Geometrical Optics

E26-1

Q.1 The image of a real object formed by a plane mirror is-

(1) Erect, real and of equal size (2*) Erect, virtual and of equal size

(3) Inverted, real and of equal size (4) Inverted, virtual and of equal size

Q.2 An object is at a distance of 0.5 m in front of a plane mirror. Distance between the object and image is

(1) 0.5 m (2*) 1 m (3) 0.25 m (4)1.5 m

Sol. Distance between object and image = 0.5 + 0.5 = 1 m

Q.3 The light reflected by a plane mirror may form a real image

(1) If the ray incident on the mirror are diverging (2*) If the rays incident on the mirror are converging

(3) If the object is placed very close to the mirror (4) Under no circumstances

Sol.

Q.4 Aclock hung on a wall has marks instead of numbers on its dial. On the opposite wall there is a mirror, and the image of the

clock in the mirror if read, indicates the time as 8 : 20. What is the time in the clock-

(1*) 3 : 40 (2) 4 : 40 (3) 5 : 20 (4) 4 : 20

Sol. 11: 60 – 08 : 20 = 3 : 40

Sol.

= 120º Anticlockwise = (360º – 120º) clockwise

Geometrical Optics

Plane mirror

Q.5 A ray of light incident on a plane miror at an angle of incidence of 30º. The deviation produced by the mirror

is-

(1) 30º (2) 60º (3) 90º (4*) 120º

EXERCISE-I

Geometrical Optics

E26-2

Q.6 Two mirrors are placed perpendicular to each other. A ray strikes the first mirror and after reflection from the

first mirror it falls on the second mirror. The ray after reflection from second mirror will emerge:

(1) Perpendicular to the original ray (2*) Parallel to the original ray

(3) At 450 to the original ray (4) At 600 to the original ray

Q.7 Two plane mirrors are inclined to one another at an angle of 60º. A ray is incident on mirror M1 at an angle i.

The reflected ray from mirror M2 is parallel to mirror M1 as shown in figure. The angle of incidence i is

))

ii)60º

M2

M1O

(1) 20º (2) 10º (3*) 30º (4) 40º

Sol.

i + 2l = 90º ...(i)

i + 2l + 90º = 180º ...(ii)

i + 2l = 90º

i + l = 60º

r = 30º

Q.8 A thick plane mirror shows a number of images of the filament of an electric bulb. Of these, the brightest image

is the-

(1) First (2*) Second (3) Last (4) Fourth

Sol. A thick mirror forms a number of images. Image is formed by front surface which is unpolished and hence, reflects

only a small part of light, while second image is formed by polished surface which reflects most of intensity. Hence

second image is brightest.

Q.9 At what angle must two plane mirrors be placed so that incident and resulting reflected rays are always parallel

to each other

(1) 0º (2) 30º (3) 60º (4*) 90º

Q.10 A ray gets succesively reflected from two mirrors inclined at an angle of 40º. If the angle of incidence on the

first mirror is 60º, then the net deviation of this ray is -

(1) 40º (2) 280º (3*) 80º (4) 100º

Sol. 1

= 180º – 2 × 60º = 60º

2

= 180º – 40º = 140º

Net = r1– r

2= 140º – 60º

80º or (–80º) = 280º

Geometrical Optics

E26-3

Q.11 When a plane mirror is placed horizontally on a level ground at a distance of 60m from the foot of a tower, the top of the

tower and its image in the mirror subtend an angle of 90° at the eye. The height of the tower will be

(1)30 m (2*) 60 m (3)90 m (4)120 m

Sol. m60h60

h45tan

Q.12 A man of height 160 cm. stands in front of a plane mirror. His eyes are at a height of 150 cm. from the floor.

Then the minimum length of the plane mirror for him to see his full length image is

(1) 85 cm (2) 170 cm (3*) 80 cm (4) 340 cm

Sol. Mirror height = man height

16080

2

Q.13 A man 180 cm high stands in front of a plane mirror. His eyes are at a height of 170 cm from the floor. Then the minimumlength of plane mirror for him to see his full length image is-(1*) 90 cm (2) 180 cm (3) 45 cm (4) 360 cm

Sol. Lmin

=2

H=

2

180= 90 cm

Q.14 If an object is placed symmetrically between two plane mirrors, inclined at an angle of 72°, then the total number of imagesformed is-(1)5 (2*) 4 (3)2 (4) Infinite

Sol. n =

1–

72

360= 4

Q.15 An object is placed between two plane mirrors set at 60º to each other. The number of images seen will be :

(1) 2 (2) 3 (3*) 5 (4) 6

Sol. n =360º 1 360º

160

n = 5

Q.16 A convex mirror has a focal length f. A real object is placed at a distance f in front of it from the pole, then it producesan image at-(1) Infinity (2) f (3*) f/2 (4) 2f

Sol.u

1

v

1 =

f

1

f

1

v

1

=

f

1

v =2

f

Curved mirror

Geometrical Optics

E26-4

Q.17 The focal length of spherical mirror is

(1) Maximum for red light (2) Maximum for blue light

(3) Maximum for white light (4*) Same for all lights

Q.18 The image formed by convex mirror of focal length 30 cm is a quarter of the size of the object. Then the distance of theobject from the mirror, is-(1) 30 cm (2*) 90 cm (3) 120 cm (4) 60 cm

Sol. v =4

u

u

1

v

1 =

f

1

u = 90 cm

Q.19 The largest distance of the image of a real object from a convex mirror of focal length 10 cm can be-(1) 20 cm (2) Infinite(3*) 10 cm (4) Depends on the position of the object

Q.20 In case of concave mirror, the minimum distance between a real object and its real image is-(1) f (2) 2f (3) 4f (4*) Zero

Ans. (4)Sol. When objects is at the centre of curvature C then its image is also at C

Q.21 An object is placed at a distance of 40 cm in front of a concave mirror of focal length 20cm. The image produced

is :

(1) virtual and inverted (2) real and erect

(3) real inverted and diminished (4*) real, inverted and of same size as the object.

Sol.1 1 1

v u f

1 1 1

v 40 20

v = – 40 cm

40vm 1

u 40

Q.22 Sun subtends an angle of 0.5º at the pole of a concave mirror of radius of curvature 15 m. The diameter of the

image of the sun formed by the mirror is

(1) 8.55 cm (2) 7.55 cm (3*) 6.55 cm (4) 5.55 cm

Sol. d arc

angle15 / 2 radius

15d

2

= 8.7 × 7.5 × 10–3

= 6.5 × 10–2 cm

Q.23 A spherical mirror forms an erect image three times the linear size of the object. If the distance between the

object and the image is 80 cm, the focal length of the mirror is

(1) 15 cm (2) – 15 cm (3*) – 30 cm (4) 40 cm

Sol.v v

m 3u u

u = 3u ...(i)

u + v = 80 ...(ii)

Geometrical Optics

E26-5

1 1 1

v u f ....(iii)

from (i) and (ii)

v = 60, u = 20

Putting in equation (iii)

1 1 1

f 60 20

f = – 30 cm

Q.24 Which of the following can form erect, virtual, diminished image?(1) plane mirror (2) concave mirror (3*) convex mirror (4) none of these

Sol.

Q.25 The focal length of a concave mirror is 20 cm. Determine where an object must be placed to form an image magnifiedtwo times when the image is real-(1*) 30cm from the mirror (2) 10cm from the mirror(3) 20cm from the mirror (4) 15cm from the mirror

Sol. v = 2u

u

1

v

1 =

f

1

u = – 30 cm

Q.26 A point source is placed 15 cm away from a convex mirror. A virtual image is formed at a distance of 6 cm. The

radius of curvature of the mirror is :

(1) 4.3 cm (2) 8.6 cm (3) 10 cm (4*) 20 cm

Sol.1 1 1

u v f

1 1 1

6 15 f

f = 10 cm

R = 2×10 = 20 cm

Q.27 Which of the following could not produce a virtual image

(1) Plane mirror (2) Convex mirror

(3) Concave mirror (4*) All the above can produce a vritual image

Q.28 The field of view is maximum for

(1) Plane mirror (2) Concave mirror (3*) Convex mirror (4) Cylindrical mirror

Q.29 The focal length of a concave mirror is f and the distance from the object to the principle focus is x. The ratio of the

size of the image to the size of the object is

(1)f

xf (2*)

x

f(3)

x

f(4) 2

2

x

f

Sol.uf

f

O

1

; where u N= f + x

x

f

O

1

Geometrical Optics

E26-6

Q.30 A person sees his virtual image by holding a mirror very close to the face. When he moves the mirror away from his face,

the image becomes inverted. What type of mirror he is using

(1) Plane mirror (2) Convex mirror (3*) Concave mirror (4) None of these

Sol. Plane mirror and convex mirror always foms erect images. Image formed by concave mirror may be erect or inverted

dependinbg on position of object.

Q.31 Time taken to cross a 4 mm window glass of refractive index 1.5 will be-

(1) 2 × 10-8

sec (2) 2 × 108

sec (3*) 2 × 10-11

sec (4) 2 × 1011

sec

Sol. t =v

x=

c

xµ

Q.32 The wavelength of monochromatic blue light in air is 420 nm. What will be its wavelength in water ?

( of water is 4/3)

(1) 280 nm (2) 560 nm (3*) 315 nm (4) 400 nmSol. = 420

w

=u

=

4203

4 = 315 nm

Q.33 The wavelength of light in vacuum is 6000 Å and in a medium it is 4000 Å. The refractive index of the medium is:(1)2.4 (2*) 1.5 (3)1.2 (4)0.67

Sol. = 4000

6000

m

V

= 1.5

Q.34 V1 is velocity of light in first medium, V2 is velocity of light in second medium, then refractive index of second

medium with respect to first medium is

(1*) V1/V2 (2) V2/V1 (3) 21 V/V (4) 12 V/V

Sol. 1 1 2 2n sin n sin

1 2

2 1

n sin

n sin

1 2 2

2 1 1

n v

n v

1 1

2 2

n v

n v

Q.35 The refractive index of a certain glass is 1.5 for light whose wavelength in vacuum is 6000 Å. The wavelenghth of this light

when it passes through glass is

(1*) 4000 Å (2) 6000 Å (3) 9000 Å (4) 15000 Å

Sol. Å40005.1

6000airmedium

Q.36 Velocity of light in water, glass and vacuum have the values vw, vg and vc respectively. Which of the following

relations is true ?

(1) vw = vg = vc (*2) vw > vg but vw < vc

(3) vw = vg but vw < vc (4) vc = vw and vw < vg

Sol. velocity in medium1

Refactive index

g w c

W g W cv v , v v

Refraction from plane surface

Geometrical Optics

E26-7

Q.37 When light travels from one medium to the other of which the refractive index is different, then which of the following will

change

(1) Frequency,wavelength and velocity (2) Frequency and wavelength

(3) Frequency and velocity (4*) Wavelength and velcity

Sol. Velocity and wavelength change but frequency remains same.

Q.38 Monochromatic light is refracted from air into the glass of refractive index . The ratio of the wavelength of incident

and refracted waves is

(1) 1 : (2) 1 : 2 (3*) : 1 (4) 1 : 1

Sol. gan

glass glass

air air

n

n

nair = 1

: 1

Q.39 A monochromatic beam of light passes from a denser medium into a rarer medium. As a result

(1*) Its velocity increases (2) Its velocity decreases

(3) Its frequency decreases (4) Its wavelength decreases

Sol. denserrarer

1

Q.40 The refractive index of water and glass with respect to air are 1.3 and 1.5 respectively. What will be the refractive

index of glass with respect to water -

(1*)13

15(2)

15

13(3) 2 (4) 3

Sol. nwg

= 7n

wa= 1.3

ngw

=1.5

1.3

Q.41 Refractive index for a material for infrared light is

(1) Equal to that of ultraviolet light (2*) Less than that for ultraviolet light

(3) Equal to that for red colour of light (4) Greater than that for ultraviolet light

Sol.

1

Q.42 If ij represents refractive index when a light ray goes from medium i to medium j, then the product

21 × 32 × 43 is equal to

(1) 31 (2) 32 (3*) (14)–1 (4) 42

Sol.j

i j

i

2 1 3 2 4 3

31 2

2 3 4

= 41.

Q.43 When a small lamp is held 1.5m above the surface of water in a tank, its image formed by reflection at the surface

appears to coincide with the image of the bottom of the tank. (µ of water = 4/3). The depth of the tank is

(1*) 2 m (2) 1.5 m (3) 1 m (4) 4 m

Geometrical Optics

E26-8

Sol.rel

d'

n

rel

ngincident median

ngrefrected media

51.5 4d

10 3

2m

Q.44 An object is placed at 24 cm distance above the surface of a lake. If water has refractive index of 4/3, then at what distancefrom lake surface, a fish will sight the object-(1*) 32 cm above the surface of water (2) 18 cm over the surface of water(3) 6 cm over the surface of water (4) 6 cm below the surface of water

Sol. x =

3

4/

1

1

24=

4

3

24=

3

424= 32 cm

Q.45 An air bubble in a glass slab (µ = 1.5) is 6 cm deep as viewed from one face and 4 cm deep as viewed from the

other face. The thickness of the glass slab is

(1) 6.67 cm (2) 10 cm (3*) 15 cm (4) Data is incomplete

Sol. 6 =2t

, t

2= 9cm

4 =1t

1.5, t

1= 6cm

t1

+ t2

= 15cm

Q.46 A bubble in glass slab [ = 1.5] when viewed from one side appears at 5 cm and 2 cm from other side then thickness of

slab is-(1) 3.75 cm (2) 23 cm (3*) 10.5 cm (4) 1.5 cm

Sol. 5 + 2 =5.1

t1+

5.1

t2

7 × 1.5 = t1 + t210.5 = t1 + t2

Q.47 A vessel of depth 2d cm is half filled with a liquid of refractive index 1 and the upper half with a liquid of

refractive index 2. The apparent depth of the vessel seen perpendicularly is

(1)

21

21d (2*)

21

11d (3)

21

112d (4)

21

12d

Sol.2

1

u d

2

2

1

1

d u 'd

1 2

1 1u ' d

Geometrical Optics

E26-9

Q.48 A beam of light is converging towards a point. A plane parallel plate of glass of thickness t , refractive index

is introduced in the path of the beam. The convergent point is shifted by (assume near normal incidence):

(1*) t 11

away (2) t 1

1

away (3) t 1

1

nearer (4) t 1

1

nearer

Q.49 A rectangular tank of depth 8 meter is full of water (= 4/3), the bottom is seen at the depth

(1*) 6m (2) 8/3 m (3) 8 cm (4) 10 cm

Sol. m63/4

8'h

'h

h

Q.50 The wavelength of light in two liquids 'x' and 'y' is 3500 Å and 7000 Å. Then the critical angle of x relative to y will be(1) 60º (2) 45º (3*) 30º (4) 15º

Sol. x y3500Å, 7000Å

sin (x) nx

= sin (y) ny

y xE

x y

nsin xsin

sin y n

x

y

1n

2

6

Q.51 Total internal reflection occurs in waves, when wave enters-(1) Glass from air (2)Air from vaccum (3) Water from air (4*)Air from water

Sol. For TIR medium at refraction must be rarer.

Q.52 A light wave travels from glass to water. The refractive index for glass and water are2

3and

3

4respectively. The value

of the critical angle will be:

(1) sin–1

2

1(2) sin–1

8

9(3*) sin–1

9

8(4) sin–1

7

5

Sol.2

3sin C =

3

4sin 90

C = sin–1

9

8

Q.53 Just before the time of sun set or shine the sun appears to be oval because

(1) The sun changes its shape at that time (2) Of the scattering of light

(3*) Of the effects of refraction (4) Of the effects of diffraction

Sol. Because of refrecting ray from sun

Q.54 To an observer on the earth the stars appears to twinkle. This can be ascribed to

(1) The fact that stars do not emit light continuosly

(2) Frequent absorption of star light by their own atmosphere

(3) Frequent absorption of star light by the earth's atmosphere

(4*) The refractive index fluctuations in the earth's atmosphere

Total Internal Reflection

Geometrical Optics

E26-10

Q.55 Mirages are observed in deserts due to phenomenon of

(1) Interference of light (2*) Total internal reflection

(3) Scattering of light (4) Double reflection of light

Sol. Mirages are observed due to total internal reflection

Q.56 The wavelength of light diminishes times ( = 1.33 for water) in a medium. A driver from inside water looks at an

object whose natural colour is green. He sees the object as

(1*) Green (2) Blue (3)Yellow (4) Red

Sol. Colour of light is determined by its frequency and an frequency does not change, colour will also not change and will

remains green.

Q.57 We see the sun a little before it rises on the horizon and a little after it sets below the horizonal. This is a

consequence of the phenomenon of

(1*) Total internal reflection (2) Refraction

(3) Dispersion (4) Scattering of light

Sol. Reflection of light

Q.58 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

(1) He must direct the beam vertically upwards

(2) He has to direct the beam horizontally

(3*) 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

(4) He has to direct the beam at an angle to the vertical which is slightly more than the critical angle of incidence for the

total internal reflection

Q.59 When a light ray travels into a single medium then refractive index is

(1) directly proportional to wavelength of light

(2) inversely proportional to wavelength of light

(3*) inversely proportional to the square of wavelength of light

(4) directly proportional to the square of wavelength of light

Sol.1 1

nv

Q.60 A glass prism of refracting angle 60º gives a minimum deviation of 30º. What is the refractive index of the

glass -

(1) 1.5 (2) 5.1 (3*) 2 (4) can not be determined

Sol.

minsin A / 2

sin A / 2

60 30sin

1 222

sin 30 2

Q.61 The ratio of the refractive index of red light to blue light in air is

(1*) Less than unity

(2) Equal to unity

(3) Greater than unity

(4) Less as well as greater than unity depending upon the experimental arrangement

Sol. blue

> red

Prism

Geometrical Optics

E26-11

Q.62 The refractive index of a piece of transparent quartz is the greatest for

(1) Red light (2*) Violet light (3) Green light (4) Yellow light

Sol.

r,1

Q.63 A ray of light is incident normally on one of the faces of a prism apex angle 30º and refractive index 2 . The angle of

deviation of the ray is:(1*) 15º (2)22.5º (3) 0º (4)12.5º

Sol. 2 sin 30° = sin e

e = 45°

Deviation = 45° – 30° = 15°

Q.64 The refractive index of the material of prism of 60º angle is 2 . At what angle the ray of light be incident on it so that

minimum deviation takes place?(1*) 45º (2)60º (3)30º (4)75º

Sol. i = e

r1= r

2=

2

A= 30°

sin i = 2 sin 30° i = 45°

Q.65 A ray of light is incident at angle of 60o on one face of a prism which has an apex angle of 30o. The ray emergingout of the prism makes an angle of 30o with the incident ray. The refractive index of the material of the prism is -

(1) 2 (2*) 3 (3)1.5 (4)1.6

Sol. r1+ r

2= 30°

2

3= sin r

1

sin r2=

2

1 = 3

Q.66 If the critical angle for the medium of prism is C and the angle of prism is A, then there will be no emergent ray when-

(1) A < 2C (2) A = 2C (3*) A > 2C (4) A 2C

Q.67 A ray of monochromatic light is incident on one refracting face of a prism of angle 750. It passes through the prism and

is incident on the other face at the critical angle. If the refractive index of the material of the prism is 2, the angle of

incidence on the first face of the prism is(1)30º (2*) 45º (3)60º (4) 0º

Sol. r2= sin–1

1= 45º

r1=A– r

2= 75º – 45º = 30º

2rsin

isin

1

sin i = 2 sin 30 =2

12 i = 45º.

Q.68 A ray of light is incident at angle i on a surface of a prism of small angle A and emerges normally from the

opposite surface. If the refractive index of the material of the prism is , the angle of incidence i is nearly equal

to :(1) A/ (2) A/(2) (3*) A (4) A/2

Geometrical Optics

E26-12

Sol.

Asin

isin

since i and A are small angle. A

i

Q.69 The critical angle between an equilateral prism and air is 45°. If the incident ray is perpendicular to the refractingsurface, then(1) After deviation it will emerge from the second refaracting sufrace(2*) It is totally reflected on the second surface and emerges out perpendicularly from third surface in air(3) It is totally reflected from the second and third refracting surfaces and finally emerges out from the first surface(4) It is totally reflected from all the three sides of prism and never emerges out

Sol. (b)

Q.70 When light rays are incident on a prism at an angle of 45°, the minimum deviation is obtained. If refractive index of the

material of prism is 2 , then the angle of prism will be

(1)30° (2)40° (3)50° (4*) 60°

Sol. (d)

2

Asin

2

mAsin

= , But2

A m = i = 45°

So

2

Asin

2

12

)2/A(sin

45sinA= 60°

Q.71 The refractive indedx of a prism for a monochromatic wave is 2 and its refracting angle is 60°. For minimum

deviation, the angle of incidence will be(1)30° (2*) 45° (3)60° (4)75°

Sol. (b)

2

60sin

isin2

2/Asin

isin

2 x sin 30 = sin i i = 45°

Q.72 An object is placed at a distance of 20 cm, in rarer medium, from the pole of a convex spherical refracting surface ofradius of curvature 10 cm. If the refrective index of the rarer medium is 1 and of the refracting medium is 2, then theposition of the image is at-(1) (40/3) cm from the pole & inside the denser medium(2*) 40 cm from the pole & inside the denser medium.(3) (40/3) cm from the pole & outside the denser medium(4) 40 cm from the pole & outside the denser medium.

Refraction at curved surface

Geometrical Optics

E26-13

Sol.R

n–n iR=

v

nR–

u

ni

10

1–2=

v

2–

20–

1 v = 40 cm

Q.73 There is a small black dot at the centre C of a solid glass sphere of refractive index . When seen from outside,the dot will appear to be located:(1) away from C for all values of (2*) at C for all values of

(3) at C for = 1.5, but away from C for 1.5 (4) at C only for 2 1.5.

Sol.R

–

u–

V1212

R–

–

R––

V1212

V = – R for all values of .

Q.74 The image for the converging beam after refraction through the curved surface is formed at:

(1*) x = 40 cm (2) x =40

3cm (3) x =

40

3cm (4) x =

180

7cm

Sol.20

2

3–1

302

3–

V

1

40

1

20

1

40

1–

V

1 V = 40 cm.

Q.75 An object is placed at a distance of 5 cm from a convex lens of focal length 10 cm, then the image is

(1) real, diminished and at a distance of 10 cm from the lens

(2) real, enlarged and at a distance of 10 cm from the lens

(3*) virtual, enlarged and at a distance of 10 cm from the lens

(4) virtual, diminished and at a distance of 10/3 cm from the lens.

Sol.1 1 1

v u f

1 1 1

v 5 10

v = 10 cm

|m| = 2 (magnified)

Sol.f

1= )1–µ( rel

21 R

1–

R

1  (R1 = R, R2 = – R )

Lens

Q.76 A biconvex lens with equal radii of curvature has refractive index 1.6 and focal length 10 cm. Its radius of curvature willbe:(1) 20 cm (2) 16 cm (3) 10 cm (4*) 12 cm

Geometrical Optics

E26-14

Q.77 An object is placed at a distance of 5 cm from a convex lens of focal length 10 cm, then the image is-

(1) Real, diminished and at a distance of 10 cm from the lens.

(2) Real, enlarged and at a distance of 10 cm from the lens.

(3*) Virtual, enlarged and at a distance of 10 cm from the lens.

(4) Virtual, diminished and at a distance of 10/3 cm from the lens.

Sol. u < f

so image is virtual, enlarged and at a distance of 10 cm from the lens.

u

1–

v

1

f

1

Q.78 Inside water, an air bubble behave-(1) Always like a convex lens (2*) Always like a concave lens(3) Always like a slab of equal thickness (4) Sometimes concave and sometimes like a convex lens

Sol. nrel

< 1So, f is negative

Q.79 A convex lens of focal length f will form a magnified real image of an object if the object is placed

(1) anywhere beyond 2f (2) anywhere beyound f

(3*) between f and 2f (4) between lens and f

Sol. Between (f1

2f)

The image will be real and magnified

Q.80 A lens behaves as a converging lens in air and a diverging lens in water. The refractive index of the material is (refractiveindex of water = 1.33)(1) equal to unity (2) equal to 1.33(3*) between unity and 1.33 (4) greater than 1.33

Sol. Lens changes its behaviour if R.I. of surrounding becomes greater than R.I. of lens.

lens< 1.33

Q.81 In the figure given below, there are two convex lens L1 and L2 having focal length of f1 and f2 respectively. The distancebetween L1 and L2 will be

(1) f1 (2) f2 (3*) f1 + f2 (4) f1 - f2

Sol.

Distance between lens is = f1

+ f2

,

Q.82 The ray diagram could be correct

(1) If n1= n

2= n

g(2) If n

1= n

2and n

1< n

g(3*) If n

1= n

2and n

1> n

g(4) Under no circumstances

Geometrical Optics

E26-15

Q.83 A glass lens is placed in a medium in which it is found to behave like a glas plate. Refractive index of the medium will be :

(1) Greater than the refractive index of glass (2) Smaller than the refractive index of glass

(3*) Equal to refractive index of glass (4) No case will be possible from above

Q.84 A divergent lens will produce

(1*) Always a virtual image (2)Always real image

(3) Sometimes real and sometimes virtual (4) None of above

Q.85 The radius of curvature of convex surface of a thin plano-convex lens is 15 cm and refractive index of its material is

1.6. The power of the lens will be

(1)+1D (2) –2D (3)+3D (4*) +4D

Q.86 A convex lens of glass ( = 1.5) is immersed in water. Compared to its power in air, its power in water-

(1) increases (2*) decreases (3) remain same (4) nothing can be predicted

Sol. g 1

air 1 2

1 1

f R R

g

water w 1 2

1 1 11

f R R

water

air

31

f 2

3 / 2f1

4 / 3

fwater

= 4 fair

P =1

,f

Power decrase

Q.87 A convex lens of focal length 25 cm and a concave lens of focal length 20 cm are mounted coaxially separated by adistance d cm. If the power of the combination is zero, d is equal to(1)45 (2)30 (3)15 (4*) 5

Sol. 0ff

d

f

1

f

1

f

1

2121

= 0500

d

20

1

25

1

=500

2520 =

500

d

d = 5 cm.

Q.88 The radius of curvature of convex surface of thin plane convex lens is 15cm and refractive index of the material

is 1.6 . The power of the lens will be

(1) + 1 D (2) – 2D (3) + 3D (4*) + 4 D

Sol. 1 1

1.6 1 0f 15

1 0.6 1

f 150 25

f = 25 cm

1 100P 2D

f 25

Geometrical Optics

E26-16

Q.89 A convex lens of power 4D and a concave lens of power 3D are placed in contact, the equivalent power of combination:

(1*) 1D (2)4

3D (3)7D (4)

3

4D

Sol. P = P1 + P2

= + 4 + (– 3)

= + 1

Q.90 Two thin lenses of power +5D and –2D are placed in contact with each other. Focal length of the combination will behave

like a-

(1) Convex lens of focal length 3m (2) Concave lens of focal length 0.33m

(3*) Convex lens of focal length 0.33m (4) None of the above

Sol. PL

= P1+ P

2

PL

=Lf

1

Q.91 What is the power of a diverging lens of focal length 40 cm ?

(1) 2.5 dioptre (2) 4.0 dioptre (3) – 3.5 dioptre (4*) – 2.5 dioptre

Sol. f’ of diverging = –ve

P =1

f

P =1

40

P = –2.5 D

Q.92 The focal length of a plano-concave lens is –10 cm, then its focal length when its plane surface is polished is

(n = 3/2):

(1) 20 cm (2) 5 cm (3*) 5 cm (4) none of these

Sol.F

1=

mf

1–

Lf

2=

10

20

F = 5

Q.93 A lens of power + 2 diopters is placed in contact with a lens of power – 1 diopter. The combination will behave

like

(1) a convergent lens of focal length 50 cm (2) a divergent lens of focal length 100 cm

(3*) a convergent lens of focal length 100 cm (4) a convergent lens of focal length 200 cm.

Sol. P = +1D Resulted

100 100100cm

100 1

Q.94 When the two thin lenses are put in contact, the focal length of the combination is

(1) the geometric mean of the two focal lengths (2) the same as the larger focal length

(3) greater than either focal length (4*) smaller than either focal length

Sol.1 2 e

1 1 1

f f f

Let f1 = f2 = f

1 2 e

1 1 1

f f f

e

ff

2

(f1 = f2) 2fe

Geometrical Optics

E26-17

Q.95 Dispersive power of a prism depends on-

(1*) Material (2) Prism angle (3) Shape of prism (4) Angle on incidence

Sol. =

1–µ

µ–µ

y

rv

Q.96 When light is passed through a prism, the colour which deviates least is:

(1*) Red (2) violet (3) Blue (4) Green

Sol. µred = minimum

Q.97 If refractive index of red, violet and yellow lights are 1.42, 1.62 and 1.50 respectively for a medium, its dispersive power

will be -

(1*) 0.4 (2)0.3 (3)0.2 (4)0.1

Sol. =

1–5.1

42.1–62.1

=5.0

2.0=

10

4= 0.4

Q.98 Two thin lenses, one convex of focal length 30 cm and the other concave of focal length 10cm are put into contact. If this

combination is equivalent to an achromatic lens then the ratio of dispersive powers (1/

2) of above two lenses is -

(1) 1/3 (2) – 3 (3*) 3 (4) – 1/3

Sol.2

2

1

1

ff

= 0 for achromatic lens

2

1

=

)10(

30

= 3

Q.99 A medium has nv = 1.56, nr = 1.44. Then its dispersive power is:

(1) 3/50 (2*) 6/25 (3) 0.03 (4) none of these

Sol. =1

2

nn

nn

rv

rv

=

25

6.

Q.100 All the listed things below are made of flint glass. Which one of these have greatest dispersive power ().

(1) prism (2) glass slab (3) biconvex lens (4*) all have same Sol. depends only on material property.

Q.101 Light of wavelength 4000 Å is incident at small angle on a prism of apex angle 4º. The prism has nv = 1.5 & nr = 1.48.

The angle of dispersion produced by the prism in this light is:

(1) 0.2º (2) 0.08º (3) 0.192º (4*) None of these

Sol. Dispersion will not occur for a light of = 4000 Å.

Q.102 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

(1) Violet ray (2) Green ray (3*) Red ray (4) Yellow ray

)1( is least so R is least.

Q.103 A spectrum is formed by a prism of dispersive power ‘ ’. If the angle of deviation is ‘ ’, then the angular dispersion is

(1) / (2) / (3) /1 (4*)

Dispersion

Sol. (3) R

Geometrical Optics

E26-18

Sol. (d) We know that

mean

r

Angular dispersion = v–

r= =

mean

Q.104 When white light passes through the achromatic combination of prisms, then what is observed(1*) Only deviation (2) Only dispersion(3) Deviation and dispersion (4) None of the above

Q.105 A shortsighted person can read a book clearly at a distance of 10 cm from the eyes. The lenses required to read the bookkept at 60cm are :(1) Convex lenses of focal length 30 cm (2) convex lenses of focal length 30 cm(3) convex lenses of focal length 12 cm (4*) concave lenses of focal length 12 cm

Sol.u

1–

v

1

f

1

10–

1–

60–

1

f

1 =

60

61– =

60

5=

12

1

f = 12

Q.106 A person can't see the objects clearly placed at a distance more than 40 cm. He is advised to use a lens of power -(1) + 2.5 D (2*) – 2.5 D (3) + 0.4 D (4) – 0.4 D

Sol.v

1–

u

1=

f

1

)40(–

1– 0 =

f

1

f = – 40 cm

PL

=Lf

1= – 2.5 D

Q.107 A person can see clearly only upto a distance of 25cm. He wants to read a book placed at a distance of 50cm. What kindof lens does he required for his spectacles and what must be its power ?(1) concave, – 1.0 D (2) Convex, + 1. 5 D (3*) Concave, – 2.0 D (4) Convex, + 2.0 D

Sol.v

1–

u

1=

f

1

25–

1–

50–

1=

f

1

f = –50 cm

PL

=Lf

1= – 2D

Q.108 Astigmatism (for a human eye) can be removed by using(1) Concave lens (2) Convex lens (3*) Cylindrical lens (4) Prismatic lens

Q.109 Circular part in the centre of retina is called(1) Blind spot (2*) Yellow spot (3) Red spot (4) None of the above

Q.110 A person cannot see distinctly at the distance less than one metre. Calculate the power of the lens that he should use toread a book at a distance of 25 cm(1*) +3.0 D (2)+0.125D (3) – 3.0 D (4) + 4.0 D

Q.111 Indicate the correct statement in the following(1) The dispersive power depends upon the angle of prism(2*) The angular dispersion depends upon the angle of the prism(3) The angular dispersion does not depend upon the dispersive power(4) The dispersive power in vacuum is one

Sol. Angular dispersim does not depends upon dispersive power

Sol. (1) Because achromatic combination has same for all wavelengths.

Geometrical Optics

E26-19

Q.112 The refractive indices of red, violet and yellow light are respectively 1.42, 1.62 and 1.50. The dispersivepower of the medium will be(1*) 0.4 (2) 0.3 (3) 0.2 (4) 0.1

Sol.v r

y 1

=1.62 1.42

0.41.50 1

Q.113 The focal length of the objective of a microscope is-

(1) Greater than the focal length of eye piece

(2*) Lesser than the focal length of the eye piece

(3) Equal to the focal length of the eye piece

(4) Any of (1) (2) and (3)

Sol. By constitution of simple microscope we can observe it

Q.114 A simple microscope has a focal length of 5 cm. The magnification at the least distance of distinct vision is-(1)1 (2)5 (3)4 (4*) 6

Sol. MP =

f

D1 =

5

251 = 6

Q.115 In a compound microscope, the intermediate image is -(1) virtual, erect and magnified (2) real, erect and magnified(3*) real, inverted and magnified (4) virtual, erect and reduced

Q.116 The resolving power of a telescope is more when its objective lens has(1) greater focal length (2) smaller focal length(3*) greater diameter (4) smaller diameter

Q.117 A Galileo telescope has an objective of focal length 100 cm & magnifying power 50. The distance betweenthe two lenses in normal adjustment will be(1) 150 cm (2) 100 cm (3*) 98 cm (4) 200 cm

Sol. In normal adjustment

m = –e

0

f

f

so 50 = –ef

100 f

e= –2 cm

( eyepiece is concave lens)and L = f

0+ f

e= 100 – 2 = 98 cm

Q.118 In a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens,

(focal length = F) then the magnifying power is

(1) 25/F (2*) 1 + 25/F (3) F/25 (4) F/25 + 1

Sol. = magnifying power

=D

1F

=25

1f

Optical Instrument

Geometrical Optics

E26-20

Q.119 The convex lens is used in-(1) Microscope (2) Telescope (3) Projector (4*) All of the above

Q.120 The magnifying power of a simple microscope can be increased if an eyepiece of :(1*) shorter focal length is used (2) longer focal length is used(3) shorter diameter is used (4) longer diameter is used

Sol. m = 1 +f

D

Q.121 Resolving power of a microscope depends upon(1) the focal length and aperture of the eye lens (2) the focal lengths of the objective and the eye lens(3) the apertures of the objective and the eye lens (4*) the wavelength of light illuminating the object

Q.122 An astronomical telescope has an eyepiece of focal-length 5 cm. If the angular magnification in normal adjustment is10, when final image is at least distance of distinct vision (25cm) from eye piece, then angular magnification will be :(1)10 (2*) 12 (3)50 (4)60

Sol. For normal adjustment

m = –e

0

f

f

When final image is at least distance of distinct vision from eyepiece,

m' = –

d

f1

f

f e

e

0=

25

5110 = 12

Q.123 An astronomical telescope has a magnifying power 10. The focal length of the eye piece is 20 cm. the focal

length of the objective is-

(1) 2 cm (2*) 200 cm (3) (1/2) cm (4) (1/200) cm

Sol.0

e

fm

f

m = 10 × 20 = 200 cm

Q.124 A person with a defective sight is using a lens having a power of +2D. The lens he is using is(1) concave lens with f = 0.5 m (2) convex lens with f = 2.0 m(3) concave lens with f = 0.2 m (4*) convex lens with f = 0.5 m

Sol. f =2

1

p

1 metre

f = 0.5 m this is positive so lense is convex lense.

Q.125 The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm respectively. If the magnifying

power for the relaxed eye is 45, then the length of the tube is :

(1) 30 cm (2) 25 cm (3*) 15 cm (4) 12 cm

Sol. By using m

=e0

e0

ff

D).ffL(

45 =51

25)51L(

L

= 15 cm

Q.126 In a compound microscope magnification will be large, if the focal length of the eye piece is :

(1) Large (2*) Smaller

(3) Equal to that of objective (4) Less than that of objective

Sol. For a compound microscope meoff

1

Geometrical Optics

E26-21

Q.127 Microscope is an optical instrument which :

(1) Enlarges the object

(2*) Increases the visual angle formed by the object at the eye

(3) Decreases the visual angle formed by the object at the eye

(4) Brings the object nearer

Sol. In microscope final image formed is enlarged which in turn increases the visual angle.

Q.128 If in compound microscope m1

and m2

be the linear magnification of the objective lens and eye lens respectively, then

magnifying power of the compound microscope will be

(1) m1– m

2(2) 21 mm (3) (m

1+ m

2)/2 (4*) m

1× m

2

Sol. Magnification of a compound microscope is given by

m =e0

0

u

D

u

v m= m

0× m

e

Q.129 For which of the following colour, the magnifying power of a microscope will be maximum :

(1) White colour (2) Red colour (3*) Violet colour (4)Yellow colour

Sol. Magnifying power of a microscope m f

1

Since fviolet

< fred

; mviolet

> mred

Q.130 The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of

eye lens is 5 cm, then the object distance for objective lens will be :

(1*) 1.8 cm (2) 1.5 cm (3) 2.1 cm (4) 2.4 cm

Sol. L

= v0+ f

e 14 = v

0+ 5 v

0= 9 cm

Magnifying power of microscope for relaxed eye

m =e0

0

f

D.

u

vor 25 = 5

25.

u

9

0or u

0=

5

9= 1.8 cm

Q.131 If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from

the objective lens and the final image is formed at infinity. The magnifying power of the microscope is :

(1)150 (2*) 200 (3)250 (4)400

Sol. m

=e0

0

f

D

u

v

From000 u

1

v

1

f

1

)25.1(

1

v

1

)2.1(

1

0

v0= 30 cm

m

3

25

25.1

30 = 200

Q.132 the focal length of objective and eye lens of a stronomical telescope are respectively 2m and 5 cm. Final image is

formed at (i) least distance of distinct vision (ii) infinity. The magnifying power in both cases will be:

(1*) –48, –40 (2) –40, –48 (3) –40, 48 (4) –48, 40

Sol. When the final image is at the least distance of distinct vision, then

Geometrical Optics

E26-22

m =55

6200

25

51

5

200

D

f1

f

f e

e

0

= –48

When the final image is at infinity, then

m =e

0

f

f =

5

200= –40

Q.133 For observing a cricket match, a binocular is preferred to a terrestrial telescope because :

(1*) The binocular gives the proper three dimensional view

(2) The binocular has shorter length

(3) The telescope does not give erect image

(4) Telescope have chromatic aberrations

Sol. In terrestrial telescope erecting lens absorbs a part of light, so less constant image. But binocular lens gives the proper

three dimensional image.

Q.134 An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The seperation between the

objective and the eye piece is 36 cm and the final image is formed at infinity. The focal length f0

of the objective and the

focal length feof the eye piece are :

(1) f0= 45 cm and f

e= –9 cm (2) f

0= 7.2 cm and f

e= 5 cm

(3) f0= 50 cm and f

e= 10 cm (4*) f

0= 30 cm and f

e= 6 cm

Sol. In this case m =e

0

f

f= 5 ...(i)

and length of telescope = f0

+ fe= 36 ...(ii)

Solving (i) and (ii), we get fe= 6 cm, f

o= 30 cm

Q.135 In Gallilean telescope, if the powers of an objective and eye lens are respectively +1.25 D and –20D, then for relaxed vision,

the length and magnification will be :

(1) 21.25 cm and 16 (2) 75 cm and 20 (3*) 75 cm and 16 (4) 8.5 cm and 21.25

Sol. fo=

25.1

1= 0.8 m and f

e=

20

1

= –0.05 m

L = f – fe= 0.8 – 0.05 = 0.75 m = 75 cm

and m =

e

0

f

f=

05.0

8.0= 16

Geometrical Optics

E26-23

Q.1 A point source of light is placed in front of a plane mirror.(1) Only the reflected rays close to the normal meet at a point when produced backward.(2) All the reflected rays meet at a point when produced backward.(3) Only the reflected rays making a small angle with the mirror, meet at a point when produced backward.(4) Light of different colours make different images.

Ans. (2)Sol. All the reflected rays meet at a point, when produced backwards.

O I

Q.2 When light is reflected from a mirror a change occurs in its :(1) phase, (2) frequency, (3) wavelength, (4) speed

Ans. (1)Sol. There is a phase change of 180º in reflection.

Q.3 A rays is incident at an angle 38º on a mirror. The angle between normal and reflected ray is(1) 38º (2) 52º (3) 90º (4) 76º

Ans. (1)Sol. By the laws of reflection angle of incidence = angle of reflection i = r

Q.4 Mark the correct options.(1) If the incident rays are converging, we have a real object.(2) If the final rays are converging, we have a real image.(3) The image of a virtual object is called a virtual image.(4) If the image is virtual, the corresponding object is called a virtual object.

Ans. (2)Sol. An image is called a real image if the rays after reflection or refraction actually meet hence converging rays from

real image.When rays actually meet real image is formed

Q.5 A straight line joining the object point and image point is always perpendicular to the mirror(1) if mirror is plane only (2) if mirror is concave only(3) if mirror is convex only (4) none of these

Ans. (4)Sol. Irrespective of the type of mirror.

Ans. (4)Sol. Minimum distance between object and image is zero when image concides with the object i.e., object is placed at

2F.

Q.7 Two plane mirrors are inclined to each other at an angle 600. If a ray of light incident on the first mirror is parallelto the second mirror, it is reflected from the second mirror(1) Perpendicular to the first mirror (2) Parallel to the first mirror(3) Parallel to the second mirror (4) Perpendicular to the second mirror

Ans. (2)

Q.6 In case of concave mirror, the minimum distance between a real object and its real image is :(1) f (2) 2f (3) 4f (4) zero

EXERCISE-II

Geometrical Optics

E26-24

Sol.

final ray is II to first mirror.

Ans. (3)

Hint :

(i) (ii)From figure (i) and (ii) it is clear that if the mirror moves distance ‘A’ then the image moves a distance ‘2A’.

Sol. Therefore Amplitude of SHM of image = 2A

Q.9 An unnumbered wall clock shows time 04: 25: 37, where 1st term represents hours, 2nd represents minutes andthe last term represents seconds. What time will its image in a plane mirror show.(1) 08: 35: 23 (2) 07: 35: 23 (3) 07: 34: 23 (4) none of these

Ans. (3)Sol. If time in the clock is T

1& time in image clock is T

2then.

T1

+ T2

= 12 : 00 : 004 : 25 : 37 + T

2= 12 : 00 : 00

T2

= 07 : 34 : 37

Ans. (3)Sol. Taking first reflection by A.

Taking first reflection by B

Q.8 A point object is kept in front of a plane mirror. The plane mirror is performing SHM of amplitude 2 cm. Theplane mirror moves along the x-axis and x- axis is normal to the mirror. The amplitude of the mirror is such thatthe object is always infront of the mirror. The amplitude of SHM of the image is(1) zero (2) 2 cm (3) 4 cm (4) 1 cm

Q.10 Two plane mirrors are parallel to each other and spaced 20 cm apart. An object is kept in between them at 15 cmfrom A. Out of the following at which point(s) image(s) is/are not formed in mirror A (distance measured from mirrorA):(1) 15 cm (2) 25 cm (3) 45 cm (4) 55 cm

Geometrical Optics

E26-25

Ans. (4)

Sol.

ED = AD – AE = 1.4 – 0.8 = 0.6 m

Ans. (2)Sol. Perpendicular distance between object & mirror is equal to perpendicular distance between image & mirror.

Initially the separation between object and image is 200 cm. After 6s the mirror has moved 30 cm towards the object.Hence object-mirror separation is 70 cm. So object image separation is 140 cm.

Ans. (3)Sol. All the images formed by two plane.

Mirror inclined to each other form images which lie on a circle.

Ans. (3)Sol. Only a portion of incident light is reflected by mirror and rest is transmitted in mid water. So intensity of reflected

light is less than intensity of incident light & hence image formed is less bright.

Q.15 A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it.When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instantis(1) 6, towards the mirror (2) 6, away from the mirror(3) 9, away from the mirror (4) 9, towards the mirror.

Ans. (3)

Sol.

Q.11 A person’s eye is at a height of 1.5 m. He stands infront of a 0.3m long plane mirror which is 0.8 m above the ground.The length of the image he sees of himself is :(1) 1.5m (2) 1.0m (3) 0.8m (4) 0.6m

Q.12 An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speedof 5 cm/s. Then after 6 s the distance between the object and its image will be :(1) 60 cm (2) 140 cm (3) 170 cm (4) 150 cm

Q.13 Images of an object placed between two plane mirrors whose reflecting surfaces make an angle of 90° with oneanother lie on a :(1) straight line (2) zig-zag curve (3) circle (4) ellipse

Q.14 The images of clouds and trees in water always less bright than in reality –(1) because water is forming the image dirty(2) because there is an optical illusion due to which the image appears to be less bright(3) because only a portion of the incident light is reflected and quite a large portion goes mid water(4) because air above the surface of water contains a lot of moisture

Geometrical Optics

E26-26

u

1

v

1

f

1

20–

1

v

1

12–

1

12

1–

20

1

v

1

v = – 30 cmvelocity of image

dt

du

u

V–

dt

dV2

2

= –

2

20–

30–

4. = – 9 cm/sec.

9 cm/sec away from mirror.

Q.16 A particle is moving towards a fixed spherical mirror. The image:(1) must move away from the mirror (2) must move towards the mirror(3) may move towards the mirror (4) will move towards the mirror, only if the mirror is convex.

Ans. (3)

Sol.

only in above two cases image moves towards mirror.

Q.17 A point object on the principal axis at a distance 15 cm in front of a concave mirror of radius of curvature 20 cmhas velocity 2 mm/s perpendicular to the principal axis. The magnitude of velocity of image at that instant will be:(1) 2 mm/s (2) 4 mm/s (3) 8 mm/s (4) 16 mm/s

Ans. (2)

Sol.

u

1

v

1

f

1

15–

1

v

1

10–

1

30

3–2

v

1 v = – 30 cm.

m =15–

30––

u

v–

0

= –2

dt

d2

dt

d θ

= 4 mm/sec.

Q.18 A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principalaxis with amplitude 2 mm. The amplitude of its image will be(1) 2 mm (2) 4 mm (3) 8 mm (4) 16 mm

Ans. (3)Sol. Using mirror formula,

15–

1

v

1

10–

1 v = – 30 cm.

| Axial magnification | =2

2

2

15

30

u

V

= 4

amplitude of image = 4 × 2 = 8 mm.

Geometrical Optics

E26-27

Q.19 In image formation from spherical mirrors, only paraxial rays are considered because they :(1) are easy to handle geometrically (2) contain most of the intensity of the incident light(3) form nearly a point image of a point source (4) show minimum dispersion effect.

Ans. (3)Sol. Paraxial rays are considered because they form nearly a point image of a point source.

Q.20 The distance of an object from the focus of a convex mirror of radius of curvature ' a ' is' b '. Then the distance of the image from the focus is:(1) b2 / 4a (2) a / b2 (3) a2 / 4b (4) 4b / a2

Ans. (3)Sol. Using Newtons formula. x distance of object from focus

xy = f2 y distance of object from focusf focal length.

by = (a/2)2 , y =b4

a2

.

Q.21 The largest distance of the image of a real object from a convex mirror of focal length 20 cm can be :(1) 20 cm (2) infinite(3) 10 cm (4) depends on the position of the object

Ans. (1)Sol. It is created at focus ie + 20 cm, when object is at infinity

Q.22 Which of the following can form erect, virtual, diminished image?(1) plane mirror (2) concave mirror (3) convex mirror (4) none of these

Ans. (3)

Sol.

Q.23 is the image of a point object O formed by spherical mirror, then which of the following statements is incorrect:(1) If O and are on same side of the principal axis, then they have to be on opposite sides of the mirror.(2) If O and are on opposite side of the principal axis, then they have to be on same side of the mirror.(3) If O and are on opposite side of the principal axis, then they can be on opposite side of the mirror as well.(4) If O is on principal axis then has to lie on principal axis only.

Ans. (3)

Sol.O

I= –

u

v

If O and I are on same sides of PA .O

Iwill be positive which implies v and u will be of opposite signs.

Similarly if O and I are on opp. sides,O

Iwill be -ve which implies v and u will have same sign.

If O is on PA, I =

u

V(O) = 0 I will also be on. P.A.

Q.24 An object is placed at a distance u from a concave mirror and its real image is received on a screenplaced at adistance of v from the mirror. If f is the focal length of the mirror, then the graph between1/v versus 1/u is

(1) (2) (3) (4)

Geometrical Optics

E26-28

Ans. (2)

Sol.u–

1

v–

1

f–

1

f

1

u

1–

v

1

Slope = – 1 intercept =f

1(positive)

(1) 2/9 (2) /9 (3) 20 (4) /18Ans. (4)

Sol.

10cm

So diameter of the image = f

= 10 ×18180

1

Q.26 A convex mirror has a focal length f. A real object, placed at a distance f in front of it from the pole, producesan image at(1) 2 f (2) f /2 (3) f (4)

Ans. (2)Sol. Using mirror formula

1 1

v u =

1

f

f

f

OHere u = –f, f = +f

f

1

)f(

1

v

1

v =2

f

Q.27 A convex mirror has a focal length = 20 cm. A convergent beam tending to converge to a point 20 cm behind convexmirror on principal axis falls on it. The image if formed at(1) infinity (2) 40 cm (3) 20 cm (4) 10 cm

Ans. (1)Sol. Using mirror formula

1 1

v u =

1

f

Here we have a virtual object so sign of u is positive.

O

f=20cm

20cm

Heref = +20

Q.25 A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtends an angle1° on the earth. Then the diameter of the image is (in cm) :

Geometrical Optics

E26-29

u = 20

20

1

20

1

v

1

v

1= 0

v =

Q.28 A candle is kept at a distance equal to double the focal length from the pole of a convex mirror. its magnificationwill be :(1) –1/3 (2) 1/3 (3) 2/3 (4) –2/3

Ans. (2)

Sol.

2Ff=+f

Taking u = – 2f & f = +f

f

1

u

1

v

1

f

1

f2

1

v

1

v

1=

f

1+

f2

1=

f2

12

m = –u

v=

f2

3/f2

=

3

1

Q.29 A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For theimage to be real, the focal length should be :(1) 10 cm (2) 15 cm (3) 20 cm (4) 30 cm

Ans. (2)Sol. Magnification is –3 because image is real & inverted.

m =v

u

–3 =v

u

v = 3u.given u = –20 cmv = –60 cm

20cm

By using mirror formula

60

1–

20

1=

f

1

f = –15 cm

Q.30 If an object is 30 cm away from a concave mirror of focal length 15 cm, the image will be(1) erect (2) virtual (3) diminished (4) of same size

Ans. (4)

Geometrical Optics

E26-30

Sol. Here u = –30 cm, f = –15 cmobject is at centre of curvatureimage will be real and of same size.

Q.31 The largest distance of the image of a real object from a convex mirror of focal length 20 cm canbe :(1) 20 cm (2) infinite(3) 10 cm (4) depends on the position of the object.

Ans. (1)Sol. When object is real then image move from focus to pole.

So maximum distance f = 20 cm.

Q.32 A boy 2 m tall stands 40 cm in front of a mirror. He sees an erect image, 1 m high. The mirror is :(1) Concave, f = 40 cm (2) Convex, f = 40 cm(3) Plane (4) Either convex or concave

Ans. (2)

Sol. Using mirror formulai

0

h

h =v

u

Given0

i

h

h=

2

1=

u

v

hence v =2

u

givenu = –40 v = 20Using mirror formula

f

1

20

1

40

1

f

1=

40

1

f = 40convex mirror with focal length = 40 cm

Q.33 What is the distance of a needle from a concave mirror of focal length 10 cm for which a virtual image of twice itsheight is formed ?(1) 2.5 cm (2) 5 cm (3) 8 cm (4) 9.1 cm

Ans. (2)Sol. Given

m = +2 = –u

v

v = –2uUsing mirror formula

u2

1+

u

1=

10

1

10

1

u2

1 u = 5 cm Ans.

Q.34 Which one of the following statements are incorrect for spherical mirrors.(1) a concave mirror forms only virtual images for any position of real object(2) a convex mirror forms only virtual images for any position of a real object.(3) a convex mirror forms only a virtual diminished image of an object placed between its pole and the focus(4) a concave mirror forms a virtual magnified image of an object placed between its pole and the focus.

Ans. (1)

Geometrical Optics

E26-31

Sol. Incorrect statementA concave mirror forms only virtual image for any position of real object.

Ans. (4)

Sol. Given 2u

v

v = ±2

fromf

1

u

1

v

1

±f

1

u

1

u2

1

after solving u = –30, –10 cm

(1) + 1 (2) – 2 (3) + 2 (4) – 1Ans. (3)

Sol. For M1

: v1

=)20(––)30(–

)20(–)30(–

f–u

uf = – 60

M = –u

v1= – 2.

For M2

: u = + 20. f = 10

v

1+

20

1=

10

1 v = 2 0

m2

= –20

20= – 1 m = m

1× m

2= + 2

Ans. (2)

Sol. = 4000

6000

m

V

= 1.5

Q.38 A ray of light passes from vacuum into a medium of refractive index n. If the angle of incidence is twice the angleof refraction, then the angle of incidence is:(1) cos–1 (n/2) (2) sin–1 (n/2) (3) 2 cos–1 (n/2) (4) 2 sin–1 (n/2)

Ans. (3)Sol. i = 2r

1 sin i = n sin r 2 sin i/2 cos i/2 = n sin i/2 cos i/2 = (n/2) i = 2 cos–1 (n/2)

Q.35 The image (of a real object) formed by a concave mirror is twice the size of the object. The focal length of themirror is 20 cm. The distance of the object from the mirror is (are)(1) 10 cm (2) 30 cm (3) 25 cm (4) both (1) and (2)

Q.36 In the figure shown find the total magnification after two successive reflections first on M1

and thenon M

2.

Q.37 The wavelength of light in vacuum is 6000 Aº and in a medium it is 4000 Aº. The refractive index of the mediumis :(1) 2.4 (2) 1.5 (3) 1.2 (4) 0.67

Geometrical Optics

E26-32

Ans. (1)Sol. r + r’ = 90º r’ = (90 – r)

1

sin r = 2

cos r

r r

r'

tan r =2

1

Critical angle = sin–12

1

= sin–1 (tan r)

Q.40 Given that, velocity of light in quartz = 1.5 108 m/s and velocity of light in glycerine = (9/4) 108 m/s. Nowa slab made of quartz is placed in glycerine as shown. The shift of the object produced by slab is

(1) 6 cm (2)3.55 cm (3) 9 cm (4) 2 cmAns. (1)

Sol. nquartz

= 2 ; nglycerine

=3

4

glycerine

quartz

n

n=

3/4

2=

2

3= µ

rel

shift = t

rel

11 = 18

2/3

11 = 6 cm

Q.41 A plane glass slab is placed over various coloured letters. The letter which appears to be raised the least is:(1) violet (2) yellow (3) red (4) green

Ans. (3)

Sol. Apparent shift

1–1t is least for red shift in least for red.

Q.42 A plane glass slab is placed over various coloured letters. The letter which apears to be raised the least is :(1) violet (2) yellow (3) red (4) green

Ans. (3)Sol. From the formula

Q.39 A ray of light from a denser medium strike a rarer medium. The angle of reflection is r and that of refraction is r’.The reflected and refracted rays make an angle of 90º with each other. The critical angle will be(1) sin–1 (tan r) (2) tan–1 (sin r) (3) sin–1 (tan r’) (4) tan–1 (sin r’)

Geometrical Optics

E26-33

glass

air

n

n

depthalRe

depthAppartent

Apparent depth = Real depthglass

air

n

nx

The letter which appear least raised has maximum Apparent depth and hence it has minimum for glass.

1

for to be minimum should be maximum which is for Red.

Q.43 An under water swimmer is at a depth of 12 m below the surface of water. A bird is at a height of 18 m from thesurface of water, directly above his eyes. For the swimmer, the bird appears to be at a distance X from the surfaceof water. (Refractive index of water is 4/3.) The value of X is(1) 24 m (2) 12 m (3) 18 m (4) 9 m

Ans. (1)

Sol.18

x

d

'd =

x = 18 ×3

4= 24m

Q.44 A bird is flying 3 m above the surface of water. If the bird is diving vertically down with speed = 6 m/s, his apparentvelocity as seen by a stationary fish underwater is(1) 8 m/s (2) 6 m/s (3) 12 m/s (4) 4 m/s

Ans. (1)

Sol.y

x3

4

x

y

dt

dx

3

4

dt

dy = 8 m/sec

Q.45 The critical angle of light going from medium A to medium B is . The speed of light in medium A is v. The speedof light in medium B is:

(1)v

sin(2) v sin (3) v cot (4) v tan

Ans. (1)

Sol. sin =B

A

C

C1

CB

=sin

V

m s–1 and in medium B is 2.5 × 108 ms–1. The critical angle for which a ray of light going from A to B is totally internallyreflected is

(1)

2

1sin 1–

(2)

5

2sin 1–

(3)

5

4sin 1–

(4)

3

1sin 1–

Ans. (3)

Q.46 Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is 2.0 × 108

Geometrical Optics

E26-34

Sol. 25.1102

105.2

V

V8

8

A

B

B

A

C

= sin–1

25.1

1= sin–1

5

4

[As C

= sin–1rarer

denser

]

Q.47 A small source of light is 4m below the surface of a liquid of refractive index 5/3. In order to cut off all the lightcoming out of liquid surface, minimum diameter of the disc placed on the surface of liquid is(1) 3m (2) 4m (3) 6m (4)

Ans. (3)Sol. In order to find the minimum diameter to block all the light we need to find the maximum radius of the circle formed.

r

4m

tan =4

r sin–1

5

3=

tan–1

4

3=

4

r=

4

3

[For radius to be maximum = C] r = 3m

Diameter = 6 m

Q.48 A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive indexof water is 4/3 and fish is 12 cm below the surface, the radius of the circle in cm is

(1) 12 × 3 × 5 (2) 4 × 5 (3) 12 × 3 × 7 (4) 12 × 3 / 7

Ans. (4)

Sol.R

= |43

Fish

12cm

tan C

=R

12...... (1)

A ray of light intering at 90º from rarer medium makes an angle of refaction equal to critical angle in the denser mediumand critical angle is given by

C

= sin–13

4

C

= tan–13

7...... (2)

Equation (1) & (2)

3

7=

R

12 R =

12 3

7

Geometrical Optics

E26-35

Q.49 Critical angle of light passing from glass to air is minimum for(1) red (2) green (3) yellow (4) violet

Ans. (4)

Sol. We know that1

Cglass

1sin

and glass

depends on wavelength of light

1glass

When is minimum the will be maximum & hence C

will be minimum. is minimum for voilet hence

Cis minimum for voilet light.

Q.50 A prism having refractive index 2 and refracting angle 30º, has one of the refracting surfaces polished. A beam

of light incident on the other refracting surface will retrace its path if the angle of incidence is:(1) 0º (2) 30º (3) 45º (4) 60º

Ans. (3)

Sol.

2º30sin

isin sin i =

2

1

2

12 i = 45º .

(1)A

(2)

A

( )2(3) A (4)

A

2

Ans. (3)Sol. We know that formula for deviation

= i + e – A & r1

+ r2

= Ai = i r

2= 0 r

1+ 0 = A

e = 0 r1

= AA = A1 sin i = sin ABecause angles

ri

A

are small i = A

(1) 30° (2) < 30° (3) 30° (4) 30°Ans. (2)

Q.51 A ray of light is incident at angle i on a surface of a prism of small angle A & emerges normally from the oppositesurface. If the refractive index of the material of the prism is , the angle of incidence i is nearly equal to :

Q.52 A beam of monochromatic light is incident at i = 50° on one face of an equilateral prism, the angle of emergenceis 40°, then the angle of minimum deviation is :

Geometrical Optics

E26-36

Sol.

50º 40º

60º

From the formula = i + e – A = 50 + 40 – 60 = 30º

min< 30º.

40º

30º

i = e 50º

1 and the angle of emergence at the second face is I2

1 = 0 (2) I1 > I2 (3) I1 < I2 (4) I1 = I2Ans. (4)Sol. Given angle of incidence I

1

Given angle of emergence I2

Condition for minimum devidationi = e I

1= I

2

Q.54 A prism having refractive index 2 and refracting angle 30°, has one of the refracting surface polished. A beam

of light incident on the other refracting surface will retrace its path if the angle of incidence is :(1) 0° (2) 30° (3) 45° (4) 60°

Ans. (3)Sol. Using A = r

1+ r

2

r1

= 30ºr

2= 0

1.sin i = 2 sin 30º

i = 45º

Q.55 A ray of light is incident at an angle 60° on the face of a prism having refractive angle 30°.The ray emerging outof the prism makes an angle 30° with the incident ray . through which it emerges from the surfece .(1) 0° (2) 30° (3) 45° (4) 60°

Ans. (1)

Sol.30º

eii=60º

= 30º = i + e – A60 + e – 30 = 30e = 0

, then the prism produces minimum deviationwhen(1) I

Q.53 The angle of a prism is 60º and the index of refraction of glass with air is 1.5. If the angle of incidence on thefirst face is I

Geometrical Optics

E26-37

Q.56 A prism of refractive index 2 has refracting angle 60°. In order that a ray suffers minimum deviation it should

be incident at an angle(1) 45° (2) 90 (3) 30° (4) none

Ans. (1)Sol. For minimum deviation

r1

= r2

= r 2r = A

r =2

A= 30º

Now from Snell’s law

1 sin i = 2 sin 30º

i = 45º

Q.57 A ray incident at a point at an angle of incidence of 60° enters a glass sphere of = 3 and it is reflected and refractedat the farther surface of the sphere. The angle between reflected and refracted rays at this surface is(1) 50° (2) 90° (3) 60° (4) 40°

Ans. (2)Sol.

Applying Snell's law on surface of incidence = sin–1sin60

3

60º60º

3 = 180 – [60 + ]

= 180 –

3

º60sinsinº60 1

= 180º – [60 + 30] = 90ºQ.58 The image for the converging beam after refraction through the curved surface (in the given figure) is formed at:

(1) x = 40 cm (2) x =40

3cm (3) x =

40

3cm (4) x =

180

7cm

Ans. (1)

Sol.20

2

3–1

302

3–

v

1

40

1

20

1

40

1–

v

1 v = 40 cm.

Q.59 In the given figure a plano-concave lens is placed on a paper on which a flower is drawn. How far above its actualposition does the flower appear to be ?

(1) 10 cm (2) 15 cm (3) 50 cm (4) none of theseAns. (1)

Geometrical Optics

E26-38

Sol.

Considering refraction at the curved surface,u = – 20 ; µ

2= 1

µ1

= 3/2 ; R = + 20

applyingv2

–u1

=R

12

v

1–

20

2/3

=

20

2/31v = – 10

i.e. 10 cm below the curved surface or 10 cm above the actual position of flower.

Q.60 A convexo - concave diverging lens is made of glass of refractive index 1.5 and focal length 24 cm. Radius of curvaturefor one surface is double that of the other. Then radii of curvature for the two surfaces are (in cm):(1) 6, 12 (2) 12, 24 (3) 3, 6 (4) 18, 36

Ans. (1)

Sol.

21 R

1–

R

1)1–(

f

1

R

1–

R2

1)1–5.1(

24–

1

R2

1–

2

1

24–

1

R = 6 cm 2R = 12 cm

(1) P

0

1(2)

1

0P (3)

0

1.

P

0(4) none of these

Ans. (3)

Sol. P =

21 R

1–

R

1)1–( ....(i)

P0

=

21o R

1–

R

11– .....(ii)

)–(

)1–(

P

P

0

0

0

P

0= )1–(

)–(P

0

0

0

Ans. (4)Sol. Image of sun is formed in the focal plane. So,

Diameter of image = f =º180

º5.0100 × × 10 mm = 9.

Q.61 When a lens of power P (in air) made of material of refractive index is immersed in liquid of refractive index

0. Then the power of lens is:

at the surface of the earth. A converging lens of focal length100 cm is used to provide an image of the sun on to a screen. The diameter in mm of the image formed is nearly(1) 1(2) 3 (3) 5 (4) 9

Q.62 The diameter of the sun subtends an angle of 0.5

Geometrical Optics

E26-39

Ans. (2)

Sol.

10

1=

v

1–

)15(

1

v = + 30 cm

for small object |dv| = 2

2

u

v|du| = 1

15

302

= 4 mm

Q.64 A biconvex lens is used to project a slide on screen. The slide is 2 cm high and 10 cm from the lens. The imageis 18 cm high. What is the focal length of the lens?(1) 9 cm (2) 18 cm (3) 4.5 cm (4) 20 cm

Ans. (1)

Sol.v

1–

u

1=

f

1...(1)

m =u

v...(2)

from (1) and (2) m =uf

f

here m = –2

18= – 9 {only real images can be formed on the screen, which is inverted}

– 9 =)10(f

f

– 9f + 90 = f10f = 90f = 9 cm

Q.65 Two symmetric double convex lenses A and B have same focal length, but the radii of curvature differ so that RA

= 0.9 RB. If n

A= 1.63, find n

B.

(1) 1.7 (2) 1.6 (3) 1.5 (4) 4/3Ans. (1)Sol. Given R

A= 0.9 R

B

BA f

1

f

1

(1.63 – 1)AR

2= (x

B– 1)

BR

2

xB

= 1.7

Q.63 A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The objectis at 15 cm from the lens. The length of the image is:(1) 1 mm (2) 4 mm (3) 2 mm (4) 8 mm

Geometrical Optics

E26-40

(1) 10 cm (2) 20 cm (3) 5 cm (4) none of theseAns. (1)

Sol.

Q.67 Parallel beam of light is incident on a system of two convex lenses of focal lengths f1

= 20 cm and f2

= 10 cm.What should be the distance between the two lenses so that rays after refraction from both the lenses pass undeviated

(1) 60 cm (2) 30 cm (3) 90 cm (4) 40 cmAns. (2)Sol. The rays coming from infinity parallel to principal axis and paraxial meet on focus after refraction and the rays originating

from focus of the lens originate parallel to principal axis afer refraction.

30 cm

Q.68 A thin symmetrical double convex lens of power P is cut into three parts, as shown in the figure. Power of A is:

(1) 2 P (2)P

2(3)

P

3(4) P

Ans. (4)Sol. f

A= f

B= f

C= f

net P

A= P

B= P

C= P

net= P

Q.69 A thin lens of focal length f and its aperture diameter d, forms a real image of intensity I. Now the central part

of the aperture upto diameter (d

2) is blocked by an opaque paper. The focal length and image intensity would change

to

(1)f I

2 2, (2) f

I,

4(3)

3

4 2

f I, (4) f,

3

4

I

Ans. (4)

Q.66 What should be the value of distance d so that final image is formed on the object itself. (focal lengths of the lensesare as given in the figure).

Geometrical Optics

E26-41

Sol. We know that P = IA & P × t = E

Hence IA =E

t

Initially energy/sec = I × 4

Id

2

d 22

Now energy/sec = I

2 2d d

2 4

= Id2

16

3

So, Now

2

2

Final Ιntensity I d 3 /16 3

Initial Intensity 4I d / 4

Focus will not change.

Q.70 A lens of power + 2.0 D is placed in contact with another lens of power – 1.0 D. The combination will behavelike(1) a converging lens of focal length 100 cm (2) a diverging lens of focal length 100 cm(3) a converging lens of focal length 50 cm (4) a diverging lens of focal length 50 cm.

Ans. (1)

Sol. Using the formula P = 1

f inm

p1

= 2D

f1

=2

100= +50 cm

f2

= –10f

2= –100 cm

21eq f

1

f

1

f

1

=

100

1

50

1=

100

12=

100

1

feq

= 100 cm

Q.71 A biconvex lens has a focal length of 10 cm. It is cut in half and two pieces are placed as shown. The focal lengthof the final combination is :

P/2 each. Let initial power of lens be P.Then (P

1)

f= (P

2)

f= P/2

Pf

= (P1)

f= (P

2)

f= P P

i= P

f

No change in power hence no change in focal length.

(1) 10 (2) 20 (3) 40 (4) Not a lensAns. (1)Sol. We know that on cutting the lens into two parts perpendicular to its principal axis power of the two parts will be

Geometrical Optics

E26-42

Q.72 Two plano-convex lenses each of focal length 10 cm & refractive index3

2are placed as shown in the figure. In the

space left, water R I. .

4

3is filled. The whole arrangement is in air. The optical power of the system is (in dioptre):

(1) 6.67 (2) - 6.67 (3) 33.3 (4) 20Ans. (1)Sol. Focal length of plano–convex lens is 10cm.

10

1=

1

2

3

R

11

10

1=

R

1

2

1 R = 5 cm

Let focal length of water lens is fw

wf

1=

1

3

4

5

1

5

1

wf

1=

5

2

3

1=

15

2

Optical power of system

P =wf

1

f

1

f

1 =

3

40

1.0

1

1.0

1= 6.67 D.

Q.73 In the figure shown 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.

(1) equivalent focal length of the combination is -18 cm(2) equivalent focal length of the combination is +36 cm(3) the system behaves like a concave mirror(4) both (1) and (3)

Ans. (4)

Sol.

1f

1=

1

2

3

15

1

10

1= –

12

1;

2f

1=

1

3

4

15

2=

45

2;

mf

1= –

15

2

21 f

1

f

1

f

1

f

2

f

1

f

1

meq

= –18

1 feq = – 18 cm

So, the combination behaves as a concave mirror

Geometrical Optics

E26-43

Q.74 A plano-convex lens, when silvered at its plane surface is equivalent to a concave mirror of focal length 28 cm. Whenits curved surface is silvered and the plane surface not silvered, it is equivalent to a concave mirror of focal length10 cm, then the refractive index of the material of the lens is :(1) 9/14 (2) 14/9 (3) 17/9 (4) none

Ans. (2)

Sol.

10

1

=

f

2

R

2

56

2

10

1

R

2 =

560

2056 =

560

36

R

1=

560

18

( – 1)560

18=

56

1

– 1 =18

10

=18

101 =

18

28=

9

14.

(1)3

50(2)

6

25(3) 0.03 (4) none

Ans. (2)Sol. Using formula

= 1n

nn

y

Rv

n

y=

2

nn Rv

=15.1

44.156.1

n

y=

2

44.156.1 = 1.5

=5.0

12.0= 0.24

Q.76 The refractive index of flint glass for blue line is 1.6333 and red line is 1.6161, then dispersive power of the glassis :(1) 0.0276 (2) 0.276 (3) 2.76 (4) 0.106

Ans. (1)Sol. 1.6333 - 1 = 1.6161 = 0.0172

ny-1

276.016247.1

6161.16333.1

Q.75 A medium has nv

= 1.56, nr

= 1.44. Then its dispersive power is :

Geometrical Optics

E26-44

(1) (2) (3) (4)

Ans. (2)Sol. Ray of Red light bends minimum because it has maximum & minimum .

Q.77 Which of the following diagrams shows correctly the dispersion of white light by a prism?