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CPP-11 Class - XI Batches - PHONON

LENS

1. A thin concavo-convex lens has two surfaces of radii of curvatureRand 2R. The material of the lens has a refractive

index . When kept in air, the focal length of the lens :

(A) Will depend on the direction from which light is incident on it

(B*) Will be the same, irrespective of the direction from which light is incident on it

(C) Will be equal to1

R(D*) Will be equal to

1

R

Sol. From left to right 1

1

Lf= ( 1)

1 1

2R R

= ( = 1)1 1

2R R

concavesurface

R

convexsurface

2R

1

1

Lf=

1

2

R

From right to left 2

1

Lf= ( 1)

1 1

2 ( )R R

= ( 1)1 1

2R R

= 1

2

R

1

1

Lf=

2

1

Lf =

2

( 1)

R

Option (B) and (D) are correct.

2. An object is placed 10 cm away from a glass piece (n= 1.5) of length

20 cm bound by spherical surfaces of radii of curvature 10 cm. Find the

position of the final image formed after twice refractions.

20 cmair

B

n = 1.5

ROC = 10cm10 cm

Aobject

ROC = 10cm

air

Ans. [50 cm]

Sol. Refraction at first surface :

1

1.5

V

1

(10)

=1.51

( 10)

V1= 30 cm fromA

Refraction at second surface.

2

1

V

1.5

(50)=

11.5

(10)V

2= +50 cm fromB

Hence final image will be 50 cm fromB.

3. A concave mirror of radiusRis kept on a horizontal table (figure). Water (refractive

index = ) is poured into it upto a height h. What should be distance of a pointobject from surface along principal axis so that its final image is formed on itself.

Consider two cases.

(i) h 0(ii) in terms ofh

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Ans. [(i)R

; (ii)

( )R h

]

Sol. Object should appear to be at distanceRfrom mirror.

(d) + h =R

d=R h

c

d

hif h

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For surface S1

2

1V

1

=

2 1

R...(1)

For surface S2

3

V

2

1

V=

3 2

R...(2)

From equation (1) and (2)

3

V=

2 1 32

R

V =3

2 1 32

R

Similarls

2

1

V

3

=

2 3

R...(1)

321

1

V

2

1

V

=1 2

R...(2)

V=1

2 1 32

R

7. An equiconvex lens of refractive index n2is placed such that the refractive :

(A*) Must be diverging if n2is less than the arithmetic mean of n1and n3(B*) Must be converging if n

2is greater than the arithmetic mean of n

1and n

3

(C) May be diverging if n2is less than the arithmetic mean of n

1and n

3

(D*) Will neither be diverging nor converging if n2is equal to arithmetic mean of n

1and n

3

Sol.for surface S1

2

1

n

V=

1n

u=

2 1

( )

n n

fR...(1)

for surface S2

3n

V

2

1

n

V=

3 2

( )

n n

R...(2) n2

n1n3

S2S1

From (1) and (2)

3 1n n

V u=

2 1n n

R

3 2n n

R=

=2 1 32 n n n

R

=1 3

2

2

2

n nn

R

8. In the figure given below, there are two convex lensL1andL

2having focal length of

f1andf

2respectively. The distance betweenL

1andL

2will be :

(A)f1

L1 L1

(B)f2(C*)f

1+f

2

(D)f1f

2

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Sol.

| |f2| |f1

| |+f1 | |f2

9. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQand

RSparallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm.

An upright objectABof height 1.2 cm is placed on the optic axis PQof the lens at a distance of 20 cm from the lens.

IfABis the image after refraction from the lens and reflection from the mirror, find the distance AB(in cm) fromthe pole of the mirror and obtain its magnification. Also locate positions ofA'andB' with respect to the optic axisRS.

0.6cm

P

R

A

B Q

S

20 cm30cm

Ans. [A' B'at 15 cm to the right of mirror.B'is 0.3 cm aboveRS&A'is 1.5 cm belowRS. Magnification is 1.5]

Sol. Reflection from lens

1 1

V u=

1

f

1 1

20V =

1

15

V = 60 cm

m =V

u=

60

20=

0

Ih

h

hI= 3h0= 3 1.2 = 3.6 cm

For mirror

A1

C1

B1

A'

C'

B'

1.5cm

0.3cmQ

5

30 cm 15 cm

0.6cm

3cm

(Image after reflectionfrom lens)

(Final image after reflectionfrom lens and mirror)

1 1

V u =

1

f

1 1

30V =

1

30V= 15 cm

m= V

u=

(15)

30=

1

2

10. A convexo-concave diverging lens is made of glass of refractive index 1.5 and focal length 24 cm. Radius of curvature

for one surface is double that of the other. Then radii of curvature for the two surfaces are (in cm) :

(A*) 6, 12 (B) 12, 24 (C) 3, 6 (D) 18, 36

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Sol.1

f=

2 1

1

1 2

1 1

R R

1

24=

1.5 1

1

1 1

2R R

=

1

2

3

2R

1

24=

1

2R

R = 6 cm, 2R= 12 cm

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1. When a lens of powerP(in air) made of material of refractive index is immersed in liquid of refractive index 0. Then

the power of lens is :

(A)0

1

P (B)0

1

P (C*) 01

.0

P

(D) None of these

Sol. P =1

f=

1

1

1 2

1 1

R R

P' =0

0

1 2

1 1

R R

'P

P=

0

0

1

( 1)=

0

1

0

1

P' =0

1

0

P

2. What will the paths of the ray be after refraction in the lenses.

[F1 First focus, F2 Second focus]

(a) (b)

Ans. [(a) ; (b) ]

Sol. (a)

f1 f2

(b)

f1 f2

3. A thin symmetrical double convex lens of power Pis cut into three parts, as shown in the figure. Power ofA

is :

(A) 2 P (B)2

P

(C)3

P(D*) P


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Sol. If we cut lens along principal axis, power of lens remain unchanged.

4. Lenses are constructed by a material of refractive index 1.50. The magnitude of the radii of curvature are 20 cm and

30 cm. Find the focal lengths of the possible lenses with the above specifications.

Ans. [ 24 cm, 120 cm]

Sol.1

f= ( 1)

1 2

1 1

R R

Case I - convex lens

1

f= (1.5 1)

1 1

20 30

= 1

2

1

12

f= 24 cm

Case II

1

f

= (1.5 1)1 1

20 30

f= 120 cm

5. A quarter cylinder of radiusRand refractive index 1.5 is placed on a

table. A point object Pis kept at a distance of mRfrom it. Find the value

of mfor which a ray fromPwill emerge parallel to the table as shown in

the figure.mR R

P

Ans.[m= 4/3]

Sol. for plane surface

1

1.5

V

1

mR=

1.51

A B

mR RV1= 1.5mR

for curved surface

1

1.5

(1.5 )mR R =11.5

R

1.5

1.5mR R=

0.5

R

3 = 1.5 m + 1 m = 4/3

6. A meniscus lens is made of a material of refractive index 2. Both its surfaces have radii

of curvatureR. It has two different media of refractive indices 1and

3respectively, on

its two sides (shown in the figure). Calculate its focal length for 1<

2<

3, when light

is incident on it as shown.

1

2

3

Ans.[f=3

3 1( )

R

]

Sol. For Istsurface 1

2

3

2

u

1

=

2 1

R

u =2

2 1

R

[u= v2]

For IIndsurface

3 2

2

rv v

=3 2rR

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3

v

2

2R(

2

1) =

3 2 r

R

3

v=

3

R

1

R

v=

3

3 1

R

7. A thin concave-concave lens is surrounded by two different liquids Aand Bas shown in figure. The system is

supported by a plane mirror at the bottom. Refractive index ofA, lens andBare 9/5, 3/2 and 4/3 respectively. The radius

of curvature of the surfaces of the lens are same and equal to 10 cm. Where should an object be placed infront of this

system so that final image is formed on the object itself.

Ans. [75 cm]

Sol.For image to form on object itself rays should fall perpendicularly on plane mirror/Focal length of combination will be :

P = P1+P

2+P

3

1

qf=

1

1

f+

2

1

f+

3

1

f

f1= 12.5 cm

f2= 10 cm

f3= 30 cm

Thus we get

feq= 75 cm

Hence object should be placed at 75 cm.

So, that light rays becomes parallel to principal axis.

8. The radius of curvature of the left & right surface of the concave lens are 10 cm & 15 cm respectively. 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

Sol. Here

Peq

= 2PL1

+ 2PL2

+ PM air water

= 4

__3

Glass

=4

__

2=1

1

Lf

+ 22

1

Lf

1

Mf

Peq = 21

12

+4

45

2

15

[Mirror is converging so in power is +ve.]

Peq =1

18

1

f=

1

18

f= 18 cm

Here system acts as concave mirror.

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9. A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space

between the 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 its principal axis, the object coincides with its own image. On repeating with another

liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the

liquid.

Ans.[n= 8/5 = 1.6]

Sol. For image to form on object itself, ray should strike the mirror perpendicularly.

Here

Peq = P1+ P2

=1

1

f+

2

1

f

1

1

f=

31

2

D

R

1

1

f=R

2

1f

= 413

1 1 R

2

1

f=

1

3

1

R=

1

3R

1

eqf=

1

R

1

3R=

3 1

3R=

2

3R

feq=3

2R= 15 cm [object is at focus]

R = 10 cmSame experiment is repeated using some other liquid.

feq

=1

1

f+

2

1

f

2

1

f= ( 1)

1

R

=1

R

( 1)

R

1

eqf= 1 1

R =

2

R R= 1

25 210

10

= 0.04

10= 0.2 0.04

= (0.2 0.04) 10

= 2 .4 = 1.6

10. An object Ois kept infront of a converging lens of focal length 30 cm behind

which there is a plane mirror at 15 cm from the lens :

(A) The final image is formed at 60 cm from the lens towards right of it

(B*) The final image is at 60 cm from lens towards left of it

(C*) The final image is real

(D) The final image is virtual

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Sol. Image formed after refraction form lens.30 cm

15cm 15cm

O

1

v

1

u=

1

f

1

v

1

(15)=

1

30

1

u=

1

30

2

30

v=1

30

v= 30 cm

for this virtual image, image formed by plane mirror will be at 45 cm on light of mirror this image will be real.

And for this image thus is at distance 2f i.e. at 60 cm. Hence its real image again will be formed on 2fof lens in its left

side.

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1. A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The object is at

15 cm from the lens. The length of the image is :

(A) 1 mm (B*) 4 mm (C) 2 mm (D) 8 mm

Sol.1

v

1

u=

1

f

1

v

1

15=

1

10

2

1dv

du v

+

2

1

v= 0

1

v=

1

10

1

15

dv

du=

2

2

v

u=

3

30

2

30

dv=

2

2

v

udu

1

v=

1

30v = 30

dv =

2

2

(30)

(15) 1

= 4 1 = 4 mn.

2. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principalaxes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incident

on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the

emergent beam. Also find out the ratio of emergent and incident intensities.

Ans. [1.0 cm if the light is incident from the side of concave lens and 2.5 mm if it is incident from the side of the convex

lens and the corresponding ratio of intensities are 1/4 and 4]

Sol. Case I:- 1stconvex t hen concave Image from convex long is at focus of concave so emergent light ray becomesparallel to principal axis.

A

B

C

Dt

F

0.5cm

10cm 10cm

f= 20 f= 10

From geometry :AB

BF=

CD

DF[CD width of emergent beam]

0.5

20=

10

CD

CD=10 0.5

20

CD= 2.5 mm

Intensity area beam croos section.


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I emergent

I incident=

2

2

(2.5)

(5)=

1

4= 1 : 4 ans

Case II :- 1stconcave, then convex

A

B D

C

10cm 10cm

0.5cmF

Since image of concave lens is at focus of convex lens so final emergent ray becomes parallel to principal axis.

From geometry CD

DF=

AB

BF

20

CD

cm=

0.5

10

cm

cm

CD= 1 cm

I emergent

I incident =

21

0.5

cm

cm

= 4 Ans.

3. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5

has a radius of curvature 20 cm. The concave surface has a radius of curvature 60

cm. The convex side is silvered and placed on a horizontal surface as shown in

figure. (a) Where should a pin be placed on the axis so that its image is formed at

the same place ? (b) If the concave part is filled with water (= 4/3), find thedistance through which the pin should be moved so that the image of the pin

again coincides with the pin.

Ans. [(a) 15 cm from the lens on the axis; (b) 1.14 cm towards the lens]

Sol. (a) Net focal length.

1

eqf=f

eq=

1

Lf+

1

Lf

1

Mf

R=60cm

=3/2

R=20cm

=2

Lf

1

Mf= 2

1 1 1 2

2 60 20 20

1

eqf=

4

30cm

feq= 7.5 cm

feq= 7.5 2 = 15 cmHence object should be placed at centre of curvature of equivalent mirror i.e. at 15 cm from mirror.

(b) 1

eqf= Peq=

2

wf+

2

Lf

2

Mf

= 21

3

11

60

+ 2

1

2

1 1

60 20

+

2

20

=2

3

1

60

+1 1

20 60

+2

20

=1

90+

6020

60 20+

2

20

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=1

90+

40

60 20+

2

20

=1

90+

4

120+

1

10

=1

90+

1

30+

1

10=

1

90+

3

90+

9

90=

13

90

feq

=13

90

Peq =180

13= 13.85 cm

Shift of object 15 13.85 = 1.15 cm

4. An insect at point 'P' sees its two images in the water-mirror system as shown

in the figure. One image is formed due to direct reflection from water surface

and the other image is formed due to refraction, reflection & again refraction

by water mirror system in order. Find the separation between the two images.

Mhas focal length 60 cm. (nw= 4/3).

Ans. [Distance P'P"= 36 12 = 24 cm]

Sol. Image (1)By direct reflection from top water surface. 12 cm below water furpace.Image (2)1strefraction from top water surface.

1

4 / 3

v

1

(12)=

(4/31)

v1 16cm

reflection from convex mirror

2

1 1

(40)v =

1

( 60) v2= +24 cm

2ndrefraction from top trasfer surface.

3

1 4 / 3

(48)v=

1 4 / 3

v

3= 36 cm

Image (2) is 36 cm below top water surface

Separation of two image = 36 12 = 24 cm

5. A symmetrical converging convex lens of focal length 10 cm & diverging concave

symmetrical lens of focal length 20 cm are cut from the middle & perpendicularly

and symmetrically to their principal axis. The parts thus obtained are arranged as

shown in the figure. The focal length of this arrangement will be :

(A) (B) 20 cm(C) 40 cm (D*) 80 cm

Sol. ForL1P

1= ( 1)

1

1 1

R

L1

L2 L3

R1R2 R2

2c

= +1

( 1)

R

ForL2P2 = ( 1) 2

1 1

( )R

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= 2

( 1)

R

ForL3P

3= ( 1)

2

1 1

( )R

= 2

( 1)

R

For converging 10 = 12( 1)R

R1= 20 ( 1)

For diverging 20 =2

2( 1)

R

R2= 40 ( 1)

P1

= +1

( 1)

R= +

1

20

P2

= 2

( 1)

R=

1

40= P

3

P12 = P1+ P2 = + 120 1

40= + 1

40

Combined P123= P12+ P3 dP12P3

=1

40

+1

40

201

40

1

40

= +1

40

1

40+

1

80

P123

= +1

80

F123= +80 cm

6. A hollow sphere of glass of R.I. n has a small markMon its interior

surface which is observed by an observer Ofrom a point outside the

sphere. Cis centre of the sphere. The inner cavity (air) is concentric

with the external surface and thickness of the glass is everywhere equal

to the radius of the inner surface. Find the distance by which the mark

will appear nearer than it really is, in terms ofnandRassuming paraxial

rays.

glass

air

CMO

2R

4RAns. [(n 1)R/(3n 1)]

Sol. Refraction from surface CD:

1

n

v

1

(2 )R=

( 1)

( )

n

R

v1 =2

(1 2 )

nR

nfrom surface (1)

CMO

(1)(2)

Refraction from surface (2)

2

1

v

(4 1)

(2 1)

n

R n

n

=1

(2 )

n

R

2

1

v+

(2 1)

(4 1)

n n

R n=

( 1)

2

n

Rv

2=

2 (4 1)

(3 1)

R n

n

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Shift = 3R2 (4 1)

(3 1)

R n

n

=(3 1)

R

n[3(3n 1) 2(4n 1)]

Shift =

( 1)

(3 1)

R n

n

7. Two media each of refractive index 1.5 with plane parallel boundaries are

separated by 100 cm. A convex lens of focal length 60 cm is placed midway

between them with its principal axis normal to the boundaries. A luminous

point object Ois placed in one medium on the axis of the lens at a distance

125 cm from it. Find the position of its image formed as a result of refraction

through the system.

Ans. [200 cm, w.r.t. lens]

Sol. Distance of object as seen by lens

= 50 +

75

1.5

4=1

50

75

= 100 cm

Now1 1

v u

=1

f

1 1

(100)v

=1

60

1 1

100v=

1

60

v=6000

40= 150

= 50 +100Apparent distnace

50 + 100 = 200 cm

8. A point object is placed at distance of 20 cm from a thin planoconvex lens of focal length

15 cm. The plane surface of the lens is now silvered. The image created by the system is at :

(A) 60 cm to the left of the system

20(B) 60 cm to the right of the system

(C*) 12 cm to the left of the system (D) 12 cm to the right of the system

Sol. Image after 1strefraction from lens

1

v

1

(20)=

1

15

1

v=

5

20 15v= 60 cm

After reflation from mirror

v' = 60 cm.

again after refraction from lens

1

v

1

( 60) =1

15

1

v

=1

15

+1

60

v=45

60 15

v= 12 cm, to the left.

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9. An object Ois kept in air and a lens of focal length 10 cm (in air) is kept at the bottom of

a container which is filled upto a height 44 cm by water. The refractive index of water is

4/3 and that of glass is 3/2. The bottom of the container is closed by a thin glass slab of

refractive index 3/2. Find the position of the final image formed by the system.

Ans. [90 cm]

Sol. Focal length of lens is water =

3 1 410

2 8 33 9

2 3

=

1 410

2 3

2

6

= (30) 4

3= 40 cm

Object will appear to lens at 44 + 12 4

3

= 60 cm

Lens image will be at

1 1

(60)v=

1

40 v=

40 60

20

= 120 cm

After refraction from water, image will be at

120 3

4= 90 cm

10. A stationary observer Olooking at a fish F(in water of, = 4/3) through a converging

lens of focal length 90 cm, The lens is allowed to fall freely from a height 62.0 cm with

its axis vertical. The fish and the observer are on the principal axis of the lens. The

fish moves up with constant velocity 100 cm/s. Initially it was at a depth of 44.0 cm.

Find the velocity with which the fish appears to move to the observer at

t= 0.2 sec. (g= 10 m/s2)

Ans. [91

4m/s = 2275 cm/s (upwards)]

Sol.

20cm

200cm/sec42cm

24cmF '

F

u=75cm/sec

After .2 sec [ lens fall by 20 cm in 0.2 sec]

Image of fish F1after 0.2 sec will be at 18 cm from water air surface so distance of fish from lens will be

(42 + 18) = 60 cm

Image of this fish due to lens is

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1

v

1

u=

1

f

v = 180 cm

We know that

dv

dt=

2

2

v

u

du

dt[Here uis ve and udecrease with time as fish is coming near lens]

dv

dt=

2

2

v

u(200 + 75)

dv

dt=

2180

60

(275)

dv

dt= 2475 cm/sec

So this is the speed of image wrto lens

vIL= vI vL=2475

vI = 2475 +vL [vL= 200] v

I= 2275 cm/sec.

lens-d

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