01.

A ray of light in passing from one transparent medium to the other medium having different
optical density, bends.
(a) Name the phenomenon. Give reason for it.
(b) How do the following quantities change : speed, wavelength, frequency and amplitude if
second medium is denser than the first medium.
(c) State whether the ray of light will bend or not, if both medium have same optical densities.

Sol. (a) Refraction.

The reason of bending (or refraction) is the change in speed of light in passing from one medium to
other of different optical density. Since light always travels the path of least time, it get bend to reach
from one point to another point of different medium. Though straight line connecting two points
situated in two different medium in least distance path but this is not least time path. Least time path
follows laws of refraction which can be proved mathematically.

(b) Speed will decrease. Wavelength will decrease. Frequency will remain unchanged. Amplitude will
decrease (due to partial reflection the boundary of two media).

(c) The ray will not bend. Because if speed is not changed then least time path is the least distance
path.

02. ‘The refractive index of water is 4/3'. Explain the meaning of this statement.

Sol. It means that the speed of light in water is ¾ th the speed of light in vacuum (or air) or the speed
of light in air is 4/3 times the speed of light in water.

03. The refractive index of water is 4/3 and of glass is 3/2. What is the refractive index of glass
with respect to water ?

𝜇𝑤 4
Sol. 𝑟. 𝑖 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 𝜇𝑤 = 𝑟. 𝑖 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑤. 𝑟. 𝑡. 𝑎𝑖𝑟 = 𝑎 𝜇𝑤 = =3
𝜇𝑎

𝜇 3
𝑟. 𝑖 𝑜𝑓 𝑔𝑙𝑎𝑠𝑠 = 𝜇𝑔 = 𝑟. 𝑖 𝑜𝑓 𝑔𝑙𝑎𝑠𝑠 𝑤. 𝑟. 𝑡. 𝑎𝑖𝑟 = 𝑎 𝜇𝑔 = 𝜇𝑔 = 2
𝑎
𝜇𝑔 3
𝜇𝑔 𝜇𝑎 9
So, Refractive index of glass with respect to water = 𝑤 𝜇𝑔 = 𝜇 = 𝜇𝑤 = 2
4 =8
𝑤 𝜇𝑎 3

04. Orange light of wavelength 6600 Å travelling in air gets refracted in water. If the speed of light
in air is 3 x 108 m/s and refractive index of water is 4/3, find : (i) the frequency of light in air, (ii) the
speed of light in water, and (iii) the wavelength of light in water.

Sol. Given : 𝜆𝑎𝑖𝑟 = 6600 Å = 6600 x 10-10 m,

Velocity of light = c = 3 x 108 ms-1, 𝜇 = 4/3

i) From relation c= 𝑓𝑎𝑖𝑟 𝜆𝑎𝑖𝑟 (in air)

𝑐 3×108
Frequency of light in air = 𝑓𝑎𝑖𝑟 = 𝜆 = 6600×10−10 = 4.54 × 1014 𝐻𝑧
𝑎𝑖𝑟

𝑐 3×108
(ii) Speed of light in water = 𝜇 = 4/3
= 2.25 × 108 𝑚𝑠 −1

(iii) Since the frequency of light remains unchanged in refraction 𝑓𝑤 = 𝑓𝑎𝑖𝑟 = 4.54 × 1014𝐻𝑧

𝑣𝑤 2.25×108 𝑚𝑠 −1
wavelength of light in water = = = 4.95 × 10−7 𝑚 𝑜𝑟 4950 Å
𝑓𝑤 4.54×1014 𝐻𝑧
05. The diagram shows a glass block suspended in a
liquid. A beam of light of single colour is incident from
liquid on one side of the block.

(a) Draw diagrams to show how does the light bend when
it travels from liquid to glass and then to liquid if
(i) the light slows down in glass, and
(ii) the light speeds up in glass.

(b) State two conditions under which the light ray passing from liquid to glass travels straight
without bending. Will the glass be visible then?

Sol. (a) (i) If light slows down in going from liquid to glass,
it means that 𝜇𝑔𝑙𝑎𝑠𝑠 > 𝜇𝑙𝑖𝑞𝑢𝑖𝑑 . So it will bend towards the
normal at the point of incidence in passing from liquid to
glass at the first surface, while it will bend away from the
normal at the second surface in passing from glass to
liquid. The ray diagram is shown in Fig. The light beam
suffers lateral displacement.

(ii) If light speeds up in going from liquid to glass,

it means that 𝜇𝑔𝑙𝑎𝑠𝑠 < 𝜇𝑙𝑖𝑞𝑢𝑖𝑑 . So it will bend away from
the normal at the point of incidence on the first surface
in passing from liquid to glass, while it will bend towards
the normal at the second surface in passing from glass
to liquid. The ray diagram is shown in Fig. The light beam
suffers lateral displacement in direction opposite to
that in case (i).

(b) A light ray passing from liquid to glass travels straight without bending under the following two
conditions:
(1) When the light ray falls normally on glass from liquid.

(2) When refractive index of liquid is same as that of glass (or speed of light in glass is same as in
liquid).
In both the above conditions, the glass block will not be visible inside the liquid.

06. The diagram shows two parallel rays A and B of red

and violet light respectively incident from air on air-glass
boundary. Complete the diagram showing the refracted
rays for them in the glass.

(i) How do the speeds of the rays differ in glass ?

(ii) Are the two refracted rays in glass parallel ? Give a reason for your answer.
(iii) How does the refractive index of glass differ for the two rays ?

Sol.(i) In glass, the speed of violet light is less than that

of the red light.

(ii) The two refracted rays inside glass are not parallel.
The reason is that the speed of red light in glass is
less than violet light, so the red ray bends less than violet ray (i.e., the angle of refraction r1 for red ray
is more than the angle of refraction r2 for the violet ray).

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚 𝑐 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚

(iii) For red ray, 𝜇𝑟 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑟𝑒𝑑 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑔𝑙𝑎𝑠𝑠 = 𝑣 , for violate light, 𝜇𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑣𝑖𝑜𝑙𝑒𝑡 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑔𝑙𝑎𝑠𝑠 =
𝑟
𝑐
𝑣𝑣

Since 𝑣𝑟 > 𝑣𝑣 𝑠𝑜, 𝜇𝑟 < 𝜇𝑣

07. Fig. shows A ray of light AO incident on a rectangular

glass P block PQRS, which is silvered at the surface RS.
The ray is partly reflected and partly refracted.

(a) Draw the path of the reflected and refracted rays.

(b) Show at least two rays emerging from the surface PQ
after reflection from the surface RS.
(c) How many images are formed in the above case ? Which
image is the brightest ?

Sol. Object
(a) The completed diagram is shown in Fig. in which OB
is the reflected ray and OC is the refracted ray for the
incident ray AO from object. 0.2𝐼
(b) Two rays emerging from the surface PQ after 𝐼 0.64𝐼
reflection from RS are labelled as 1 and 2.
(c) Multiple (or infinite) images are formed. The
second image formed due to first reflection at C at 0.8𝐼 0.16𝐼
the silvered surface RS, is the brightest. It is seen in 0.8𝐼
the direction of ray 1(image 𝐼2 ).

𝐼1

𝐼2

𝐼3

08. A ray of light passes from medium 1 to medium 2. Which of the following quantities of
refracted rays will differ from that of incident ray and how: speed, wavelength, frequency and
intensity?

Sol. Speed: increases from denser to rarer. Wavelength: increases from denser to rarer.
Frequency: not changed. Intensity: always decreases.

09. A ray of green light enters a liquid from air, as

shown in Fig. The angle 1 is 45° and angle 2 is 30°.

(a) Find the refractive index of liquid.

(b) Show in the diagram the path of the ray after

strikes the mirror and re-enters in air. Mark in
the diagram the angles wherever necessary.
(c) Redraw the diagram if plane mirror become normal to the refracted ray inside the liquid. State
the principle used.

𝑠𝑖𝑛𝑖 𝑠𝑖𝑛45𝑜 1
Sol. (a) 𝜇 = = = × 2 = √2
𝑠𝑖𝑛𝑟 𝑠𝑖𝑛30𝑜 √2

(b)

45°

30°

(c)

Principle of reversibility of light: Light retraced its path if its direction is reversed by normal incidence
on a mirror.

10. Name the factors on which refractive index of a medium depends and how?

Sol. Colour : For red r.i. is minimum and for violate r.i. is maximum.
Optical density : r.i. increases with the increase of optical density.
Wavelength: r.i. decreases with the increase of wavelength .
Temperature : r.i. decreases with the increase in temperature.

11. Light of a single colour is passed through a liquid having a piece of glass suspended in it. On
changing the temperature of liquid, at a particular temperature the glass piece is not seen.

(i) When is the glass piece not seen ? (ii) Why is the light of a single colour used ?

Ans. (i) The glass piece is not seen when the refractive index of liquid becomes equal to the refractive
index of glass. Image is formed by regular deviation of light by reflection or refraction. Here no bending
will occur and glass will be invisible.
(ii) Light of a single colour is used instead of white light because white light is combination of all
colours. For a particular temperature r.i. of liquid and glass will be equal for a single colour. Due to
other colours glass will be visible.
12. Fig. alongside shows an equilateral prism ABC and
the ray QR emerging out from the prism after suffering
minimum deviation. Complete the diagram to show the
refracted ray PQ inside the prism and the incident ray
OP on the prism. State in words how have you completed
the diagram.
Sol. (1) The refracted ray PQ is drawn parallel to the
base BC of the prism since the emergent ray QR has
suffered minimum deviation.

(2) After drawing normal on face AB of prism at P and

on face AC of prism at Q, the incident ray OP is
drawn such that the angle of incidence i1 is equal to
the angle of emergence i2.

13. In Fig., a monochromatic point source of

light S is viewed by an observer O through
a prism P. Complete the diagram to show the
image formed by the prism and as seen by the
observer O. Label the image by the letter I.

Sol.

14. Fig. shows two identical prisms P and Q placed

with their faces parallel to each other. A light ray of
yellow colour AB is incident at the face of the prism P.
Complete the diagram to show the path of the ray till
it emerges out of the prism Q.

Sol. CD Emerging out of prism Q is parallel

to incident ray AB on prism P.

15. Fig. below shows a light ray of

green colour incident normally on
the prisms A, B and C. In each case,
draw the path of the ray of light as it
enters and emerges out of the prism.
Mark the angle wherever necessary.
Sol.

16. What do you understand by the deviation produced by a prism ? Why is it caused ? State
three factors on which the angle of deviation depends.

Sol. In a prism incident ray bends twice towards the base by two refracting surfaces. Angle between
incident ray and emergent ray is called deviation.

Total deviation occurs because due to the shape of the prism, rays bend twice towards base
consecutively in the same sense by both the refracting surfaces unlike parallel glass slab.

Factors effecting deviation:

i) the angle of incidence
ii) the refractive index of prism
iii) the angle of prism
iv) colour or wavelength of light

17. Derive the deviation produced by the prism.

Sol. All rays are shown in figure.

Here OP is incident ray, PQ is refracted ray

Through prism and QR is emergent ray.

Here 𝑖1 is angle of incidence, 𝑟1 is angle of

refraction, 𝑟2 is angle of incidence for 2nd
surface, and 𝑖2 is angle of emergence.

Deviation 𝛿 = 𝛿1 + 𝛿2 = (𝑖1 − 𝑟1 ) + (𝑖2 − 𝑟2 )

= (𝑖1 + 𝑖2 ) − (𝑟1 + 𝑟2 )

From quadrilateral APNQ, ∠𝐴𝑃𝑁 = ∠𝐴𝑄𝑁 = 90°

𝑠𝑜, ∠𝑃𝑁𝑄 + ∠𝐴 = 180°

Again from Δ𝑃𝑁𝑄, (𝑟1 + 𝑟2 ) + ∠𝑃𝑁𝑄 = 180°

So, (𝑟1 + 𝑟2 ) = ∠𝐴

So, 𝛿 = (𝑖1 + 𝑖2 ) − ∠𝐴 𝑜𝑟, 𝛿 = 𝑖 + 𝑒 − 𝐴 (𝑒 𝑖𝑠 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑒𝑚𝑒𝑟𝑔𝑒𝑛𝑐𝑒)

( Only practise the above derivation if it is done in your school. )
18. (a) How does the angle of deviation produced by a prism change with increase in the angle of
incidence. Draw a curve showing the variation in the angle of deviation with the angle of
incidence at a prism surface.
(b) Using the curve in part (a) above, how do you infer that for a given prism, the angle of
minimum deviation is unique for the given light.

Sol. (a) Deviation in a prism is 𝛿 = 𝑖 + 𝑒 − 𝐴 ,

where 𝑖 is angle of incidence, e is angle of
emergence and A is angle of prism.

(b) From curve minimum deviation occurs when slope is zero. This will occur at a particular point
when i = e and so ray inside prism parallel to base. So this will happen for a particular angle of
incidence and 𝛿𝑚𝑖𝑛 is unique. 𝛿𝑚𝑖𝑛 = 2𝑖 − 𝐴

Not in our class

𝑑𝛿 𝑑𝑒
𝛿 = 𝑖+𝑒−𝐴 𝑜𝑟, = 1 + 𝑑𝑖
𝑑𝑖
𝑑𝛿 𝑑𝑒
Now, at minimum deviation, slope =0 𝑜𝑟, 1 − 𝑑𝑖 = 0 [ negative as change of i and e is opposite]
𝑑𝑖
𝑑𝑒
So, 1 = 𝑜𝑟, 𝑑𝑒 = 𝑑𝑖 𝑜𝑟 𝑖 = 𝑒
𝑑𝑖

19. A ray of light incident at an angle of incidence 48° on a prism of refracting angle 60° suffers
minimum deviation. Calculate the angle of minimum deviation.

Sol. We know, 𝛿𝑚𝑖𝑛 = 2𝑖 − 𝐴 = 2 × 48° − 60° = 36°

20. What should be the angle of incidence for a ray of light which suffers minimum deviation of
36°
through an equilateral prism ?

𝛿𝑚𝑖𝑛 +𝐴 36°+60°
Sol. We know, 𝛿𝑚𝑖𝑛 = 2𝑖 − 𝐴 𝑜𝑟, 𝑖 = = = 48° (A=60° as equilateral prism)
2 2

21. Derive the relation between apparent depth and real depth of a submerged body in water
seen from vertically above from air.

Sol. Rays from object O after refraction from

Water surface appear from I shown in figure.
If it is seen from vertically above then A and B
are very closed and we can say

𝐴𝐵
𝑖 = sin 𝑖 = 𝑡𝑎𝑛 𝑖 = 𝐴𝑂
𝐴𝐵
and 𝑟 = 𝑠𝑖𝑛 𝑟 = 𝑡𝑎𝑛 𝑟 = 𝐴𝐼
 𝐴𝐵
𝑟𝑒𝑎𝑙 𝑑𝑒𝑝𝑡ℎ 𝐴𝑂 𝑠𝑖𝑛 𝑟
So, 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ = = 𝐴𝐼
𝐴𝐵 = = 𝜇𝑤𝑎𝑡𝑒𝑟
𝐴𝐼 𝑠𝑖𝑛 𝑖
𝐴𝑂

22. Derive the relation between apparent height and real height of a body in air seen from
vertically below from water.

Sol. Rays from object O after refraction from r

water surface appear from I shown in figure. i
If it is seen from vertically below then A and B i
are very closed and we can say

𝐴𝐵
𝑖 = sin 𝑖 = 𝑡𝑎𝑛 𝑖 = 𝐴𝑂 r
𝐴𝐵
and 𝑟 = 𝑠𝑖𝑛 𝑟 = 𝑡𝑎𝑛 𝑟 = 𝐴𝐼
𝐴𝐵
𝑟𝑒𝑎𝑙 ℎ𝑒𝑖𝑔ℎ𝑡 𝐴𝑂 𝑠𝑖𝑛 𝑟 1
So, 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 = = 𝐴𝐼
𝐴𝐵 = =𝜇
𝐴𝐼 𝑠𝑖𝑛 𝑖 𝑤𝑎𝑡𝑒𝑟
𝐴𝑂

23. A star appears twinkling in the sky. Explain.

Sol. Due to temperature difference r.i. of air in various layer are different. So refraction occurs through
it when light comes from star. Since r.i. of layers varies as temperature changes, the direction of
refracted rays changes continuously. Since amount of light is very small, sometime it coming off our
eye and twinkling of star occurs.

24. The sun is seen a few minutes before it rises above the horizon in the morning while in the
evening few minutes longer after it sets. Explain.

Sol. Temperature of air layer is less at higher height, so density and r.i. is less. So rays coming from Sun
at horizon bend towards earth while travelling from rarer to denser medium through atmosphere and
appear above its original position. Thus sun is seen a few minutes before it rises above the horizon in
the morning while in the evening few minutes longer after it sets.

25. A coin kept in a vessel and not visible when seen from just below the edge of the vessel, can
be viewed from the same position when water is poured into the vessel. Explain.

Sol. When we see from just below the edge, light rays cannot reach directly from coin to eye. But after
pouring water light rays bend away from normal at water surface and reach to eye. Thus coin is appear
at a height and viewed.

26. A print appears to be raised when a glass slab is placed over it. Explain.

Sol. Diverging rays coming from print refracted away from the normal at the upper surface of the glass
slab and diverge more. So print appears from a point above it.

27. A piece of paper stuck at the bottom of a glass slab appears to be raised when seen from
above.
Sol. Diverging rays coming from paper piece refracted away from the normal at the upper surface of
the glass slab and diverge more. So it appears from a point above the object.

28. A tank appears shallow than its actual depth.

Sol. Diverging rays coming from a point of the bottom surface refracted away from the normal at the
upper surface of the glass slab and diverge more. So the point appears above the actual depth and
thus whole surface appears shallow.

29. A person's legs appear to be short when standing in a tank.

Sol. Diverging rays coming from a point of the leg inside water surface refracted away from the normal
at the upper surface of the glass slab and diverge more. So the point appears above the actual depth
and thus legs appear to be short seen from a point above surface.

30. A point source of light O of single colour is seen through a

rectangular glass slab PQRS. The paths of two rays, in and outside
the slab, are shown in Fig.
(i) In the diagram, label the position 𝑰 of the source O where it will
appear when seen through the surface RS.
(ii) Does the source O appear to be nearer or farther with respect to
the surface PQ?
(iii) How does the shift depends on the thickness PS or QR of the slab?
(iv) Justify your answer in (ii) with the help of an appropriate ray diagram.
(v) For the same rectangular glass slab, which colour from the visible
spectra (violet to red) will produce the maximum shift ? Show with diagram.

(i) 𝐼 is the position of image shown in figure i.e. position of object O

when it is seen through the surface RS. O

1
We can prove that shift 𝑂𝐼 = 𝑡 (1 − 𝜇) P 𝐼 Q
where t is width of the slab PR or QS

(ii) From diagram it is clear that source O appears at 𝐼 nearer to PQ

when viewed through RS. R S

(iii) Shift decreases with the decrease of thickness of the slab.

O
(iv) From diagram it is clear that shift decreases to 𝑂𝐼′ from 𝑂𝐼
when thickness decreases to PR’ from PR. 𝐼′
P 𝐼 Q
1
This can be said also using shift formula 𝑂𝐼 = 𝑡 (1 − 𝜇)
R’ S’

R S

(v) For violet shift is maximum which is shown in diagram

as violet bends maximum.
O
1
From shift formula 𝑂𝐼 = 𝑡 (1 − 𝜇) we can say this also where 𝐼𝑟
𝜇 of violet is maximum. 𝐼𝑣
31. A glass slab is placed over a piece of paper on which VIBGYOR İs printed with each letter into
its corresponding colours.
(i) Will the image of all the letters be in the place?
(ii) The letter of which colour will appear to be raised (a) maximum, and (b) minimum ? Explain
your answer.

Sol. (i) No. The image of all the letters will not be in the same place.

(i) (a) The letter of violet colour (V) appears to be raised maximum.
(b) The letter of red colour (R) appears to be raised minimum.
𝑟𝑒𝑎𝑙 𝑑𝑒𝑝𝑡ℎ
Reason : 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ = 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑑𝑒𝑥 and refractive index of glass is most for the violet light
while least for the red light, therefore the apparent depth is least for the violet letter and most for the
red letter.

32. An object placed in one medium when seen from the other medium, appears to be vertically
shifted. Name two factors on which the magnitude of shift depends and state how does it
depend on them.

Sol. Vertically shift depends on (i) refractive index of two mediums and (ii) distance of the object from
the separating surface of two mediums.

Shift will increase if r.i. of denser medium is increased or rarer medium is decreased.

33. A coin kept inside water (𝝁 = 4/3) when viewed From air in a vertical direction, appears to be
raised by 2.0 mm. Find the depth of coin in water.

𝑥 3𝑥
Sol. Let real depth = 𝑥 so, apparent depth = 𝜇 = 4
3𝑥 𝑥
By the problem, shift = 𝑥 − = 4 = 2 𝑚𝑚. 𝑠𝑜, 𝑥 = 8 𝑚𝑚.
4

34. A postage stamp kept below a rectangular glass slab of refractive index 1.5 when viewed
from vertically above it, appears to be raised by 7.0 mm. Calculate the thickness of the glass
slab.

Sol. Thickness = real depth = 𝑥 (𝑙𝑒𝑡)

𝑥 𝑥
So, apparent depth = =
𝜇 1.5
𝑥
So, appear to be raised = 𝑥 − 1.5 = 7 𝑐𝑚. 𝑠𝑜, 𝑥 = 21 𝑐𝑚.

35. Explain the term critical angle with the aid of a labelled diagram.

Sol. Critical angle is the angle of incidence in the denser

medium corresponding to which the angle of refraction rarer medium
in the rarer medium is 90°. In diagram ic is critical angle. r r=90°

denser medium
i ic
36. How is the critical angle related to the refractive index of a medium?

𝑠𝑖𝑛 𝑖 𝜇 1
Sol. 𝑠𝑖𝑛 90°
𝑐
= 𝜇 𝑟𝑎𝑟𝑒𝑟 𝑜𝑟, 𝑠𝑖𝑛 𝑖𝑐 = 𝜇 , where rarer medium is air and 𝜇𝑑𝑟 is r.i of denser w.r.t. rarer
𝑑𝑒𝑛𝑠𝑒𝑟 𝑑𝑟
medium.

37. State the approximate value of the critical angle for (a) glass-air surface (b) water-air surface
(c) diamond-air surface.

Sol. (a) 42° (b) 49° (c) 24° .

38. Name two factors which affect the critical angle for a given pair of media. State how do the
factors affect it.

Sol. Critical angle depends on i) refractive index of two mediums ii) wavelength or frequency of light.

i) Critical angle decreases with the increase of refractive index of denser medium w.r.t. rarer medium.
ii) Since r.i. is greater for lower wavelength (𝜇𝑣 > 𝜇𝑟 ), so, critical angle is less for lower wavelength
(𝑖𝑐−𝑣 < 𝑖𝑐−𝑟 ).

39. The critical angle for glass-air is 45° for the light of yellow colour. State whether it will be less
than, equal to, or more than 45° for (i) red light, and (ii) blue light?

Ans. Greater wavelength, lesser 𝜇 and so greater critical angle. (i) so for red more than 45° (ii) for violet
less than 45°.

40. (a) What is total internal reflection ? (b) State two conditions necessary for the total internal
reflection.

Sol. When a ray of light travelling in a denser medium, is incident at the surface of a rarer medium at
the angle of incidence greater than the critical angle for the pair of media, the ray is totally reflected
back into the denser medium. This phenomenon is called total internal reflection.

There are following two necessary conditions for the total internal reflection :

(1) The light must travel from a denser to a rarer medium.

(2) The angle of incidence must be greater than the critical angle for the pair of media.

41. An empty test tube placed in water in beaker with mouth outside the water surface, shines
like a mirror when seen at certain angles.

Sol. When an empty test tube placed in water light rays travel from water to glass and then glass to air.
During travelling from glass to air if they incident at angle greater than critical angle (42°), they totally
reflected back through glass and then water. If we see along that rays, test tube shines like mirror.

42. A crack in a glass vessel often shines like a mirror.

Sol. Air is present inside the crack. During travelling of light rays from glass to air inside crack, if they
incident at angle greater than critical angle (42°), they are totally reflected back through glass and if
we see along that rays, crack shines like mirror.

43. On a hot sunny day, a driver may see a pool of water (or wet road) in front of him at some
distance. It is the phenomenon of mirage which is often observed in desert. Explain.

Sol.

In desert sand is hot during day time. So lower air layers are hotter and rarer than upper layers. So ray
coming downward from the top of the tree travels from denser to rarer medium, bends continuously
away from the normal and angle of refraction increases continuously and after reaching 90o total
internal reflection occurs. So ray then moves upward bending continuously towards normal and
reaches the eye of observer and an inverted image of tree is seen which gives a false impression of
water in front of the tree.

44. A piece of diamond sparkles when viewed from certain directions.

Sol. Critical angle of diamond is low. It is cut such that light rays after entering it from different
directions experience total internal reflection multiple times and finally exist from a certain directions
and thus it sparkles from that direction. This is happening for maximum amount of rays as diamond’s
critical angle is low.

45. Show with diagram how can a prism erect an inverted image with 180° deviation.

Sol. This Is done by a rectangular isosceles prism.

46. Show with diagram how can a prism erect an inverted image without deviation.

Sol. This Is done by a rectangular isosceles prism.

47. Fig. below shows a light ray of green colour incident normally on the prisms A, B and C. In
each case, draw the path of the ray of light as it enters and emerges out of the prism. Mark the
angle wherever necessary.

(a) (b) (c)

30° 30° 30°

Sol. (a) (b)

(c)
Lens:

01. Draw by convex lens, plano- convex lens concavo convex lens.

Sol.

01. Draw by concave lens, plano- concave lens convexo concave lens.

Sol.

03. Explain the converging action of a convex lens.

Sol. We can consider a convex lens the combination

of two of prisms on ether side and one rectangular
slab in middle shown in figure. Base of upper prism
is downward and base of lower prism is upward.
Thus upper prism bends rays downwards towards
its base and lower prism bends rays upwards and
middle part slab passes rays undeviated. Thus all
rays are converge at a point after refraction through
lens.

04. Explain the diverging action of a concave lens.

Sol. We can consider a concave lens the combination

of two of prisms on ether side and one rectangular
slab in middle shown in figure. Base of upper prism
is upward and base of lower prism is downward.
Thus upper prism bends rays upwards towards
its base and lower prism bends rays downwards and
middle part slab passes rays undeviated. Thus all
rays are diverge and appear from a point after refraction
through lens.

05. Define Centre of curvature, Principal axis, Optical centre and Radius of curvature.
Centre of curvature: A lens has two surfaces. Each surface of the lens is a part of a sphere. The centre
of the sphere whose part is the lens surface, is called the centre of curvature of that surface of the
lens. Since a lens has two spherical surfaces, so there are two centres of curvature of a lens.
the centres of curvature of the two surfaces are C1 and C2 .

Principal axis : It is the line joining the centres curvature

of the two surfaces of the lens. The line C1C2 is the principal
axis.

Optical centre : It is a point on the principal axis of the lens

such that a ray of light passing through this point emerges
parallel to its direction of incidence. It is marked by the letter O.

(2) Radius of curvature: The radius of the sphere whose part is

the lens surface, is called the radius of curvature of that surface
of the lens. For a thin lens, the radius of curvature of a surface
of lens is equal to the distance of centre of curvature of that
surface from the optical centre i.e. PC1=OC1 and PC2=OC2
in figure (a).

06. Is optical centre equidistant from the two surfaces?

Sol. No, it is towards the surface of small radius

of curvature for a by convex lens or by concave
lens shown in figure.
O
For equi convex or equi-concave lens it is C1 C2
at the middle position.

07. Can optical centre be outside the lens?

Sol. Yes, for convexo concave, outside the concave surface and for concavo convex, outside the
convex surface.

O C2 C1 O C1 C2
08. Can optical centre be on one surface of a lens?

Sol. Yes, for plano convex or plano concave lens it is on

the curved surface. O

09. Define 1st principal focus of a lens.

Sol. For a lens, the first principal focal point is a point F1 on the principal axis of the lens such that the
rays of light coming from it (for convex) or appearing to meet at it (for concave), after refraction
through the lens, become parallel to the principal axis of the lens (Fig.).

10. Define 2nd principal focus of a lens.

Sol. For a lens, the second principal focal point is a point F2 on the principal axis of the lens such that
the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it (for
convex lens) or appear to be diverging from it ( for concave lens).[Fig.].

11. Define focal plane.

Sol. Focal plane : A plane normal to the principal axis, passing through the focus, is called the focal
plane. A lens has two focal planes; first focal plane, passing through the first focal point and second
focal plane passing through the second focal point.
12. Define 2nd secondary focal point.

Sol. When parallel light rays but not parallel to principal axis incidence on a lens, after refraction they
meet at point (for convex lens) or appear to diverge from a point (for concave lens) on 2nd focal plane
called second secondary focal point. It can be any point on 2nd focal plane.

13. Focal length of which lens is more, thick or thin?

Sol. For thin lens focal length is more.

14. Are first and second focal length equal for all lens except equi convex or equi-concave lens?

Sol. Yes, for all types of lens first focus and second focus are equal for all lens.

15. When 1st and 2nd focus are not equal?

Sol. When refractive index of medium of both outside lens are not equal.

16. What will happen when a convex lens is immersed in water?

Sol. Focal length increases

17. What will happen when a convex lens is immersed in a liquid of refractive index greater than
the r.i. of lens?

Sol. It will behave like a diverging lens.

18. Draw the image formation by a convex lens when object is at infinity. State the nature and
position of the image. State one application.

Sol. Image is formed at 2nd focal plane.

It is real, inverted and highly diminished.

Application: Burning glass, camera lens.

19. Draw the image formation by a convex lens when object is beyond 2F1. State the nature and
position of the image. State one application.

Sol. Image is real, inverted and diminished.

It is formed between F2 and 2F2 .
Application: camera lens.

20. Draw the image formation by a convex lens when object is at 2F1. State the nature and
position of the image. State one application.

Sol. Image is real, inverted and same size.

It is formed at 2F2 .

Application: terrestrial telescope.

21. Draw the image formation by a convex lens when object is between 2F1 and F1. State the
nature and position of the image. State one application.

Sol. Image is real, inverted and magnified.

It is formed beyond 2F2 .

Application: slide projector.

22. Draw the image formation by a convex lens when object is at F1. State the nature and position
of the image. State one application.

Sol. Image is real, inverted and highly

magnified.

It is formed at infinity. .

Application: collimator of spectrometer.

23. Draw the image formation by a convex lens when object is between optical centre and F1.
State the nature and position of the image.

Sol. Image is virtual, erect and magnified.

It is on the same side of the object and

behind it.

Application: Magnifying glass.

24. Draw the image formation by a concave lens when object is at infinity. State the nature and
position of the image. State one application.

Sol. Image is virtual, erect and highly diminished.

It is formed at F2. .

Application: Galilean telescope.

25. Draw the image formation by a concave lens when object is at finite distance. State the
nature and position of the image. State one application.

Sol. Image is virtual, erect and diminished.

It is formed between optical centre and F2

Application: Spectacles for short sightedness.

26. (i) Draw the outline of the lens L

used and name it.
(ii) Locate the final image formed.

Sol. (i) Concave lens.

(ii) It is formed between optical centre and F2 ( diagram to be shown)

27. In the above question answer (i) and (ii) if F1 and F2 are interchanged.

Sol. (i) convex lens.

(ii) beyond 2F2. ( diagram to be shown)

28. State the sign convention to measure the distances for a lens.

Sol. All positions are taken from optical centre O. Leftward from O is negative and rightward from O is
positive. Above principal axis is positive and bellow it is negative.

29. The focal length of a lens is (i) positive, (ii) negative. In each case, state the kind of lens.

Sol. (i) convex (ii) concave

30. What information about the nature of image (i) real or virtual, (iii) erect or inverted, do you get
from the sign of magnification + or -?

Sol. + sign : virtual and erect and - sign :real and inverted

31. Define the term power of a lens. In what unit is it expressed?

Sol. Deviation power of a lens is it's power. So more deviation, smaller focal length and more power.
1
Thus power 𝑃 = 𝑓 , it’s unit is dioptre when focal length is in metre.
Converging power is taken as positive (convex lens) and diverging power is taken as negative (concave
lens).

32. Power of a lens is – 0.5 D. What information you get from it?

1 1
Sol. 𝑓 = = 𝑚 = −2 𝑚 = −200 𝑐𝑚.
𝐷 −0.5
So lens is concave with focal length 200 cm.

33. What is a magnifying glass ? State its two uses.

Sol. Magnifying glass is a convex lens which is used to see tiny object distinctly by increasing the
angle subtended by the object placed at 25 cm distance before naked eye.
To get angular magnification object should be placed less than 25 cm (D) within focus before lens. To
get maximum magnification object is placed such that image is formed at D distance.

So, here object distance = - u , image distance = - D , focal length = +f

1 1 1 1 1 1 𝐷 𝐷
𝑠𝑜, 𝑓𝑟𝑜𝑚 𝑣 − 𝑢 = 𝑓 𝑤𝑒 𝑔𝑒𝑡, −𝐷 − −𝑢 = 𝑓 𝑜𝑟, =1+𝑓
𝑢

𝑣 −𝐷 𝐷 𝐷
Now, magnification 𝑚 = 𝑢 = = 𝑢 =1+𝑓
−𝑢

Use: i) As a simple microscope or magnifying glass.

ii) Objective lens of compound microscope, telescope.

34. Find the maximum magnifying power of a convex lens of focal length 5 cm.

𝐷 25
Sol. 𝑚 = 1 + 𝑓 = 1 + =6
5

35. Draw a neat labelled ray diagram to show the formation of image by a magnifying glass. State
three characteristics of the image.

Sol. Image is erect, virtual and magnified.

36. Where is the object placed in reference to the principal focus of a magnifying glass, so as to
see its enlarged image ? Where is the image obtained ?

Sol. Between 1st principal focus and optical centre. Image is obtained at 25 cm.

37. The diagram in Fig shows the experimental set

up for the determination of focal length of a lens
using a plane mirror.

(i) Draw two rays from the point O of the object to

show the formation of image I at O itself.
(ii) What is the size of the image I?
(iii) State two more characteristics of the image I.
(iv) Name the distance of the object O from the
optical centre of the lens.
(v) To what point will the rays return if the mirror is moved away from the lens by a distance equal
to the focal length of the lens.

Sol. i)

ii) Same size as object.

iii) real, inverted.

iv) Focal length of the lens.

v) Rays will return at the same point and it is independent of the mirror distance from lens.

Numerical

38. The focal length of a camera lens is 20 cm. Find how far away from the film must the lens be
set in order to photograph an object located at a distance 100 cm from the lens.

Sol. Here focal length f = 20 cm, u = - 100 cm

1 1 1 1 1 1
𝑠𝑜, 𝑓𝑟𝑜𝑚 𝑣 − 𝑢 = 𝑓 , − −100 = 20 𝑜𝑟, 𝑣 = 25 𝑐𝑚
𝑣
So lens must be set at a distance 25 cm from film on which real image is formed.

39. A convex lens forms an image 16.0 cm long of an object 4.0 cm long kept at a distance 6 cm
from the lens. The object and the image are on the same side of lens.
(a) What is the nature of image ?
(b) Find : (i) the position of image, and (ii) the focal length of lens.

(a) Since the image is magnified and on same side of the lens as the object, so the image is virtual.

(b) Given : I = 16.0 cm (positive), O = 4.0 cm (positive), u = - 6 cm (negative)

𝐼 𝑣 16 𝑣
(i) From relation, 𝑚 = 𝑂 = 𝑢 𝑜𝑟, = −6 𝑜𝑟, 𝑣 = −24 𝑐𝑚.
4

So image is 24 cm in front of the lens.

1 1 1 1 1 1
ii) 𝑠𝑜, 𝑓𝑟𝑜𝑚 𝑣 − 𝑢 = 𝑓 , − −6 = 𝑓 𝑜𝑟, 𝑓 = 8 𝑐𝑚
−24

40. (a) At what position a candle of length 3 cm be placed in front of a convex lens so that its
image of length 6 cm be obtained on a screen placed at distance 30 cm behind the lens?
(b) What is the focal length of lens in part (a)?

Sol. (a) Since image is formed behind the lens, it is real, inverted.

So, 𝑂 = +3 𝑐𝑚, 𝐼 = −6 𝑐𝑚, 𝑣 = +30 𝑐𝑚,

𝐼 𝑣 −6 30
𝑠𝑜, 𝑓𝑟𝑜𝑚 𝑚 = 𝑂 = 𝑢 𝑜𝑟, 3 = 𝑢 𝑜𝑟, 𝑢 = −15 𝑐𝑚
Candle to be placed at a distance 15 cm in front of the lens.

1 1 1 1 1 1
(b) 𝑓𝑟𝑜𝑚 𝑣 − 𝑢 = 𝑓 , − −15 = 𝑓 𝑜𝑟, 𝑓 = 10 𝑐𝑚
30

41. The focal length of a convex lens is 25 cm. At what distance from the optical centre of the
lens an object be placed to obtain a virtual image of twice the size?

𝑣 𝑣
Sol. 𝐹𝑟𝑜𝑚 𝑚 = 𝑢 𝑜𝑟, |𝑚| = |𝑢| = 2 𝑜𝑟, |𝑣| = 2|𝑢 |

𝑙𝑒𝑡 𝑢 = −𝑥, 𝑡ℎ𝑒𝑛 𝑣 = −2𝑥 (𝑠𝑖𝑛𝑐𝑒 𝑖𝑚𝑎𝑔𝑒 𝑖𝑠 𝑣𝑖𝑟𝑡𝑢𝑎𝑙, 𝑖𝑡 𝑖𝑠 𝑖𝑛𝑓𝑟𝑜𝑛𝑡 𝑜𝑓 𝑙𝑒𝑛𝑠)

1 1 1 1 1 1 1 1 1 1
𝑓𝑟𝑜𝑚 𝑣 − 𝑢 = 𝑓 , − −𝑥 = 25 𝑜𝑟, −2𝑥 + 𝑥 = 2𝑥 = 25 𝑜𝑟, 𝑥 = 12.5
−2𝑥

So, 𝑢 = −𝑥 = −12.5 𝑐𝑚 i.e. 12.5 cm in front of the lens.