# CBSE Class 10 Science Light Reflection and Refraction Assignment Set B

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## Assignment for Class 10 Science Chapter 10 Light Reflection And Refraction

Class 10 Science students should refer to the following printable assignment in Pdf for Chapter 10 Light Reflection And Refraction in standard 10. This test paper with questions and answers for Grade 10 Science will be very useful for exams and help you to score good marks

### Chapter 10 Light Reflection And Refraction Class 10 Science Assignment

Question : Four students set up the glass slab experiment as shown below: The correct fixing of the pins P1 and P2 , for 'defining the incident ray', and of pins P3 and P4, for finding the emergent ray', has been done by student:
a. A
b. C
c. B
d. D

Explanation: Emergent light is parallel to the incident light and laterally displaced.

Question : As incident angle is increased for a given pair of medium, the refraction angle will
a. decrease
b. always remains same
c. may increase or decrease
d. increase

Question : A ray of light is travelling from a rarer medium to a denser medium. While entering the denser medium at the point of incidence, it
a. goes straight into the second medium
b. bends towards the normal
c. bends away from the normal
d. does not enter at all

Question : Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
a. A convex lens of focal length 50 cm
b. A concave lens of focal length 50 cm
c. A convex lens of focal length 5 cm
d. A concave lens of focal length 5 cm

Question. What is the magnification of the images formed by plane mirrors and why?
Answer: The magnification of the image formed by a plane mirror is 1 because size of the image is equal to the size of the object.

Question. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer: As we can see in the figure given, when the lower half of the convex lens is covered with a black paper, it still forms the complete image of the object as that formed with uncovered lens. However the intensity of the image is reduced when the convex lens is covered with black paper.

Question. What is meant by power of a lens?
Answer: The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it. The power of a lens is defined as the reciprocal of its focal length in metres.
power of a lens = 1/Focal lenth (in m)
The SI unit of power is dioptre denoted by the letter D. One dioptre is the power of a lens whose focal length is 1 metre.

Question. The magnification produced by a plane mirror is +1. What does this mean?
Answer: Magnification m = 1 implies u = v
Therefore the image is formed at the same distance from the mirror as the object is from the mirror.
h' = h (size of the image is same as the object)
m = +ve implies that image formed by plane mirror is virtual and erect.

Question. Why is a concave mirror of large focal length often used as a shaving mirror?
Answer: While shaving, we keep our face between the pole and focus of a concave mirror to obtain an erect and enlarged image of the face. This enlarged image of the face helps us in having a better and a closer shave.

Question. What happens to the ray of light after reflection from a concave mirror, if it (i) is parallel to the principal axis? (ii) passes through the center of curvature?
Answer: (i) After reflection, the ray of light passes through the focus of a concave mirror in this case.
(ii) In this case, the ray of light gets reflected back along the same path, i.e., through centre of curvature.

Question. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer: The object should be placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. The image formed will be virtual, erect and larger than the object as shown in the ray diagram.

Question. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer: Here P = +1.5 D,   f = ?
∴ p = 1/f   ⇒  1.5 = 1/f    ∴ f (Focal length) = 1/1.5 = 10/15 = 0.67 m
∴ The prescribed lens is converging in nature.

Question. Find the focal length of a lens of power -2.0 D. What type of lens is this?
Answer: Here P = –2.0 D, f = ?
∴ p = 1/f  ∴  -2.0 = 1/f   ⇒  f = -1/2.0 = 0.50 m
∴ The lens is concave lens.

Question. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer: Here u = -10 cm, f = 15 cm, v = ?
Using mirror formula,
1/f = 1/v + 1/u     ⇒  1/15 = 1/u + 1/-10
1/15 + 1/10 = 1/v  ⇒ 2 + 3 = 1/v
5/30 = 1/v ∴ v = 6 cm

Question : What is a ray of light?
Answer : The path along which light travels is called a ray of light.

Question. List four properties of the image formed by a concave mirror when object is placed between focus and pole of the mirror.
Answer: Image is virtual, erect, magnified, i.e., bigger than the object and behind the mirror.

Question : What is an optically denser medium?
Answer : A medium in which light travels comparatively slower than the other medium is called an optically denser medium.

Question : State two factors which determine lateral displacement of a ray of light passing through a rectangular glass slab.
Answer :  2 Factors which determine lateral displacement are as given below:
i. Lateral displacement is directly proportional to the thickness of glass slab.
ii. Lateral displacement is directly proportional to the angle of incidence.

Question : A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image.
Answer : Since the lens is concave, hence f is negative
Given: u = - 30 cm; f= - 10 cm; h = 2.5 cm; v = ?; h' = ? The positive sign with the magnification indicates that the image formed erect.
The size of the image is determined by h'.
h'/h = m
h' = h x m
= 2.5 x  0.25
= 0.625 m
Thus the image formed is virtual and erect. It is at a distance of 7.5 cm from lens and its size is 0.625 cm.

Question. State the condition when a concave mirror forms an image
(i) larger than the actual object. (ii) smaller than the actual object.
Answer: (i) When object is placed anywhere between pole and centre of curvature.
(ii) When object is placed anywhere between infinity and centre of curvature.

Question. A student takes a mirror which is depressed at the centre and mounts it on a mirror stand. An erect and enlarged image of his face is formed. He places the mirror on a stand along a metre scale at 10 cm mark. In front of this mirror, he mounts a white screen and moves it back and forth along the metre scale till a highly sharp, well-defined image of a distant building is formed on the screen at 25.5 cm mark.
(i) Name the mirror and find its focal length.
(ii) Why does the student get sharp image of the distant building at 25.5 cm mark?
Answer: (i) The mirror is a concave mirror. The focal length is the difference between mirror and screen, so, f = 25.5 cm – 10 cm = 15.5 cm
(ii) The rays of light coming from a distant object are parallel to principal axis and meets at the focus of concave mirror where the screen is placed. Thus, a sharp image of the distant building is formed.

Question. For finding the focal length of a concave mirror, where do we keep the object? What is the position of image formed? On which structure we get the image? What is the nature of the image formed?
Answer: A large sized object placed at a far off (Infinity) place, seen from the window of the lab, is taken as object.
The image is formed at F point of the mirror. This image is formed on the screen.
The distance between the mirror and the screen is measured; which is equal to focal length of the mirror.
The image formed is real and inverted.

Question. A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations as given below:
S.No. Distance of flame Distance of the screen
from the lens (cm) from the lens (cm)
1. 60 20
2. 40 24
3. 30 30
4. 24 40
5. 12 70
(a) What is the focal length of a given convex lens?
(b) Which set of observations is incorrect and why?
Answer: (a) When the position of the object is at 2F, the image formed at 2F on the end other side of the lens is real, inverted and of same size as that of the object. Therefore,
Object distance = Image distance
2F = 30 cm
F = 15 cm
(b) For the last observation, we find that the object lies between the optical center and focus of the lens, so virtual image is formed and does not fall on the screen. Hence last observation is incorrect.

Question. If we cover one half of a convex lens while focussing on a distant object, in what way will it affect the image formed? Can this method be used to find the approximate focal length of a concave lens?
Answer: A full size image will be obtained but only the intensity or brightness of the image will reduce.
No, because a concave lens always forms a virtual image wherever the object may be.

Question. Explain how a converging lens acts as a magnifying glass. For this purpose, why a lens having a short focal length is chosen instead of lens having a long focal length?
Answer: To use a converging lens or convex lens as a magnifying glass, the object should be placed within its focus to get a magnified and erect image of the object as shown in the figure given. A convex lens of short focal length has a greater magnifying power in comparison to a convex lens of large focal length.
Hence, the object will appear much larger when seen through a convex lens having short focal length.

Question. To find the image-distance for varying object-distances in case of a convex lens, a student obtains on a screen a sharp image of a bright object placed very far from the lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen.
(a) In which direction-towards or away from the lens, does he move the screen to focus the object?
(b) What happens to the size of image—does it increase or decrease?
(c) What happen when he moves the object very close to the lens?
Answer: (a) He moves the screen away from lens to focus the object.
(b) The size of the image increases.
(c) When the object is placed very close to the lens, then no image will be formed on the screen. As now image formed is erect and virtual.

Question. A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?
Answer: (a) The screen should be moved towards the lens to get a sharp image of the object again.
(b) Magnification of the image decreases on moving the object away from the lens.

Question. A student focuses the image of a well illuminated distant object on a screen using a convex lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen by adjusting the lens.
(a) In which direction—towards the screen or away from the screen, does he move the lens?
(b) What happens to the size of the image—does it decrease or increase?
(c) What happens to the image on the screen when he moves the object very close to the lens?

Answer: (a) He should move the lens away from the screen.
(b) Size of the image increases.
(c) No image will be formed on the screen.

Question. In an experiment with a rectangular glass slab, a student observed that a ray of light incident at an angle of 55° with the normal on one face of the slab, after refraction, strikes the opposite face of the slab before emerging out into air making an angle of 40° with the normal. Draw a labelled diagram to show the path of this ray. What value would you assign to the angle of refraction and angle of emergence?
∠i is angle of incidence = 55°
Given, ∠r2 = 40°
∠r1 and ∠r2 are alternate interior angles.
∠r1 = ∠r2 = 40°
So, Angle of refraction = 40°
Since, the emergent ray is parallel to the incident ray, the angle of emergent must be equal to angle of incidence, i.e., –e = –i = 55°.

Question. When a ray of light passes through a glass slab how many times does it change its path and why?
Answer: The ray of light bends twice.
First time when it enters from air to the glass slab, it bends towards the normal, i.e., from rarer medium to denser medium.
Second time, when the ray moves out from the glass slab to air, it bends away from the normal, i.e., it moves from denser medium to rarer medium.

Question : Define the principal focus of concave mirror.
Answer : Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.
Principal focus (p.f.) is a point on principal axis of a concave mirror where the rays parallel to principal axis meet after reflection from the mirror. Additional reading. For convex mirror, principal focus is a point on principal axis of a convex mirror where rays parallel to principal axis appear to diverge from after reflection from the mirror. Question : Write two different uses of concave mirrors.
Answer :  Uses of concave mirrors:
(1) It is used in torches, headlights of vehicles and search-lights to get powerful parallel beams of light.
(2) It is used as a shaving mirror to see a larger image of the face when held close to the face.
(3) It is used by dentists and ENT specialists to see larger images of teeth and inner parts of ear and throat.
(4) Large concave mirrors or parabolic mirrors are used in solar furnaces to concentrate sunlight.

Question : A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent if instead of water we use liquids like kerosene or turpentine? Support your answer with reason.
Answer :  A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water due to the phenomenon of refraction of light. No, the pencil will not appear to be bent to the same extent if instead of water we use liquids like kerosene or turpentine. This is because bending will be different in different liquids, since velocity of light at the interface separating two media depends on the relative refractive index of the medium.

Question : An object 5 cm high is placed at a distance of 2f from a convex lens. What is the height of the image formed?
Answer :  We know that when an object is placed at a distance of 2f in front of a convex lens, then the image formed is of the same size as the object. Hence, the height of the image formed will be 5 cm.

Question : To find the image distance for varying object distances in case of a convex lens of focal length 15 cm, a student obtains on a screen a sharp image of a bright object by placing it at 20 cm distance from the lens. After that he gradually moves the object away from the lens and each time focuses the image on the screen.
(A) In which direction-towards or away from the lens does he move the screen to focus the object?
(B) How does the size of image change?
(C) Approximately at what distance does he obtain the image of magnification -1?
(D) How does the intensity of image change as the object moves farther and farther away from the lens?
Answer :  (A) He move the screen towards the lens.
(B) The size of the image decrease gradually.
(C) Nearly 30 cm from the lens he obtain m = –1.
(D) As the object moves farther and farther away from the lens, intensity of the image gradually increases.

Question : The absolute refractive indices of glass and water are 3/2 and 4/3 respectively. If the speed of light is 2 × 108 m/s, calculate the speed of light in
(i) vacuum, (ii) water.
(i) Given: vg = 2 × 108 m/s (Speed of light in glass)
We know, Absolute Refractive Index of a Medium = Speed of light in Vacuum c. / Speed of light in the Medium Question :  An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
: Question : An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer : Focal length of convex mirror, f= +15 cm
Object distance, u= – 10 cm
According to the mirror formula, Question : Name the type of mirror used (i) by dentists and (ii) in solar furnaces. Give two reasons why such mirrors are used in each case.
Answer : (i) Concave mirrors are used by dentists to see the large images of the teeth of patients because when a tooth is within the focus of a concave mirror, then an enlarged image of the tooth is seen in the concave mirror. Thus, it becomes easier to locate the defect in the tooth.
(ii) Large concave mirrors are used in solar furnaces as reflectors. Solar furnace is placed at the focus of the concave reflector which focusses the Sun’s heat rays on the furnace due to which the solar furnace gets very hot. Even steel can be melted in this solar furnace.

Question : A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer : The ray of light bends towards the normal. When a ray of light enters from an optically rarer medium (having low refractive index) to an optically denser medium (having high refractive index), its speed slows down and it bends towards the normal. Since water is optically denser than air, a ray of light entering from air into water will bend towards the normal.

Question : What is the principle of reversibility of light? Show that the incident ray of light is parallel to the emergent ray of light when light falls obliquely on a side of a rectangular glass slab.
Principle of reversibility of light. Refractive index for light going from medium 1 to medium 2 is equal to the reciprocal of refractive index for light going from medium 2 to medium 1. According to Snell’s law
aη sin i = 8η sin r1 …….(i)
Similarly, 8η sin r2 = aη sin e ……..(ii)
r2 = r1 because glass slab is rectangular and therefore the two normal are parallel.
r1 and r2 are alternate interior angles.
Comparing (i) and (ii), we get
aη sin i = aη sin e
sin i = sin e ⇒ i = e

Question. (a) What is meant by ‘power of a lens’?
(b) State and define the S.I. unit of power of a lens.
(c) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other. Calculate the lens power of this combination.
Answer: (a) The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it.
(b) The SI unit of the power of a lens is dioptre. One dioptre is the power of a lens whose focal length is 1 metre.
(c) Focal length of the convex lens:
f1 = +25 cm = + 25/100 m = +0.25 m
Focal length of the concave lens:
f2 = –10 cm = - 10/100 m = –0.10 m
P1 = 1/f1 = 1/0.25   = +4D;       P2 =   1/f2  = 1/-0.10 = –10D
Power of combination, P = P1 + P2 = 4 + (–10) = –6D (dioptre)

Question. (a) Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens.
(b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate
(i) the distance of the object from the lens
(ii) the magnification for the image formed
(iii) the nature of the image formed 2011OD
Answer: (a) When an object is placed any-where between infinity and optical centre of a concave lens, the image
formed is
(i) between O and F.
(ii) virtual.
(iii) erect.
(iv) diminished.
(b) Concave lens.
Focal length, f = –15 cm v = –10 cm,
u = ?, m = ?
Nature of the image = ?
According to lens formula:
1/f = 1/v - 1/u       ⇒    1/u = 1/v - 1/f      ⇒    1/u = 1/-10  - 1/(-50)
⇒   1/u = -1/10 + 1/15 = -3+2/30  = -1/30    ∴ u = –30 cm
(i) The distance of the object from the lens = 30 cm
(ii) Magnification, m = v/u = -10/-30  = 1/3
(iii) Nature: +ve sign of m shows that the image is erect.
Since the value of m is less than (1), therefore image is diminished.

Question. (a) If the image formed by a lens is diminished in size and erect, for all positions of the object, what type of lens is it?
(b) Name the point on the lens through which a ray of light passes undeviated.
(c) An object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find (i) the position (ii) the magnification and (iii) the nature of the image formed.
Answer: (a) The lens is concave.
(b) Optical centre is the point on the lens through which a ray of light passes undeviated.
(c) Convex lens: Focal length, f = +20 cm Object distance, u = –30 cm
Image distance, v = ? Magnification, m = ?
Nature of the image = ?
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-25 = 1/10 ⇒ 1/v + 1/25 = 1/10
1/v = 1/10 - 1/25      ⇒  1/v = 5-2/50  = 3/50      ⇒     v = 50/3 = +16.6 cm
h2/h1 = v/u       ⇒   h2/5 = 50/3x -50      ∴   h2 = -50 x 5/3 x 25 = -10/3 = –3.3 cm
(iii) Nature of the image:
• The +ve sign of v shows that image is real.
• The –ve sign of h2 shows that image is inverted.
∴ Size of image = 3.3 cm

Question. List the sign conventions for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three times magnified real image of an object placed 16 cm in front of it. 64
Answer: Sign conventions for reflection of light by spherical mirrors:
(i) All the distances are measured from pole of the mirror as origin.
(ii) The object is always placed to the left of the mirror.
(iii) Distances measured in the direction of incident light are considered to be positive (+ve).
(iv) Distances measured against the direction of incident light are considered to be negative (–ve).
(v) The perpendicular distances to the principal axis in the upward direction are considered to be positive (+ve).
(vi) The perpendicular distances to the principal axis in the downward direction are considered to be negative (–ve).
u = object distance v = image distance f = focal length
h = height of the object h¢ = height of the image
P = Principal axis C = Centre of curvature F = Focus
∴  PA, i.e., u = –16
Magnification is 3 times      ∴ m = –3
-v/u = m  ⇒  -v/-16 = –3   ∴  –v = 3 × 16 = 48    ⇒    v = –48 cm
According to mirror formula:
1/f = 1/u + 1/v      ⇒     1/f = 1/-16 + 1/-48     ⇒     1/f = -1/16 - 1/48  =  -3-1 /48 = -4/48   = -1/2
∴ f = –12 cm
The negative (–) sign of focal length shows that it is a concave lens of focal length 12 cm.
• Thus an inverted, magnified image is formed on the screen.

Question. With the help of a ray diagram, state what is meant by refraction of light. State Snell’s law for refraction of light and also express it mathematically.
The refractive index of air with respect to glass is 2/3 and the refractive index of water with respect to air is 4/3. If the speed of light in glass is 2×108 m/s, find the speed of light in (a) air, (b) water.
Answer: Refraction of light. The bending of light when it passes from one medium to another is called refraction of light.
Diagram and Snell’s law. The second law of refraction of light is the Snell’s Law of Refraction. It states that the ratio of sine of the angle of incidence to the sine of angle of refraction is a constant for a given pair of medium.
sin i/sin r = constant (n)
This constant (n) is called refractive index of the medium.
(a) Refractive index of air with respect to glass:
gna = Velocity of light in glass / Velocity of light in air       ⇒    gna = vg/va  = 2/3
∴ vg = 2 × 108 m/s
va = ?           ∴    2 x 108 /va = 2/3
⇒  3 x 2 x 108/va  =  va    ∴   va = 3 × 108 m/s
∴  Velocity of light in air = 3 × 108 m/s
(b) Refractive index of water with respect to air:
anw = va/vw = 4/3        ∴          va = 3 × 108 m/s
⇒  3 x 108 /vw = 4/3       ⇒   3 x 3 x 108 /4 = vw
⇒   vw   = 9/4 x 108    ∴    vw = 2.25 × 108 m/s
∴ Velocity of light in water = 2.25 × 108 m/s

Question. To construct a ray diagram, we use two light rays which are so chosen that it is easy to know their directions after refraction from the convex lens. List these two rays and state the path of these rays after refraction.
Answer: Two rays chosen for refraction are:
• A ray of light parallel to the principal axis.
• A ray of light passing through optical centre of a lens.
Path of these rays after refraction:
• A ray parallel to the principal axis, after refraction through the lens, will pass through the principal focus on the other side of the lens.
• A ray of light passing through optical centre of a lens will emerge from the lens without any deviation in its path.

Question. A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations as given below:
S.No. Distance of flame Distance of the screen
from the lens (cm) from the lens (cm)
1. 60 20
2. 40 24
3. 30 30
4. 24 40
5. 12 70
In which case
(i) is the size of image smaller than the size of object?
(ii) is the size of object and image same?
Answer: (i) When the object is placed between infinity and 2F, diminished, real and inverted image is formed. So in the first observation, size of the image is smaller than the size of object.
(ii) When the object is placed at 2F, an image of the same size as that of the object is formed at 2F on the other side of the lens. It is real and inverted as well. Hence in 3rd observation, the size of object and image is same.

Question. State the law of refraction of light that defines the refractive index of a medium with respect to the other. Express it mathematically. How is refractive index of any medium ‘A’ with respect to a medium ‘B’ related to the speed of propagation of light in two media A and B? State the name of this constant when one medium is vacuum or air. The refractive indices of glass and water with respect to vacuum are 3/2 and 4/3 respectively. If the speed of light in glass is 2×108 m/s, find the speed of light in (i) vacuum, (ii) water.
Answer: The second law of refraction gives a relationship between the angle of incidence and the angle of refraction. This law is also known as Snell’s Law of Refraction.
According to the Snell’s law, “The ratio of sine of angle of incident to the sine of refraction is constant for a given pair of media.”
sin i/sin r = constant. This constant is called refractive index.
Example: A ray of light travelling in air enters into glass and gets refracted
then, sin i/sin r = n
...where sin i = sine of angle of incidence in air
sin = sine of angle of refraction in glass
= refractive index of glass = n
The refractive index of medium ‘A’ with respect to medium ‘B’ is equal to the ratio of speed of light in medium ‘A’ to the speed of light in medium ‘B’.
BnA = Speed of light in medium B/Speed of light in medium A      BnA =  vB/vA
...where BnA = refractive index of A . . . B
vB = velocity of light in medium B
vA = velocity of light in medium A
When light is going from vacuum to another medium, then the value of refractive index is called the absolute refractive index.
nng = = 3/2 ,    nnw =  4/3 , vg = 2 × 108 m/s, vn = ?, vw = ?
Formula: nng = vv/vg
∴  3/2 = vv/2 x 103         ⇒               vv = 3/2 x 2 x 108 = 3 × 108 m/s
∴ Velocity of light in vacuum = 3 × 108 m/s
nnw = vv/vw    ⇒   4/3 = 3 x 108
⇒  rw = 3 x 100 x 106 x 3 /4 = 225 × 106 = 2.25 × 108 m/s
∴ Velocity of light in water = 2.25 × 108 m/s

Question. List the sign conventions that are followed in case of refraction of light through spherical lenses. Draw a diagram and apply these conventions in determining the nature and focal length of a spherical lens which forms a four times magnified real image of an object placed 20 cm from the lens.
Answer: Sign conventions in case of refraction of light through spherical lenses:
(i) All the distances are measured from the optical centre of the lens.
(ii) The distances measured in the same direction as that of incident light are taken as positive (+ve).
(iii) The distances measured in the opposite direction as that of incident light are taken as negative (–ve).
(iv) The perpendicular distances to the principal axis in the upward direction are taken as positive (+ve).
(v) The perpendicular distances to the principal axis in the downward direction are taken as negative (–ve).
u = object distance from O (optical centre), v = image distance from O, f = focal length
These rules conclude that:
(i) ‘u’ (object distance) is always –ve
(ii) ‘f ’ of convex lens is always +ve
(iii) ‘f ’ of concave lens is always –ve
(iv) ‘v’ in case of virtual image is always –ve
(v) ‘v’ in case of real image is always +ve.
(vi) Real image is always inverted thus h2 (size of the image) in case of real image is always –ve.
(vii) h2 (size of the image) in case of virtual image is always +ve as it is always erect.
Nature of the image = ? Focal length of the lens = ?
h2 = – 4h1 ...(Real image is always formed in downward direction)
m = +4 ...(+ve sign for real image, as the image will be formed on the right side of the lens)
u = –20 ...(–ve sign shows the object is placed on the left side of the lens)
m = h2/h1    m =  v/u  ⇒  h2/h1 = v/u      ⇒   -4h1/h1 = v/-20
∴ v = – 4 × –20 = +80 cm

Question. (a) State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in a vacuum.
(b) The absolute refractive indices of two media ‘A’ and ‘B’ are 2.0 and 1.5 respectively. If the speed of light in medium ‘B’ is 2 × 108 m/s, calculate the speed of light in: (i) vacuum, (ii) medium ‘A’.
(i) The first law of refraction of light states that the incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
(ii) The second law of refraction of light is the Snell’s Law of Refraction. It states that the ratio of sine of the angle of incidence to the sine of angle of refraction is a constant for a given pair of medium.
sin i/sin r  = constant (n)
This constant (n) is called refractive index of the medium.
• When the light is going from vacuum to another medium, then the value of refractive index is called the absolute refractive index.
• The ratio of speed of light in vacuum to the speed of light in a medium is called the absolute refractive index of that medium, i.e.,
Absolute refractive index (of a medium) =  Speed of light in vacuum (c )/Speed of light in medium (v )
(b) nA = 2.0; nB = 1.5; vB = 2 × 108 m/s
(i) Speed of light in vacuum, c = ?
nB = 1.5 = c/vB        ∴    c = nBvB = 1.5 × 2 × 108 m/s = 3 × 108 m/s
(ii) For medium ‘A’: nA = vA     ∴   vA = c/nA  = 3 X 108 m/s /2  = 1.5 × 108 m/s

Question. A student places a candle flame at a distance of about 60 cm from a convex lens of focal length 10 cm and focuses the image of the flame on a screen. After that he gradually moves the flame towards the lens and each time focuses the image on the screen.
(a) In which direction-toward or away from the lens, does he move the screen to focus the image?
(b) How does the size of the image change?
(c) How does the intensity of the image change as the flame moves towards the lens?
(d) Approximately for what distance between the flame and the lens, the image formed on the screen is inverted and of the same size?
Answer: (a) He will move the screen away from the lens to focus the image.
(b) Size of the image goes on increasing.
(c) Intensity of image goes on decreasing. (d) About 20 cms.

Question. A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.
(a) In which direction does he move the lens to focus the flame on the screen?
(b) What happens to the size of the image of the flame formed on the screen?
(c) What difference is seen in the intensity (brightness) of the image of the flame on the screen?
(d) What is seen on the screen when the flame is very close (at about 5 cm) to the lens?
Answer: (a) He moves the lens away from the screen.
(b) The size of image formed on the screen goes on increasing.
(c) Brightness of the image of flame goes on decreasing.
(d) No distinct image of flame will be formed on the screen.

Question. (a) Explain the following terms related to spherical lenses:
(i) optical center (ii) centres of curvature (iii) principal axis
(iv) aperture (v) principal focus (vi) focal length
(b) A converging lens has focal length of 12 cm. Calculate at what distance should the object be placed from the lens so that it forms an image at 48 cm on the other side of the lens.
Answer: (a) (i) Optical center. It is a point within the lens that lies on the principal axis through which a ray of light passes undeflected.
(ii) Centre of curvature. The centre of curvature of the surface of a lens is the centre of the sphere of which it forms a part. A lens has two centres of curvature because it has two surfaces.
(iii) Principal axis. It is a line through the centres of curvatures of the lens.
(iv) Aperture. The diameter of the circular boundary of the lens is called the aperture of the lens.
(v) Principal focus. A beam of light parallel to the principal axis either converges to a point or appears to diverge from a point on the principal axis after refraction through the lens, is called the principal focus. All lenses have two principal focuses.
(vi) Focal length. The distance between the optical centre and the principal focus of the lens is called its focal length.
(b) A converging lens is a convex lens.
f = +12 cm; u = ?; v = +48 cm (+ve as it is formed on other side of the object)
According to lens formula:
1/f = 1/v - 1/u      ⇒   1/12 = 1/48 - 1/u       ⇒     1/u = 1/48 - 1/12 = 1-4/48 = -3/48 = 1/16
∴ u = –16 cm