Lens-D

CPP-11

Batches - PHONON

Class - XI

LENS 1.

A thin concavo-convex lens has two surfaces of radii of curvature R and 2R. The material of the lens has a refractive index µ. 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 R (C) Will be equal to µ  1

Sol.

From left to right 

(D*) Will be equal to

R µ 1

1 1   1  1 1 – – f L1 = (µ – 1)  – R –2 R  = (µ = 1)  2R R 

µ

1  µ – 1 f L1 = – 2 R

From right to left 

1 f L2 = (µ – 1)

=–



 1 1  –   = (µ – 1)  2 R (  R)  

convex surface 2R

concave surface R

 1 1 –    2R R 

 µ – 1 2R

1 1 2R = = – f L1 f L2 ( µ –1)

Option (B) and (D) are correct. 2.

Sol.

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. Ans. [50 cm] Refraction at first surface :

20 cm

air

air B

object

A 10 cm ROC = 10cm

n = 1.5 ROC = 10cm

1.5 1 1.5 –1 – =  V1 = – 30 cm from A V1 (–10) (10)

Refraction at second surface. 1 1.5 1 – 1.5 – =  V2 = +50 cm from B V2 (–50) (–10)

Hence final image will be 50 cm from B. 3.

A concave mirror of radius R is kept on a horizontal table (figure). Water (refractive index = ) is poured into it upto a height h. What should be distance of a point object from surface along principal axis so that its final image is formed on itself. Consider two cases. (i) h 0 (ii) in terms of h

Page 1

( R  h) R ; (ii) ]  

Ans. [(i) Sol.

Object should appear to be at distance R from mirror.  µ(d) + h = R

4.

Sol.

d

 R–h  d=   µ 

 if

h << R



d=

c

h

R µ

A person's eye is at a height of 1.5 m. He stands infront of a 0.3 m long plane mirror which is 0.8 m above the ground, The length of the image he sees of himself is : (A) 1.5 m (B) 1.0 m (C) 0.8 m (D*) 0.6 m 2 × height of mirror = 2 × 0.3 = 0.6 m 0.3 m 1.5m

0.8 m

5.

The values of d1 & d2 for final rays to be parallel to the principal axis are : (focal lengths of the lenses are written above the respective lenses) (A*) d1 = 10 cm, d2 = 15 cm (B*) d1 = 20 cm, d2 = 15 cm (C*) d1 = 30 cm, d2 = 15 cm (D) None of these

Sol.

f1=10cm f2=20cm f3= –5cm

O L1

10cm

L2 d1

L3 d2

Point object O in placed on focus of lines L1 as shown in figure so there is no condition for d1. After reflection from L3, ray become parallel so ray must comes from focus of L3. So, d2 = 15 cm or d2 = 25 cm 6.

A thin lens made of a material of refractive index 2 has medium with refractive index µ1 and –µ3 on either side. The lens is biconvex and the two radii of curvature has equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed it the beam is incident from (a) the medium 1 and (b) from the medium 3 ? Ans. [(a)

Sol.

3 R 1 R ; (b) ] 2 2  1  3 2 2  1  3

S1

S2

µ1 µ2

µ3

Page 2

For surface S1 µ2 µ µ – µ1 – 1= 2 V1  R

...(1)

For surface S2 µ2 µ3 – µ2 µ3 – V = V –R 1

...(2)

From equation (1) and (2) µ3 2 µ2 – µ1 – µ3 = V R

V=

µ3 R 2µ2 – µ1 – µ3

µ1

Similarls µ2 µ3 µ2 – µ3 – = V1  R

...(1)

µ1 µ2 µ1 – µ2 – = V1 V –R

...(2)

µ2 µ3

µ1 R V = 2µ – µ – µ 2 1 3

7.

Sol.

An equiconvex lens of refractive index n2 is placed such that the refractive : (A*) Must be diverging if n2 is less than the arithmetic mean of n1 and n3 (B*) Must be converging if n2 is greater than the arithmetic mean of n1 and n3 (C) May be diverging if n2 is less than the arithmetic mean of n1 and n3 (D*) Will neither be diverging nor converging if n2 is equal to arithmetic mean of n1 and n3 for surface S1 n2 n2 – n1 n1 = = V1 ( fR) u

...(1)

for surface S2

S1

n3 – n2 n3 n2 – V = (– R ) V 1

S2

n1 ...(2)

n2

n3

From (1) and (2) n3 n1 n2 – n1 n3 – n2 – = – = V u R R n1  n3  2  2 n2 – n1 – n3  =  = R  n2 – 2  R    

8.

In the figure given below, there are two convex lens L1 and L2 having focal length of f1 and f2 respectively. The distance between L1 and L2 will be : (A) f1 (B) f2 (C*) f1 + f2 (D) f1 – f2

L1

L1

Page 3

| f1 |+| f2|

Sol.

|f1|

9.

|f2|

A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A B is the image after refraction from the lens and reflection from the mirror, find the distance A B (in cm) from the pole of the mirror and obtain its magnification. Also locate positions of A' and B' with respect to the optic axis RS. A P R

B

Q

0.6cm S

30 cm

Sol.

20 cm

Ans. [A' B' at 15 cm to the right of mirror. B' is 0.3 cm above RS & A' is 1.5 cm below RS. Magnification is – 1.5] Reflection from lens 1 1 1 – = f V u 1 1 1  = V 20 15

V = 60 cm m=

hI V 60 = = h u –20 0

hI = –3 h0 = –3 × 1.2 = – 3.6 cm For mirror A1 A'

0.6cm C1

Q 0.3cm

C'

5

1.5cm

3cm 30 cm B1 (Image after reflection from lens)

15 cm B' (Final image after reflection from lens and mirror)

1 1 1  = f V u 1 1 1  =–  V = – 15 cm V 30 30

m= –

10.

V (–15) 1 = – = u 30 2

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 Page 4

1  µ2 – µ1   1 1  =  µ  R –R  f  1  1 2 

Sol.

–

1  1.5 –1  =  24  1 

 1 1  1 –  =   R 2R   2 

 3  –   2R 

1 1 = 24 2R R = 6 cm, 2R = 12 cm

Page 5

CPP-12

Batches - PHONON

Class - XI

LENS 1.

When a lens of power P (in air) made of material of refractive index µ is immersed in liquid of refractive index µ0. Then the power of lens is :   0  1   0 P (A) P (B) P (C*) . (D) None of these  1   0   1 0 1  µ –1  P= f =    1 

Sol.

 1 1   –  R R  1 2 

 µ – µ0   1 1  P' =  µ   R – R  0 2    1

P '  µ – µ0 = P  µ0

 1  µ – µ0  1 ×   =  ( µ –1)  µ – 1  µ0

 µ – µ0  P  P' =   µ – 1  µ0

2.

What will the paths of the ray be after refraction in the lenses. [F1 – First focus, F2 – Second focus]

(a)

(b)

Ans. [(a)

Sol.

;

(a)

]

(b) f1

3.

(b)

f2

f1

f2

A thin symmetrical double convex lens of power P is cut into three parts, as shown in the figure. Power of A is : (A) 2 P (C)

P 3

(B)

P 2

(D*) P Page 6

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]  1 1 1   –  = (µ – 1) f  R1 R2 

Sol.

Case I - convex lens 1 = ± (1.5 – 1) f

1   1    20 30  

1  1  =±      2   12  f = ± 24 cm Case II 1 1   1 –  = ± (1.5 – 1)  f  20 30 

f = ± 120 cm 5.

Sol.

A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance of mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure. Ans. [m = 4/3] for plane surface 1.5 1 1.5 – 1 – = V1 mR 

V1 = 1.5 mR for curved surface

P mR

A mR

R

B R

1.5 1.5 1 1 – 1.5 0.5 – (1.5mR  R ) =  = 1.5mR  R  –R R

3 = 1.5 m + 1  m = 4/3 6.

A meniscus lens is made of a material of refractive index µ2. Both its surfaces have radii of curvature R. It has two different media of refractive indices µ1 and µ3 respectively, on its two sides (shown in the figure). Calculate its focal length for µ1 < µ2 < µ3, when light is incident on it as shown. 3 R ] (3  1 ) For Ist surface

Ans. [ f = Sol.

µ1

µ1

µ3 µ2

µ2 µ3

µ2 µ1 µ2 – µ1 – = u  R

 µ2 R  u =  µ – µ  [ u = v2] 1   2 For IInd surface µ3 r2 µ3 – r2 – = v v2 R

Page 7

µ2 µ3 µ –r – (µ2–µ1) = 3 2 µ R v R 2 µ3 µ3 µ1 = – v R R

µ3 R v= µ –µ 3 1

7.

Sol.

A thin concave-concave lens is surrounded by two different liquids A and B as shown in figure. The system is supported by a plane mirror at the bottom. Refractive index of A, lens and B are 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] For image to form on object itself rays should fall perpendicularly on plane mirror/ Focal length of combination will be : P = P1 + P2 + P3 1 1 1 1 f q = f1 + f 2 + f 3

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.

Sol.

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 Here Peq = 2PL1 + 2PL2 + PM air water 4 __ µ=  1   1   1  2 4       = f +2 f – µ= __  L1   L2   fM  3

 1 4  2  Peq = – 2   + –   12  45  –15  1 Peq = 18

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

Glass

–1 1 = f 18

f = – 18 cm Here system acts as concave mirror.

Page 8

9.

Sol.

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] For image to form on object itself, ray should strike the mirror perpendicularly. Here Peq = P1 + P2 1 1 = f + f 1 2 1 1  3   D –1    = =R  f1 f1  2  R 1 4   1 1 –1 – – = f 2  3   R   1 1 1 –1 =–   = f2 3R 3 R 1 1 3 –1 2 1 f eq = R – 3R = 3R = 3R

3 R = 15 cm [ object is at focus] 2 R = 10 cm Same experiment is repeated using some other liquid.

feq =



1 1 feq = f + f 1 2 1 = (µ – 1) f2

=

 1 – R   

1 ( µ –1) – R R

1 2–µ 1 2 µ 1 – µ 1 = = =  – = 0.04 f eq R R 25 10 10 R

µ = 0.2 – 0.04 10

µ = (0.2 – 0.04) 10 = 2 – .4 = 1.6 10.

An object O is 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

Page 9

Sol.

Image formed after refraction form lens.

30 cm

1 1 1 – = f v u 1 1 1 – = (–15) v 30

O 15cm

15cm

1 1 2 = – u 30 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 2f of lens in its left side.

Page 10

CPP-13

Batches - PHONON

Class - XI

LENS 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 1 1 1 – = f v u

Sol.

1 1 1 – = v –15 10

– dv  1  1  + =0 du  v 2  v 2

1 1 1 = – v 10 15

dv v2 = 2 du u

2.

Sol.

dv =

v2 du u2

dv =

(30) 2 ×1 (15) 2

=

3 2 – 30 30

1 1 =  v = 30 v 30

= 4 × 1 = 4 mn. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes 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] Case I:- 1st convex t hen concave  Image from convex long is at focus of concave so emergent light ray becomes parallel to principal axis.

f = 10

f = 20 0.5cm

A

C

B

D 10cm

From geometry :

AB CD = BF DF



0.5 CD = 20 10



CD =

t F 10cm

[CD — width of emergent beam]

10  0.5  CD = 2.5 mm 20

Intensity  area beam croos section. Page 11

(2.5) 2 I emergent 1 = = = 1 : 4 ans (5) 2 I incident 4

Case II :- 1st concave, then convex

C A 0.5cm F

B 10cm

D 10cm

 Since image of concave lens is at focus of convex lens so final emergent ray becomes parallel to principal axis. CD 0.5cm CD AB = Þ = 20cm 10cm DF BF CD = 1 cm

 From geometry Þ 

2



3.

Sol.

I emergent  1cm  =  = 4 Ans. I incident  0.5cm 

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 the distance 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] (a) Net focal length. –1 1 1 1 f eq = feq = f L + f L – f M

2 1 1  1 1  2 = f – f = 2   –     2  60 20   20 L M

R=60cm µ=3/2 R=20cm

–1 4 f eq = 30 cm

feq = 7.5 cm feq = 7.5 × 2 = 15 cm Hence object should be placed at centre of curvature of equivalent mirror i.e. at 15 cm from mirror. (b)

–

2 2 1 2 = Peq = + – f eq fw f L f M

1 =2  3

1  1  60  + 2  

1    2

 –1 1  2  60  20  + 20  

=

1 2 1 1 2 + –  +   3  60   20 60  20

=

1 60 – 20 2 + + 90 60  20 20

Page 12

=

1 40 2 + + 90 60  20 20

=

1 4 1 + + 90 120 10

=

1 1 1 1 3 9 13 + + = + + = 90 30 10 90 90 90 90

feq = Peq =

13 90 180 = 13.85 cm 13

Shift of object 15 – 13.85 = 1.15 cm 4.

Sol.

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. M has focal length 60 cm. (nw = 4/3). Ans. [Distance P' P" = 36 – 12 = 24 cm] Image (1) By direct reflection from top water surface. 12 cm below water furpace. Image (2) 1st refraction from top water surface. 4/3 1 (4 / 3 –1) – =  v1 – 16cm v1 (–12) 

reflection from convex mirror 1 1 1  =  v2 = +24 cm v2 (–40) (60)

2nd refraction from top trasfer surface. 1 4/3 1– 4/3 – =  v3 = 36 cm v3 (–48) 

Image (2) is 36 cm below top water surface  Separation of two image = 36 – 12 = 24 cm

5.

Sol.

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  1 1 For L1  P1 = (µ – 1)   R –    1 

=+

( µ –1) R1

L3

L2 L1

R1

R2

R2 2c

1 1  For L2  P2 = (µ – 1)   – ( R )   2 

Page 13

=–

( µ –1) R2

 1 1 ( µ –1) For L3  P3 = (µ – 1)  (– R ) –   = – R  2  2

For converging 

R1 10 = 2(µ –1)  R1 = 20 (µ – 1)

For diverging 

R2 20 = 2(µ –1)  R2 = 40 (µ – 1)



P1 = +

( µ –1) 1 =+ R1 20

P2 = –

( µ –1) 1 =– = P3 R2 40

P12 = P1 + P2 = +

1 1 1 – =+ 20 40 40

Combined  P123 = P12 + P3 – dP12 P3

 1   1   1  1  =    +  –  –20     –   40   40   40   40  =+

P123 = +  6.

Sol.

1 1 1 – + 40 40 80 1 80

F123 = +80 cm

A hollow sphere of glass of R.I. n has a small mark M on its interior surface which is observed by an observer O from a point outside the sphere. C is 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 of n and R assuming paraxial rays. Ans. [(n – 1)R/(3n – 1)] Refraction from surface CD:

M

C O

air glass 2R 4R

n 1 (n –1) – = v1 (–2 R ) (– R ) 2nR v1 = from surface (1) (1 – 2n)

(1) M

C

(2) O

Refraction from surface (2) 1 – v2

n 1– n =  R(4n – 1)  (–2 R ) –   (2n – 1) 

1 n(2n –1) 2 R (4n –1) (n –1) + =  v2 = – (3n –1) v2 R (4n –1) 2R

Page 14

Shift = 3R –

2 R (4n –1) (3n –1)

R = (3n –1) [3(3n – 1) – 2(4n – 1)] R (n –1) Shift = (3n –1)

7.

Sol.

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 O is 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] 50 Distance of object as seen by lens 4=1 75 = 50 + 1.5 75 = 100 cm Now





1 1 1 – = f v u

1 1 1 1 1 1 – =  – = v (–100) 60 v 100 60

v=

6000 = 150 40

= 50 +100 Apparent distnace 50 + 100µ = 200 cm 8.

Sol.

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 (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 Image after 1st refraction from lens 



20

1 1 1 – (–20) = v 15 1 5 =  v = 60 cm v 20  15

After reflation from mirror v' = – 60 cm. again after refraction from lens 1 1 1 – (60) = v 15





1 1 1 = + v 15 60

v=

45 60  15

v = 12 cm, to the left. Page 15

9.

An object O is 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] 3 1 4 10  –   2 8 3 Focal length of lens is water = 3 9  –  2 3

Sol.

1  4   10    2  3   = 2 6

= (30) ×

4 = 40 cm 3

4 3 = 60 cm

Object will appear to lens at 44 + 12 × Lens image will be at

1 1 1 40  60 – = v = = 120 cm v (–60) 40 20

After refraction from water, image will be at 120 ×

10.

3 = 90 cm 4

A stationary observer O looking at a fish F (in water of, µ = 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 m/s = 2275 cm/s (upwards)] 4

Sol. 20cm After .2 sec

[ lens fall by 20 cm in 0.2 sec] 200cm/sec 42cm

u=75cm/sec F'

24cm F

Image of fish F1 after 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 Page 16

1 1 1 – = f v u

v = – 180 cm We know that dv v 2 du = dt u 2 dt

[Here u is –ve and u decrease with time as fish is coming near lens]

dv v 2 = (200 + 75) dt u 2 2

dv  180  = (275) dt  60 

dv = 2475 cm/sec dt

So this is the speed of image wrto lens vIL= vI – vL = 2475 vI = 2475 + vL [ vL = – 200]  vI = 2275 cm/sec.

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