 2 a d 4.84 a d +   =    ⇒ a d 2.2 a d 1 + = − Putting a = 96 cm gives d = 36 cm 2 2 a d f 20.625 4a − ∴ = = cm .

KEY CONCEPTS

Light is said to move rectilinearly if, a>> Dλ

Mirror formula.

1 1 2 1 v u R f+ = =

Newton’s Formula. XY = f2

Optical power of a mirror (in Diopters) = − m 1 f ; f m is meter.

Laws of refraction. 2 1 1 1 2 2 n v sini sinr n v λ = = = λ

Refractive index of the medium relative of vacuum = µ ∈r r

Deviation (d) of ray incident at ∠i and refracted at

r

is given by δ = −| i r |

Principle of Reversibility of light Rays.

1 2 2 1 1 n n =

Refraction through a Parallel Slab

( )

t sin i r d cosr − =

Apparent Depth and shift of Submerged object.

Apparent shift rel 1 d 1 n   =  

Refraction through a Composite Slab

Apparent shift =

1 2 n

1rel 2rel n rel

1 1 n t 1 t 1 .... 1 t n n n       − + − + + −             

Critical Angle and Total Internal Reflection (T.I.R.). 1 r d n C sin n − =

Angle of deviation. d = i + e – A

Angle of dispersion. q = dv – dr

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Cauchy’s formula.

( )

b2 n λ = +a λ

Dispersive power v r y n n n 1 − ω = −

Dispersion without deviation

The condition for direct vision combination is.

y y n 1 A n′ 1 A′  −  = −      ⇔ v r n n 1 2 +       v r n n A 1 A 2 ′ + ′   ′ =  

Deviation without dispersion

Condition for achromatic combination is.

(

nv −n Ar

)

=

(

n′v−n Ar′

)

Spherical Surfaces 2 1 2 1 n n n n v u R − − =

Transverse magnification 2 1 v / n v R m u R u / n   − = = 

Refraction at plane Surface

2 1 n u v n =

For a spherical, thin lens having the same medium on both sides.

(

rel

)

1 2 1 1 n 1 1 1 v u R R   − = −   … (a) when lens rel medium n n n =

Transverse magnification (m) v m u =

Lens formula 1 1 1 v u f− =

The equivalent focal length of thin lenses in contact is given by

1 2 3 1 1 1 1 F f= +f +f ….

The combination of lens and mirror behaves like a mirror of focal length ‘f’ given by

m 1 1 2

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CONCEPTUAL QUESTIONS

1. A thick plane mirror shows a number of images of the filament of an electric bulb. Of these, the brightness image is the

(a) first (b) second

(c) fourth (d) last.

2. A light beam is being reflected by using two mirrors, as in a periscope used in submarines. If one of the mirrors rotates by an angle θ, the reflected light will deviate from the original path by the angle

(a) 2θ (b) 0º

(c) θ (d) 4θ.

3. A convex mirror is used to form the image of an real object. Then which of the following statements is wrong

(a) the image lies between the pole and the focus (b) the image is diminished in size

(c) the image is erect (d) the image is real.

4. All the following statements are correct except

(a) A real, inverted same sized image can be formed using convex mirror (b) The magnification produced by a convex mirror is always less than one (c) A virtual erect and same sized image can be obtained using a plane mirror (d) A virtual, erect, magnified image can be formed using a cancave mirror

5. When light travels from one medium to the other which the refractive index is different, then which of the following will change

(a) frequency, wavelength and velocity (b) frequency and wavelength (c) frequency and velocity (d) wavelength and velocity. 6. Light of different colours propagates through air

(a) with the velocity of air (b) with different velocities (c) with the velocity of sound (d) having the equal velocities. 7. A cut diamond sparkles because of its

(a) hardness (b) high refractive index

(c) emission of light by the diamond (d) absorption of light by the diamond.

8. A diver in a swimming pool wants to signal his distress to a person lying on the edge of the pool by flashing his water proof flash light

(a) he must direct the beam vertically upwards (b) he has to direct to the beam horizontally

(c) he has to direct the beam at an angle to the vertical which is slightly less than the critical angle of incidence for total internal reflection

(d) he has to direct the beam at an angle to the vertical which is slightly more than the critical angle of incidence for total internal reflection.

9. When white light passes through a glass prism, one gets spectrum on the other side of the prism. In the emergent beam, the ray which is deviating least is or deviation by a prism is lowest for

(a) violet ray (b) green ray

(c) red ray (d) yellow ray.

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(a) travel with same speed

(b) travel with same speed but deviate differently due to the shape of the prism (c) have different anisotropic properties while travelling through the prism (d) travel with different speeds.

11. A glass lens is placed in a medium in which it is found to behave like a glass plate. Refractive index of the medium will be

(a) greater than the refractive index

(b) smaller than the refractive index of glass (c) equal to refractive index of glass

(d) no case will be possible from above.

12. A biconvex lens from a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would

(a) disappear

(b) remain as real image still

(c) be virtual and of the same size as the object (d) suffer from aberrations.

KEY 1. (b) 2. (a) 3. (d) 4. (a) 5. (d) 6. (d) 7. (b) 8. (c) 9. (c) 10. (c) 11. (c) 12. (c)CINTSHI

HINTS AND SOLUTIONS 1. Several images will be formed but second image will be brightest

90% 10% 100% 9% Incident light First image Second brightest image Third image 10% 90% 10% 80% 10% .

2. When a mirror is rotated by an angle θ, the reflected ray deviate from its original path by angle 2θ. 3. The image formed by a convex mirror is always virtual.

4. Convex mirror can produce real & erect or virtual & inverted. 5. Velocity and wavelength change but frequency remains same.

6. In vacuum, the speed of light is independent of wave length. Thus vacuum (or air) is a non dispersive medium in which all colours travel with the same speed.

7. Due to high refractive index its critical angle is very small so that most of the light incident on the diamond is total internally reflected repeatedly and diamond sparkles.

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back in same medium. 9. δ ∝ µ −( 1) ⇒ µR is least so δR is least. 10. Conceptual 11. a g I a I g ( 1) f f ( 1) µ − = µ − 1 f ⇒ = ∞ if I gµ =1 a I a g ⇒ µ = µ . 12. 1 2 1 ( 1) 1 1 f R R   = µ −  

For biconvex lens R2 = −R1 1 ( 1) 2

f R

  ∴ = µ −    

Given R= ∞ ∴ f = ∞, so no focus at real distance.

LEVEL – I

MODEL QUESTIONS

SINGLE CHOICE CORRECT QUESTIONS

1. A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the

pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be

(a) real and will remain at C

(b) real and located at a point between C and ∞ (c) virtual and located at a point between C and O (d) real and located at a point between C and O.

2. A ray of light passes through four transparent media with refractive indices µ µ µ1, ,2 3 and µ4 as shown in the figure, the surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

(a) µ = µ1 2 (b) µ = µ2 3

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3. A ray of light passes through an equilateral prism such that the angle of incidence and the angle of emergence are both equal to 3/4th of the angle of prism. The angle of minimum deviation is

(a) 15º (b) 30º

(c) 45º (d) 60º.

4. A ray of light falls on a transparent glass slab with refractive index (relative to air) of 1.62. The angle of

incidence for which the reflected and refracted rays are mutually perpendicular is.

(a) tan (1.62)−1 (b) sin (1.62)−1

(c) cos (1.62)−1 (d) none of these.

5. A 4.0 cm thick water layer (µ =4 / 3) rests on a 6.0 cm layer of CCl (4 µ =1.5) in a tank. How far below

the water surface, viewed normally from above, does the bottom of the tank seem to be

(a) 7.0 cm (b) 8.0 cm

(c) 9.0 cm (d) 10.0 cm.

6. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the

surfaces of the lenses are as given in the diagrams

(a) R1 R2 2 1 R R ≠ (b) R (c) R R (d) R .

7. A lens of focal length f projects m times magnified image of an object on a screen. The distance of the

screen from the lens is (a) f (m 1)− (b) f (m 1)+ (c) f(m 1)− (d) f(m 1)+ .

8. A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 m is put in water (refractive

index

4 3 =

). Its focal length is

(a) 0.15 m (b) 0.30m (c) 0.45 m (d) 1.20 m.

9. A 16 cm long image of an object is formed by a convex lens on a screen. On moving the lens towards

the screen, without changing the positions of the object and the screen, a 9 cm long image is formed again on the screen. The size of the object is

(a) 9 cm (b) 11 cm (c) 12 cm (d) 13 cm.

10. Two lenses, one concave and the other convex of same power are placed such that their principal axes

coincide. If the separation between the lenses is x, then (a) real image is formed for x = 0 only

(b) real image is formed for all values of x (c) system will behave like a glass plate for x = 0

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MAY BE MORE THAN ONE CORRECT CHOICE ANSWERS

11. For which of the pairs of u and f for a mirror image is smaller in size.

(a) u = – 10 cm, f = 20 cm (b) u = – 20 cm, f = – 30 cm (c) u = – 45 cm, f = – 10 cm (d) u = – 60 cm, f = 30 cm

12. There are three optical media 1, 2 and 3 with their refractive indices µ > µ > µ1 2 3. (TIR – total internal reflection)

(a) when a ray of light travels from 3 to 1 no TIR will take place

(b) critical angle between 1 and 2 is less than the critical angle between 1 and 3 (c) critical angle between 1 and 2 is more than the critical angle between 1 and 3

(d) chances of TIR are more when ray of light travels from 1 to 3 as compare to the case when it travel from 1 to 2

13. A ray of light is incident in situation as shown in figure which of the following statements is/are true?

30º µ1=4 µ2 µ3=2

(I) if then angle of deviation is 60º (II) if then angle of deviation is 60º (III) if then angle of deviation is 120º (IV) if then angle of deviation is zero

(a) I and IV (b) III and I

(c) II and IV (d) none

14. In the figure light is incident at angle which is slightly greater than the critical angle. Now, keeping the incident ray fixed a parallel slab of refractive index n3 is placed on surface AB. Which of the following statements are correct .

A B n2 n <n1 2 θ (a) n1 A B n2 θ (b) n3 n1 C D

(a) total internal reflection occurs at AB for n3 < n1 (b) total internal reflection occurs at AB for n3 > n1

(c) the ray will return back to the same medium for all values of n3 (d) total internal reflection may occur at CD for n3 > n1

15. For refraction through a small angled prism, the angle of minimum deviation . (a) increases with the increases in R.I. of the prism

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(b) will be 2D for a ray of R.I. 2.4, if it is D for a ray of R.I. 1.2 (c) is directly proportional to the angle of the prism

(d) will decrease with the increase in R.I. of the prism

16. Parallel rays of light are falling on convex spherical surface of radius of curvature R = 20 cm as shown. Refractive index of the medium is µ =1.5. After refraction from the spherical surface parallel rays

(a) actually meet at some point

(b) appears to meet after extending the refracted rays backwards

(c) meet (or appears to meet) at a distance of 30 cm from the spherical surface (d) meet (or appears to meet) at a distance of 60 cm from the spherical surface

17. A point object is placed at 30 cm from a convex glass lens S 3 2 µ =

 

  of focal length 20 cm. The final image of object will be formed at infinity if

(a) another concave lens of focal length 60 cm is placed in contact with the previous lens (b) another convex lens of focal length 60 cm is placed at a distance of 30 cm from the first lens (c) the whole system is immersed in a liquid of refractive index 4/3

(d) the whole system is immersed in a liquid of refractive index 9/8

18. The radius of curvature of the left and right surface of the concave lens are 10 cm and 15 cm respec-tively. The radius of curvature of the mirror is 15 cm .

(a) equivalent focal length of the combination is – 18 cm (b) equivalent focal length of the combination is + 36 cm (c) the system behaves like a concave mirror

(d) the system behaves like a convex mirror COMPREHENSION

@ Write-up – 1

A ray of light enters a spherical drop of water of refractive index m as shown in the figure.

φ

α A 19. Select the correct statement .

(a) Incident rays are partially reflected at point A. (b) Incident rays are totally reflected at point A. (c) Incident rays are totally transmitted through A. (d) None of these.

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20. An expression of the angle between incidence ray and emergent ray (angle of deviation) as shown in the figure is φ α A (a) 0º (b) (c) (d)

21. Consider the figure of previous question the angle for which minimum deviation is produced will be given by (a) 2 2 1 cos 3 µ + φ = (b) 2 2 1 cos 3 µ − φ = (c) 2 2 1 sin 3 µ + φ = (d) 2 2 1 sin 3 µ − φ = . @ Write-up – 2

Magnification by a lens of an object at distance 10 cm from it is – 2. Now a second lens is placed exactly at the same position where first was kept, without changing the distance between the object and lens. The magnification by this second lens is – 3.

22. Now both the lens are kept in contact at the same place. What will be the new magnification?

(a) – 13/5 (b) – 12/7

(c) – 6/11 (d) – 5/7

23. What is the focal length of the combination when both lenses are in contact ?

(a) 1760 cm (b) 175 cm

(c) 12 cm7 (d) 13 cm9

24. What is the focal length for the first lens

(a) 10 cm (b) 20 cm

(c) 20/3 cm (d) 10/3 cm

ASSERTION / REASON Codes.

(a) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for State-ment – 1.

(b) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for State-ment – 1.

(c) Statement – 1 is True, Statement – 2 is False.

(d) Statement – 1 is False, Statement – 2 is True.

25. STATEMENT – 1

For all types of mirrors, object and image always move in opposite directions. STATEMENT – 2

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26. STATEMENT – 1

For a prism of refracting angle 60º and refractive index 2, minimum deviation is 30º. STATEMENT – 2 At minimum deviation, m A D sin 2 A sin 2 + µ =

; where the symbols have their usual meanings

27. STATEMENT – 1

When light travels from denser to rarer medium the critical angle of incidence have different values for different wavelengths of light.

STATEMENT – 2

Refractive index of a medium varies with wavelength of light.

28. STATEMENT – 1

White light splits into seven colours when refraction occurs even at single surface. STATEMENT – 2

For different wavelength angles of refraction are different, as their corresponding refractive indices are different.

29. STATEMENT – 1

The focal length of a thin converging lens, made of glass, for violet light is fv, while that for red light is fr, these satisfy the relation .

fr > fv.

STATEMENT – 2

The refractive index of the material of the lens (glass) satisfies mv > mr; and 1 2 1 ( 1) 1 1 f R R   = µ −

  applies to the lens.

MATRIX MATCH

Each question contains statements given in two column which have to be matched. Statements (a, b, c, d) in Column A have to be matched with statements (p, q, r, s) in Column B. The answers to these questions have to be appropriately bubbled as illustrated in the following example.

30. A real object is being seen by optical component listing in column A and nature of image of this object

is listing in column B .

Column – A Column – B

Nature of image

(a) Convex lens. (p) Real. (b) Convex mirror. (q) Virtual. (c) Concave lens. (r) Erect. (d) Concave mirror. (s) Inverted.

31. Column–B shows the optical phenomenon that can be associated with optical components given in

Column–A. Note the Column–I may have more than one matching options in Column–B.

Column – A Column – B

(a) Convex mirror. (p) Dispersion. (b) Converging lens. (q) Deviation.

(c) Thin prism. (r) Real image of real object. (d) Glass slab. (s) Virtual image of real object.

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Q

P R

SINGLE CHOICE CORRECT QUESTIONS

32. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer

(a) greater deviation (b) no deviation

(c) same deviation as before (d) total internal reflection.

n S

i

l

i

33. A diverging beam of light from a point source S having divergence angle α , falls symmetrically on a glass slab as shown. the angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is

(a) zero (b) α

(c) sin (1/ n)−1 (d) 2sin (1/ n)−1 .

34. An isosceles prism of angle 120º has a refractive index of 2. Two parallel monochromatic rays enter the prism parallel to each other in air as shown. the rays emerging from the opposite faces

a) are parallel to each other (b) are diverging

(c) make an angle 30º with each other (d) make an angle 60º with each other.

35. The velocity of light in a medium is half its velocity in air. If a ray of light emerges from such a medium

into air, the angle of incidence, at which it will be totally internally reflected, is

(a) 15º (b) 30º

(c) 45º (d) 60º.

36. White light is incident from under water on the water-air interface. If the angle of incidence is slowly

increased from zero, the energent beam coming out into the air will turn from

(a) white to violet (b) white to black (c) white to red (d) white to yellow.

37. The refractive index of diamond is 2.0. The velocity of light in diamond in cm/s is

(a) 6 10× 10 (b) 1.5 10× 10

(c) 1.5 10× 8 (d) 6 10× 8.

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lens of the same focal length is placed in contact with it. The image will now be formed at

(a) f (b) between f and 2f

(c) 2f (d) infinity.

39. A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of glass with

1.5

µ = . The focal length of the lens is

(a) 30 cm (b) 31 cm

(c) 40 cm (d) 41 cm.

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

of two liquids L1 or L2 having refracting indices n1 and n2 respectively (n2 >n 1)1> . The lens will diverge

a parallel beam of light if it is filled with

(a) air and placed in air. (b) air and immersed in (c) L1 and immersed in L2 (d) L2 and immersed in L1.

MLAY BE MORE THAN ONE CORRECT CHOICE ANSWERS

41. A plane mirror M is arranged parallel to a wall W at a distance  from it. The light produced by a point source S kept on the wall is reflected by the mirror and produces a light spot on the wall. The mirror moves with velocity n towards the wall. Then.

(a) The spot of light will move with the speed n on the wall. (b) The spot of light will not move on the wall.

(c) As the mirror comes closer, the spot of light will becomes larger and shift ways from the wall with speed larger then n.

(d) The size of the light spot on the wall remains the same.

42. Two points P and Q lie on either side of an axis XY as shown. It is desired to produce an image of p at Q using a spherical mir-ror, with XY as the optic axis. The mirror must be

(a) converging and positioned to the left of P (b) diverging and positioned to the left of P (c) converging and positioned to the right of Q (d) diverging and positioned to the right of Q. 43. The deviation produced by a prism depends on

(a) the angle of incidence of light on its face (b) the refracting angle of prism

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(c) the refractive index of the prism (d) the wavelength of the light used.

44. In the figure, a ray of light falls on face 1 of a piece of glass (refrac-tive index = m) kept in air. Faces 1 and 3 are planar, and the ray is not intercepted by faces 2 and 4. The angle(s) independent of m is/ are

(a) i + i’ + d (b) i + i’ – d

(c) i i' 2 + (d) i i' 2 + – d 45. For the refraction of light through a prism.

(a) For every angle of deviation there are two angles of incidence

(b) the light traveling inside and equilateral prism is necessarily parallel to the base when prism is set for minimum deviation

(c) There are two angles of incidence for maximum deviation

(d) Angle of minimum deviation will increase if refractive index of prism is increased keeping the outside medium unchanged and if mp > ms.

46. A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved

by approximately placing

(a) a convex mirror of suitable focal length (b) a concave mirror of suitable focal length (c) a convex lens of focal length less than 0.25 m (d) a concave lens of suitable focal length.

47. An image of a bright square is obtained on a screen with the aid of a convergent lens. The distance between the square and the lens is 40 cm. The area of the image is nine time larger than that of the square. Select the correct statement(s) .

(a) Image is formed at a distance 120 cm from lens. (b) Image is formed at a distance 360 cm from lens. (c) Focal length of lens is 30 cm.

(d) Focal length of lens is 36 cm.

48. A lens of focal length ‘f’ is placed in between an object and screen separated by a distance ‘D’. The lens forms two real images of object on the screen for two of its different positions, a distance ‘x’ apart. The two real images have magnifications m1 and m2 respectively (m1> m2).

(a) 1 2 x f m m = − (b) m 1m2 = 1 (c) 2 2 D x f 4D − = (d) D ³ 4f.

49. Consider the rays shown in the diagram as paraxial. The image of the virtual point object O formed by the lens LL is

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(a) Virtual. (b) Real.

(c) Located below the principal axis. (d) Located left of the lens.

50. A thin lens having power P is cut into three parts A, B and C as shown. The power of (a) A is twice of B (b) B is equal to A

(c) C is equal to B (d) A is equal to C

COMPREHENSION

@ Write-up – 1

A thin biconvex lens of refractive index 3

2 is placed on a horizontal plane mirror as shown in the figure. The space between lens and the mirror is then filled with water of refractive index

4 3.

It is found that when a point object is placed 15 cm above the lens on the principal axis the object coincides with its own image.

51. At what distance object should be placed before water is filled so that image coincides with object if R is radius of curvature of lens

(a) 1.5 R (b) R (c) 2R (d)

R 2 52. What is value of R from above information

(a) R = 10 cm (b) R = 15 cm (c) R = 5 cm (d) R = 20 cm

53. In the above experiment when water is present, and parallel rays are incident then it will converge at a distance

(a) 2.25 cm (b) 15 cm (c) 10 cm (d) 7.5 cm 54. If an object is placed at 30 cm from lens then image obtained will be

(a) Magnified, real (b) Magnified, virtual (c) Diminished, real (d) Diminished, Virtual

55. On repeating the above experiment in which water is replaced by a liquid of refractive index m image again coincide at a distance 25 cm from the lens then refraction index of liquid is

(a) 1.5 (b) 1.4 (c) 1.8 (d) 1.6

@ Write-up – 2

The convex surface of a thin concavo–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm. The concave surface has a radius of curvature of 60 cm. The convex side is silvered and placed on a horizontal surface as shown in the figure.

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56. The focal length of the combination has the magnitude

(a) 8.6 cm (b) 7.5 cm

(c) 1.5 cm (d) 15 cm

57. The combination behaves like

(a) a concave mirror (b) a convex mirror (c) a concave lens (d) a convex lens

58. A small object is placed on the principal axis of the combination, at a distance of 30 cm in front of the mirror. The magnification of the image is

(a) 1 3 − (b) 3 4 (c) 5 (d) none of these. ASSERTOPM / REASON Codes.

(a) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for State-ment – 1.

(b) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

(c) Statement – 1 is True, Statement – 2 is False.

(d) Statement – 1 is False, Statement – 2 is True.

59. STATEMENT - 1

Geometrical optics can be regarded as the limiting case of wave optics. STATEMENT - 2

When size of obstacle or opening is very large compared to the wavelength of light then wave nature can be ignored and light can be assumed to be traveling in straight line.

60. STATEMENT - 1

Virtual images cannot be photographed. STATEMENT - 2

Rays from virtual image are diverging.

61. STATEMENT - 1

Virtual object can’t be seen by human eye. STATEMENT - 2

Virtual object is formed by converging rays.

62. STATEMENT - 1

A convex mirror is used as rear view mirror. STATEMENT - 2

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63. STATEMENT - 1

The behavior of any lens depends on surrounding medium. STATEMENT - 2

A lens can be looked upon as a collection of small prisms with varying prism angle.

MATRIX MATCH

Each question contains statements given in two column which have to be matched. Statements (a, b, c, d) in Column A have to be matched with statements (p, q, r, s) in Column B. The answers to these questions have to be appropriately bubbled as illustrated in the following example.

64. A ray of light strikes at the boundary separating two media at angle . and are refractive indices of media with .

65. Four rays of light parallel to optic axis and their path after passing through an optical system are shown in column-A match the corresponding optical instrument from Column-B

Column-A Column-B a) p) convex lens b) q) concare lens c) r) convex mirror d) s) concare mirror LEVEL MODEL QUESTIONS

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SINGLE CHOICE CORRECT QUESTIONS

KEY

1. (d) 2. (d) 3. (b) 4. (a) 5. (a) 6. (c) 7. (d) 8. (d) 9. (c) 10. (b, c)

MAY BE MORE THAN CORRECT CHOICE ANSWERS

11. (a, c, d) 12. (a, c, d) 13. (b) 14. (a, d) 15. (a, c) 16. (a, d) 17. (a, d) 18. (a, c) COMPREHENSION

@

Write-up – 2 19. (a) 20. (d) 21. (b) @ Write-up – 2 22. (c) 23. (a) 24. (c) ASSERTION / REASON 25. (b) 26. (a) 27. (a) 28. (a) 29. (a) MATRIX MATCH 30. (a – p, q, r, s), (b – q, r), (c – q, r), (d – p, q, r, s) 31. (a – q, s), (b – p, q, r, s), (c – p, q, s), (d – s) PRACTICE QUESTIONS

SINGLE CHOICE CORRECT QUESTIONS

32. (c) 33. (b)

34. (c) 35. (b)

36. (b) 37. (b)

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40. (d)

MAY BE MORE THAN CORRECT CHOICE ANSWERS

41. (b, d) 42. (a) 43. (a, b, c, d) 44. (b) 45. (b, d) 46. (b, c) 47. (a, c) 48. (a, b, c, d) 49. (a, c, d) 50. (b, c, d) COMPREHENSION @ Write-up – 1 51. (b) 52. (a) 53. (d) 54. (c) 55. (d) @ WRITE-UP – 2 56. (b) 57. (a) 58. (d) ASSERTION / REASON 59. (a) 60. (a) 61. (a) 62. (a) 63. (b) MATRIX MATCH 64. (a – s) (b – p) (c – q) (d – r) 65. (a – r) (b – q) (c – p) (d – s)

LEVEL - 1

MODEL QUESTIONS

SINGE CHOICE CORRECT QUESTIONS

HINTS AND SOLUTIONS 1. The image will be real and between C and O.

2. According to Snell’s Law, µsinθ=constant which gives µ = µ1 4. 3. Given A = 60º 3 i i A 45º 4 ′ = = =

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 i i A+ = + δ′ or 90º = 60º +

∴ δ =30º

Note that i i′= is the condition for minimum deviation. Hence δ =30º= δmin. 4 we know that sini sinr µ = and i r 90º+ = or r = 90 - i sini tani sin(90 i) µ = = − or i tan ( ) tan (1.62)= −1 µ = −1 . 5.

( )

4 / 3 v 6 1.5 = − final 16cm; v 1 28 7 cm 3 4 / 3 3   = − = = −   . 6.

(

)

1 2 1 1 1 1 f R R   = µ −  , For no dispersion 1 d 0 f   =     Þ 1 2 1 1 d 0 R R   µ=   Þ R1=R2. 7. Image will be real.

We know that 1 1 1 f v u= − ⇒ vf = −1 uv ⇒ v 1 mf = + [ u is negative] ⇒ v f(m 1)= + . 8. 1 2 1 2 1 1 1 1 f R R µ   = µ    1 3 / 2 1 1 1 f 4 / 3 0.3 0.3    = +    or 1 9 1 2 f 8 0.3    =    or 1 1 2 f 8 0.3= ×

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or f 1.20m= . 9. y= y y1× 2 = 16 9× = × =4 3 12cm. 10. 1 2 1 2 1 1 1 x f f= +f −f f 1 1 12 1 1 1 x f f= − +f f 1 12 1 x f = f ⇒ f 0> for every x. for x 0, f= = ∞

Hence for x = 0, system will behave like a glass plate. MAY BE MORE THAN CORRECT CHOICE ANSWERS

11. For convex mirror (positive focal length) image is always smaller in size. For concave mirror (negative focal length) image is smaller when object lies beyond 2f.

12. TIR takes place when ray of light travels from denser to rarer medium. Further,

3 2

12 13

1 1

sinθ =µ and sinθ =µ

µ µ Since, 3 2 1 1 µ µ > µ µ θ > θ12 13

Smaller the value of critical angle, more are the chances of TIR.

13. If m2 < 2, because of total internal reflection, angle of deviation is 20º. If m2 < 2, ray deviates by 60º. 14. n3 <n1 and n 1 < n2 ⇒ n 3 < n2. 15. d = (m – 1)A. 16. Using, 2 1 2 1 u R µ µ µ − µ − = ν , we get 1.5 1.0 1.5 1.0 20 − − = ν ∞ or ν = ±60cm.

17. Final image is formed at infinity if the combined focal length of the two lenses (in contact) becomes 30 cm or

1 1 1 30 20 f= +

i.e., when another concave lens of focal length 60 cm is kept in contact with the first lens Similarly, let be the refractive index of a liquid in which focal length of the given lens becomes 30 cm. Then

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1 2 1 3 1 1 1 20 2 R R     =    ... (i) 1 2 1 3 / 2 1 1 1 30 R R     = µ    ... (ii)

From equations, (i) and (ii), we get

9 8 µ = . 18. eq g w m 1 1 1 1 2 F f f f   = + + ×   = 3 1 1 4 1 1 2 2 1 1 2 10 15 3 15 15 15    +  +           ⇒ eq 90 F 17 = cm. COMPREHENSION @ Write-up – 1 19.

From Snell’s law, m sin a = sin f

1 1

sinα = sinφ <

µ µ

a is less than critical angle. 20.

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21. d 4d 2 d d δ= − α + φ φ d 0 d δ= φ , d 1 d 2 α ⇒ = φ 1 1 sin−  sin  α = φ µ   2 2 1 cos 3 µ − φ = . @ Write-up – 2 22 to 24. v1=m, u = 20 cm; 1 1 1 1 f = 20−−10 V2 = m2 u = 30 cm; 2 1 1 1 f =30−−10 1 2 cm 15 − = 1 17 f 60= 1 17 1 11 v 60= +−10 60= ⇒ m=uv

(

)

60 11 10 = − =6011 . ASSERTION / REASON

References

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