Please refer to below Exam Question for Class 10 Science Chapter 10 Light Reflection and Refraction. These questions and answers have been prepared by expert Class 10 Science teachers based on the latest NCERT Book for Class 10 Science and examination guidelines issued by CBSE, NCERT, and KVS. We have provided Class 10 Science exam questions for all chapters in your textbooks. You will be able to easily learn problems and solutions which are expected to come in the upcoming class tests and exams for standard 10th.

Chapter 10 Light Reflection and Refraction Class 10 Science Exam Question

All questions and answers provided below for Exam Question Class 10 Science Chapter 10 Light Reflection and Refraction are very important and should be revised daily.

Exam Question Class 10 Science Chapter 10 Light Reflection and Refraction

Very Short Answer Type Questions

Question: Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence?
Answer: When two plane mirrors are placed at right angle with each other, then the incident ray and reflected ray will always be parallel to each other, irrespective of the angle of incidence.

Question: What is the name of the phenomenon in which the right side of an object appears
to be the left side of the image in a plane mirror?
Answer: Lateral inversion is the phenomenon in which the right side of an object appears to be the left side in its image in a plane mirror.

Question: What is the power of a converging lens having a focal length of 20 cm.
Answer: As convex lenses are converging lenses and it is given that Focal length = + 20 cm = + 0.2 m, as  per the new sign convention.
As Power = 1/f (in m)
Therefore, P = +1/0.2 D=+5D 

Question: If a magnification of -1 is to be obtained using a concave mirror of focal length 6 cm, then what should be the position of the object?
Answer: Magnification of -1 means that the image is real and size of image is equal to the size of the object. This is possible when the object is placed at C of a concave mirror. Therefore, position of object = C = 2f = 12 cm in front of the concave mirror.

Question: A man stands 10 m away in front of a large plane mirror. How far must he walk before he is 5 m away from his image?
Answer: As the distance between the man and the mirror is 10 m, the distance between man and his image is 10 + 10 or 20 m, as image distance and object distance are equal. Distanc between the man and his image is 5 m when the man is 2.5m away from the mirror. Therefore, the man has to walk 10 m – 2.5m = 7.5 m towards the mirror.

Question: Which type of mirror should a student use if he wants to project the image of a candle flame on the walls of the school laboratory?
Answer: To project the image of a candle flame on the walls of the school laboratory, he should use a concave mirror as only a concave mirror forms a real image (which can be obtained on a screen) for all positions of object except when the object is placed between the pole and focus of the mirror. A convex mirror and plane mirror,
on the other hand, always form virtual images..

Question: Find the focal length of a convex mirror whose radius of curvature is 32 cm. 
Answer: Radius of curvature (R) = 32 cm
Radius of curvature = 2 x Focal length (f)
f = R/2=32/2=16cm
Therefore, convex mirror’s focal length is 16 cm.

Question: Observe the given figure. What is the angle of reflection?

Answer: The angle of reflection is the angle between the normal and the reflected ray.
Therefore: 90 – 70 = 20⁰

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 se  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: An object O is placed in front of a plane mirror as shown below. 
At which point is the image of the object ‘O’ formed?

Answer: At point R, the image of object ‘O’ is formed.

Short Answer Type Questions: 

Question. At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side. What will be the magnification produced in this case? 
Answer: Focal length of convex lens, f = +18 cm
Image distance from the lens, v = +24 cm
By using lens formula,   

Question. The image of an object placed at 60 cm in front of a lens is obtained on a screen at a distance of 120 cm from it. Find the focal length of the lens. What would be the height of the image if the object is 5 cm high? 
Answer: Since the image can be obtained on a screen so image is real and real image is formed by convex lens so the lens in convex lens.
Given,
Object distance, u = –60 cm
Image distance, v = +120 cm
Using lens formula,     

Question. For which position of the object does a convex lens form a virtual and erect image? Explainwith the help of a ray diagram.
Answer: When an object is placed between F1 and optical centre, O of a convex lens, it forms a virtual and erect image. The ray diagram for this situation can be drawn as follows:

Question. State the laws of refraction of light. If the speed of light in vacuum is 3 × 108 m/s, find the absolute refractive index of a medium in which light travels with a speed of 1.4 × 108 m/s. 
Answer: The following are the laws of refraction of light
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media.
This law is also known as Snell’s law of refraction. sin i/sin r = constant
where i is the angle of incidence and r is the angle
of refraction.
The speed of light in vacuum = 3 × 108 m/s
The speed of light in a medium = 1.4 × 108 m/s
∴ Absolute refractive index
= Speed of light in vacuum/Speed of light in a medium
n = 3 x 108 m/s/1.4 x 108 m/s = 2.14

Question. The image of a candle flame placed at a distance of 40 cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 40 cm from the lens. Identify the type of lens and write its focal length. What will be the nature of the image formed if the candle flame is shiffed 25 cm towards the lens? Draw a ray diagram to justify your answer. 
Answer: Given : u = –40 cm, v = 40 cm
1/f = 1/40 + 1/40 = 2/40 ⇒ f = 20 cm
Type of lens : Convex lens
Focal length = 20 cm
Nature of the image will be virtual and erect if the candle flame is shiffed 25 cm towards the lens.   

Question. An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.
Answer: Focal length of given concave lens, f = – 5 cm
Distance, u = –10 cm, object size, h = 6 cm
Image distance, v = ?     

Question. Draw ray diagram to show the path of the refracted ray in each of the following cases.
A ray of light incident on a concave lens
(i) is parallel to its principal axis, (ii) is passing through its optical centre and (iii) is directed towards its principal focus. 
Answer: (i) A ray of light incident on a concave lens is parallel to its principal axis, the diagram can be drawn as follows :     

Question. What is the principle of reversibility of light?
Show that the incident of light is parallel to the emergent ray of light when light falls obliquely on a side of a rectangular glass slab.
Answer: Principle of reversibility of light states that the light will follow exactly the same path if the direction is reversed.
Using Snell’s law of refraction,
sin i/sin r = sin e/sin r2
Since r1 = r2, so i = e     

So, PQ is parallel to RS.
So, we conclude that incident ray is parallel to the emergent ray.

Question. What is understood by lateral displacement of light? Illustrate it with the help of a diagram. List any two factors on which the lateral displacement in a particular substance depends.
Answer: When a ray of light is incident obliquely on a parallel sided glass slab, the emergent ray shifts laterally. The perpendicular distance between the direction of the incident ray and emergent ray is called lateral shift.
The factors on which the lateral displacement depends are:
(i) thickness of the refracting material.
(ii) the refractive index of the material.

Question. Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follows, without doing any calculations : 

(a) What is the focal length of the convex lens? Give reason in support of your answer.
(b) Write the serial number of that observation which is not correct. How did you arrive at this conclusion?
(c) Take an appropriate scale to draw ray diagram for the observation at S. No. 4 and the approximate value of magnification.
Answer: (a) When an object placed at the 2F from a convex lens, then its image is formed on the other side of the lens at the same distance from the lens.
Thus from S.No.(3) we can say that.
∴ f = v / 2⇒ f = 30/2 = +15 cm
Thus, the focal length is + 15 cm.
(b) In this case S.No. (6) is incorrect as the object distance is between focus and pole, for such case, the image formed is virtual and on the same side as the object, hence image distance is negative.
(c) The approximate value of magnification for object distance –20 cm and image distance + 60 cm is –3. 

Question. Analyse the following observation table showing variation of image-distance (v) with object-distance (u) in case of a convex lens and answer the questions that follow without doing any calculations.   97
(a) What is the focal length of the convex lens? Give reason to justify your answer.
(b) Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion?
(c) Select an appropriate scale and draw a ray diagram for the observation at S. No. 2. Also find the approximate value of magnification.
Answer: (a) When an object is placed at 2F from the convex lens, then its image is formed on the other side of the lens at the same distance from the lens.
Thus from S.No. (3), we can say that
∴ f = v/2 = 40/2 = 20cm
(b) In this case, S.No. (6) is incorrect as the object distance is between focus and optical centre for such cases, the image formed is virtual and image distance is negative.   

The approximate value of magnication for object distance –60 cm and image distance +30 cm is –1/2.

Question. (a) Define focal length of a spherical lens.
(b) A divergent lens has a focal length of 30 cm. At what distance should an object of height 5 cm from the optical centre of the lens be placed so that its image is formed 15 cm away from the lens? Find the size of the image also?
(c) Draw a ray diagram to show the formation of image in the above situation.
Answer: (a) Distance between the optical centre and the focus of the lens is known as the focal length of the lens.
(b) Given : f = – 30 cm, v = – 15 cm, h = 5 cm
From the lens formula,   

Question. (a) Define optical centre of a spherical lens.
(b) A divergent lens has a focal length of 20 cm. At what distance should an object of height 4 cm from the optical centre of the lens be placed so that its image is formed 10 cm away from the lens. Find the size of the image also.
(c) Draw a ray diagram to show the formation of the image in above situation.
Answer: (a) Optical centre is the central part of the lens through which a ray of light passes without suffering any deviation. It is usually represented by the letter O.
(b) Given : f = – 20 cm, h = 4 cm, v = – 10 cm
From lens formula, 1/v – 1/u = 1/f   

Question. (a) Define focal length of a divergent lens.
(b) A divergent lens has a focal length of 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed.
(c) Draw a ray diagram to show the formation of image in the above situation.
Answer: (a) Distance between the principal focus and the optical centre is known as the focal length of the lens.
(b) Given, f = –30 cm, v = –15 cm, h = 6 cm
Now, from lens formula,   

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 three times magnified real image of an object placed 16 cm from the lens. 
Answer: For lenses, we follow sign convention, similar to one used for spherical mirrors. We apply the rules for signs of distances, except that all measurements are taken from the optical centre of the lens. 

Question. (a) Under what condition with a glass lens placed in a transparent liquid become invisible.
(b) Describe and illustrate with a diagram, how we should arrange two converging lenses so that a parallel beam of light entering one lens emerges as a parallel beam after passing through the second lens.
(c) An object is placed at a distance of 3 cm from a concave lens of focal length 12 cm. Find the (i) position and (ii) nature of the image formed.
Answer: (a) If the refractive index of glass lens is equal to the refractive index of liquid then of the glass lens is placed in a transparent liquid will become invisible.
(b) Suppose we have two converging lens of focal lengths f1 and f2. We will keep the two converging lens at a distance of f1 + f2 so that a parallel beam of light entering one lens emerges as a parallel beam after passing through the second lens. 

Here the focus of the two lenses should coincide.
(c) (i) Focal length of concave lens, f = –12 cm
Object distance, u = –3 cm
Image distance, v = ?
Using lens formula,   

Question. At what distance from a concave lens of focal length 20 cm a 6 cm tall object be placed so as to obtain its image at 15 cm from the lens?
Also calculate the size of the image formed.
Draw a ray diagram to justify your answer for the above situation and label it.
Answer: Focal length of concave lens, f = –20 cm
Height of the object, h = 6 cm
Image distance, v = –15 cm
From lens formula,     

Question. At what distance from a concave lens of focal length 25 cm a 10 cm tall object be placed so as to obtain its image at 20 cm from the lens.
Also calculate the size of the image formed.
Draw a ray diagram to justify your answer for the above situation and label it.
Answer: Focal length of concave lens , f = – 25 cm
Image distance, v = –20 cm
Height of the object, h = 10 cm
Now, from lens formula,   

Question. If the image formed by a lens for all positions of the object placed in front of it is always virtual, erect and diminished, state the type of the lens. Draw a ray diagram in support of your answer. If the numerical value of focal length of such a lens is 20 cm, find its power in new cartesian sign conventions.
Answer: Concave lens always forms virtual, erect and diminished image for all positions of the object. 

Focal length of the concave lens,
f = –20 cm = -20/100 cm
Power of the lens, p = 1/f(in m) = -100/20m = 5D

Question. (a) In the above ray diagram mark the object distance (u) and the image distance (v) with their proper sign (+ve or –ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the convex lens in this case.
(b) Find the power of a convex lens which forms a real and inverted image of magnification –1 of an object placed at a distance of 20 cm from its optical centre.
Answer: (a) The lens formula is given as   

Question. What is meant by power of a lens? Define its S.I. unit.
You have two lenses A and B of focal lengths +10 and –10 cm respectively. State the nature and power of each lens. Which of the two lenses will form a virtual and magnified image of an object placed 8 cm from the lens?
Draw a ray diagram to justify your answer.
Answer: Given that :
Focal length of lens A, fA = +10 cm
Focal length of lens B, fB = –10 cm
Lens A is convex lens . Lens B is concave lens,       

Question. One half of a convex lens of focal length 10 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens?
Draw a ray diagram to justify your answer.
A 4 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image.
Answer: Yes, the lens will produce an image of complete object.   

Nature of image: Virtual, erect and enlarged Position of image : In front of lens 60 cm from the lens
Size of image : Four times of object, i.e., (4 × 4) cm = 16 cm

Question. What is meant by the power of a lens ? What is its S.I. unit ? Name the type of lens whose power is positive.
The image of an object formed by a lens is real, inverted and of the same size as the object. If the image is at a distance of 40 cm from the lens, what is the nature and power of the lens? Draw ray diagram to justify your answer. 
Answer: Convex lens has positive power. Since the image of an object formed by a lens is real, inverted and of the same size as the object.
Given: Size of object = Size of image and h’= –h      

Question. (a) Explain the following terms related to spherical lenses:
(i) optical centre
(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 centre : The centre point of a lens is known as the optical centre. It always lies inside the lens. A light beam passing through the optical centre without any deviation.
(ii) Centre of curvature : It is defined as the centre of the sphere of which the lens is originally a part of. Because the spherical lens consists of two spherical surfaces, the lens has two centre of curvature.
(iii) Principal axis : A straight line passing through the optical centre and principal focus of a spherical lens. This line is called the principal axis.
(iv) Aperture : The diameter of the reflecting surface of spherical lens is called its aperture.
(v) Principal focus : A number of rays parallel to the principal axis are falling on a lens.
These rays, after refraction from the lens, are appearing to converge to or diverge from a point on the principal axis. This point on the principal axis is called the principal focus of the lens.
(vi) Focal length : The distance between the optical centre and the principal focus of a spherical lens is called the focal length. It is represented by the letter f.
(b) Given f = 12 cm, v = 48 cm, u = ?     

Question. Explain the following terms related to spherical lenses
(a) Centres of curvature
(b) Principal axis
(c) Optical centre (d) Principal focus
At what distance from a concave lens of focal length 20 cm, should a 6 cm tall object be placed so that it forms an image at 15 cm from the lens? Also determine the size of the image formed. 
Answer: Given, f = – 20 cm, v = –15 cm, h = 6 cm     

Question. (a) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it.
(b) In the above diagram mark the object distance (u) and the image distance (v) with their proper sign (+ve or –ve as per the new Cartesian sign convention) and state how these distances are related to the focal length ( f ) of the concave lens in this case.
(c) Find the nature and power of a lens which forms a real and inverted image of magnification –1 at a distance of 40 cm from its optical centre.
Answer: (a)   

Question. Draw ray diagrams to show the formation of three times magnifled (a) real, diagram.
Answer:    

Question. A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula. 
Answer: Given that : u = –16 cm, f = 24 cm, h = 4 cm     

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: (a) When travelling obliquely from one medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light.
The ratio of sine of angle of incidence to the sine of the angle of refraction is constant, for the light of a given colour and for the given pair of media.

Question. The image of an object formed by a lens is of magnification –1. If the distance between the object and its image is 60 cm, what is the focal length of the lens? If the object is moved 20 cm towards the lens, where would the image be formed? State reason and also draw a ray diagram in support of your answer. 
Answer: Magnification of –1 indicates that the image is real and inverted and is of the same size as of the object. The object must be at 2f and image also at 2f on the other side.
Total distance between image and object Also 4f = 60 cm ⇒ f = 15 cm If object is moved 20 cm towards the lens, then the object will be between focus and optical
centre of the lens and image formed will be virtual and erect and on the same side of the lens.     

Question. “A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it”. Draw ray diagram to justify this statement stating the position of the object with respect to the lens in each case.
An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.
Answer: Magnified erect image:   

Question. The image of a candle flame placed at a distance of 30 cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 60 cm from the optical centre of the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 3 cm, find the height of its image.
Answer: Given that u = –30 cm, v = 60 cm, h = 3 cm
Lens Formula :     

Question. (a) If the refractive index of glass for light going from air to glass is 3/2, find the refractive index of air for light going from glass to air. 
Answer: (b) Refractive index of glass w.r.t air is
gna = 3/2
Now, refractive index of air w.r.t glass will be 
ang = 1/gna = 1/|(3/2) = 2/3

Question. What is meant by power of a lens? Name and define its S.I. unit.
One student uses a lens of focal length +50 cm and another of –50 cm. State the nature and find the power of each lens. Which of the two lenses will always give a virtual and diminished image irrespective of the position of the object?
Answer: A convex lens has the focal length +50 cm. 
∴ power = 1/f = + 100/50 = + 2D
A concave lens has the focal length –50 cm. 
∴ power = 1/f  = – 100/50 = – 2D
Concave lens always gives a virtual, erect and diminished image irrespective of the position of the object.

Question. The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.
Answer: Given that u = –45 cm, v = +90 cm, h = 2 cm (as the image is formed on the screen, the image is real and hence image is formed by convex lens on the other side of the lens).
Type of the lens used : Convex lens     

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 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 
Answer: (a) Laws of refraction of light :
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point incidence, all lie in the same plane.
(ii) The ratio of sine of angle of incidence to the sine of the angle of refraction is constant, for the light of a given colour and for the given pair of media.
This law is also known as Snell’s law of refraction.
sin i/sin r = constant,
where i is the angle of incidence and r is the angle of refraction.
This constant value is called refractive index of the second medium with respect to the first.
⇒ constant = n21 = v1/v2 ∴ sin i/sin r = v1/v2
If n is the absolute refractive index of the medium, c is the velocity of light in vacuum and v is the speed of light in given medium, then n = c/v.
(b) Given that nA = 2.0, nB = 1.5, vB = 2 × 108 m/s           

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: 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: Draw ray diagrams for the following cases when a ray of light:
(A) passing through centre of curvature of a concave mirror is incident on it.
(B) parallel to principal axis is incident on convex mirror.
(C) is passing through focus of a concave mirror incident on it. 
Answer:

Question:  Why does a light ray incident on a rectangular glass slab immersed in any medium emerges
parallel to itself? Explain using a diagram.
Answer: When a ray of light is incident on a rectangularglass slab immersed in a medium, the emergent
ray comes out parallel to the incident ray.
This is because the extent of bending of the Light – Reflection and Refraction 243 ray of light at the opposite parallel faces (air-glass interface) and (glass-air interface) at the rectangular glass slab is equal and opposite.
The path of the light ray is as shown:  

In the given figure, EO is the incident ray, OO’is the refracted ray and O’H is the emergent ray. The light ray at point O has entered from a rarer medium to a denser medium from air to glass and so the light ray has bent towards the normal. At O’, the light ray has entered from the glass to the air from a denser medium to a rarer medium. The light here has bent away from the normal. The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray.
Hence, a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself.

Question: Under what condition in an arrangement of two plane mirrors, will the incident ray
and the reflected ray always be parallel to each other, whatever may be the angle of incidence? Show the same with the help of diagram. 
Answer: When two plane mirrors are placed at right angle i.e., at ??0?? to each other, then the incident ray and the reflected ray will always be parallel to each other, irrespective of the angle of incidence. 

Question: A convex mirror used on a bus has a focal length of 200 cm. If a scooter is located at 100 cm. from this mirror, find the position, nature and magnification of the image formed in the mirror. 
Answer: We know;
focal length ‘f’ = 200 cm
object distance = –100 cm
Since the object (scooter) is placed to the left of the mirror, it is always taken with a negative sign.  

Thus, the position of the image is 66.67 cm behind the convex mirror. Since the image is formed behind the convex mirror, its nature will be virtual and erect.

Question: An object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm. Write four characteristics of the image formed by the mirror. 
Answer: The four characteristics of image formed:
(1) Image formed by a plane mirror is always virtual and erect.
(2) The size of image is equal to that of the object.
(3) The image formed is as far behind the mirror as the object is in front of it.
(4) The image is laterally inverted.

Question: Draw a ray diagram showing the path of the rays of light when it enters with oblique incidence (1) from air into water and (2) from water into air. 
Answer: (1) Ray diagram showing the path of rays of light when it enters with oblique incidence from air into water.  

(2) Ray diagram showing the path of rays of light when it enters with oblique incidence from water into air. 

Question: (A) For the same angle of incidence of 45⁰, the refraction angle in two transparent media P and Q is 20?? and 30?? respectively. Which of the two is optically denser and why?
(B) Define 1 dioptre power of a lens.
(C) find the focal length of a lens of power +0.5 D. 
Answer: (A) Medium P is optically denser than medium
Q. 

For given angle of incidence (i), V will be more in medium Q. as angle of refraction ∠r is more than of medium P. Thus the speed of light is less in medium P and the bending of ray towards the normal will be more than the medium Q. Hence,medium P is optically denser than medium Q. We can also say since speed of light (V) is less in medium P, refractive index will be more and medium P’ will be optically denser.
(B) 1 Diopter Power of a lens: 1 Dioptre is the power of a lens whose focal length is 1 metre. It is thus a unit of reciprocal length 1 D = 1 m^–1

Question: A concave mirror is used for image formation for different positions of an object. What inferences can be drawn about the following when an object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 10 cm?
(A) Position of the image
(B) Size of the image
(C) Nature of the image
Draw a labelled ray diagram to justify your inferences. 
Answer: Object distance u = –10 cm
focal length of concave mirror f = –15 cm
Position of image v = ?
using mirror formula  

Question: Draw ray diagram in each of the following cases to show what happens after reflection to the incident ray when:
(A) it is parallel to the principal axis and falling on a convex mirror.
(B) it is falling on a concave mirror while passing through its principal focus.
(C) it is coming oblique to the principal axis and falling on the pole of a convex mirror.
Answer: (A) An incident ray parallel to the principal axis,after reflection appear to diverge from the principal focus is case of convex mirror.
Explanation: A straight line XP is principal axis. AB is incident ray which is parallel to principal axis X P of a convex mirror MM’.The ray of light gets reflected at point B on the mirror and goes in the direction B D and it appears to be coming from the focus F of convex mirror. According to the laws of reflection, ∠i = ∠r. Therefore, the incident and reflected rays make equal angles with the principal axis.
(B) An incident ray passing through the principal focus of a concave mirror, after reflection,will emerge parallel to the principal axis. 80
(C) An incident ray coming obliquely to the principal axis towards a point (Pole of the mirror) on the convex mirror is reflected obliquely.

Question: The refractive index of a medium ‘x‘ with respect to medium ‘y‘ is 2/3 and the refractive index of medium ‘y‘ with respect to medium ‘z‘ is 4/3 .find the refractive index of medium ‘z‘ with respect to medium ‘x‘. If the speed of light in medium ‘x‘ is 3 ?? 10?? ms??1, calculate
the speed of light in medium ‘y‘. 
Answer: The refractive index of medium 2 with respect to medium 1 is given by the ratio of the speed of light in medium 1 and the speed of light in medium 2.
Given Refractive index of medium ‘x‘ with respect to medium ‘y‘ = 2/3

n =Speed of light in medium y/Speed of light is medium x =2/3 …(i)
Refractive index of medium ‘y‘ with respect to medium ‘z‘ = 4/3

x_yz = Speed of light in medium z/Speed of light in medium y=4/3 …(ii) To calculate
Refractive index of medium ‘z‘ with respect to medium ‘X‘ = ?
nzx =Speed of light in mediumx/Speed of light in medium Z

From (i) and (ii) equations.  

Hence, refractive index of medium ‘z‘ with respect to medium ‘x‘ is 9/8.
speed of light in medium ‘x‘ = 3×10⁸  m/s speed of light in medium ‘y‘ = ?
n_x = Speed of light in air/Speed of light in medium X Or C/VX
n_y = Speed of light in air /Speed of light in medium Y = C/Vy
Refractive index of medium ‘x‘ with respect to ‘y‘ = 2/3 

speed of light in medium x = 3 X 10⁸ m sec–1 substituting the value we get y = 2X 3 X10/3 =2X10⁸ M sec–¹

Question: Write the relationship between absolute refractive index of a medium and the speed of light in vacuum. If the speed of light in vacuum is x 10⁸ m/s, find the absolute refractive index of the medium in which light travels with a speed of 1.4X10⁸  m/s.
Answer: The absolute refractive index (n) of a medium is the ratio of speed of light in vacuum (c) to the speed of light in that medium (v).
It can be expressed mathematically as 
n = Speed of light vacuum (c)/Speed of light in the medium (v)
n=c/v⇒ n=3×10⁸/1.4×10⁸=3/1.4=2.14
Refractive index of the medium = 2.14

Question: Redraw the following diagram on your answer-sheet and show the path of the
reflected ray. Also mark the angle of incidence (∠i) and the angle of reflection (∠r) on the diagram.
Answer: Path of reflected ray is shown below: 

Angle of incidence (i) and angle of reflection (r) are marked in the figure.

Question:Trace the path of a ray of light incident at  an angle of 45⁰ on a rectangular glass slab.
Write the measure of the angle of refraction, the angle of emergence and the lateral displacement suffered by the ray as it passes through the slab. 
Answer: Ray diagram when light is incident at an angle of 4??0 on a glass slab is given below: 

Angle of emergence = < e = ∠45°
Angle of refraction will be less than 45^o
Lateral displacement: It depends on the thickness of rectangular glass slab.
Explanation:
(1) When a light ray travels from air to glass,the angle of incidence is greater than angle of refraction as ray bends towards normal.
(2) When a light ray travels from glass to air, the angle of refraction also called angle of emergence is greater than the angle of incidence ray of light bends away from the normal.
(3) The angle of emergence will be equal to the angle of incidence. ∠i = ∠c
(4) Emergent ray is parallel to the incident ray along the original direction but it will be laterally displaced to the left of the incident ray.

Question: An object of height 2.0 cm is placed on the principal axis of a convex lens of focal length 10 cm. The distance of the object from the optical centre (O) of the lens is 20 cm. Choose a proper scale and draw a ray diagram to find the position, nature and size of the image.
Also measure and write the size of the image and its distance from the optical centre of the lens. 
Answer: The height of object AB is 2 cm. Focal length of the convex lens, f = 10 cm and object distance, u = – 20 cm.
As Object is at 2F₁, image will be formed at 2F₂. Scale used: X-axis: 10 small divisions = 2 cm
Y-axis: 10 small divisions = 2 cm Ray diagram showing formation of image is drawn below: 

So, image distance = + 20 cm. (plus sign shows image is formed behind the lens)
Position of image = 20 cm behind the lens
Height of image = –2 cm. (minus sign shows image is inverted)
Nature of image: Real and inverted

Question: List four characteristics of the images formed by plane mirrors. 
Answer: Characteristics of the images formed by plane mirrors are:
(1) The image formed by a plane mirror is virtual and erect.
(2) The image is of the same size as the object.
(3) It is formed as far behind the mirror as the object is in front of it.
(4) The image is laterally inverted (inverted sideways).

Question: Study the following ray diagram and state any two mistakes committed by the student while tracing it. Rectify these mistakes and draw the correct ray diagram to show the real position of the image corresponding to the position of the object. In the question write O as writen in the answer. 

Answer: The mistakes committed by the student while tracing the ray diagram are:
(1) The distance from optical center to F₁ is not equal to distance from optical center to F₂. Both distances OF₁ and OF₂ should be equal.
(2) OF₁ = 1/2 O₂F₁ and OF₂ = 1/2 O₂F₂ which shows that F₁ and F₂ are not at the centre of O_2F1 and O₂F₂ respectively.
(3) Since the object lies between F₁ and 2F₁, image should be magnified and lie beyond ₂F₂.
Correct ray diagram is drawn below: 

Question:  A student places a candle flame at a distance of about ??0 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) As the student moves the flame towards the lens, the image is formed farther away from the focus. So, he has to move the screen away from the lens from Focus to 2 F and beyond, to focus the image. However, when object distance is less than 10 cm, no real image is formed.
(B) The size of the image gradually increases from point size to very large size.
(C) The intensity of image formed gradually decreases.
(D) Image formed on the screen will be inverted and of the same size as object when object is placed at 2F, i.e. at 20 cm (since f = 10 cm).

Question:  A 6 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 15 cm and its sharp image is formed at a distance of 30 cm from it on a screen placed on the other side of the lens.
If the object is now moved a little away from the lens, in which direction (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: Given, u = – 15 cm
                       v = 30 cm
height object = 6 cm
Using lens formula 

When u = –15 cm, the object is placed between f and 2f and the image is formed beyond 2 f on
the other side. This a enlarged inverted image.
If the object is moved away from the lens, the image moves towards the lens, So the screen
should be moved towards the lens.
∴ magnification, m = v/u = 30 /-15 = – 2
If the denominator increases, V decreases and hence the magnification decreases. So, the image is smaller than the previous image.

Question: While tracing the path of a ray of light passing through a rectangular glass slab, a
student tabulated his observations as given below: 

The correct observations are:
(A) (I) and (II) (B) (II) and (III)
(C) (III) and (IV) (D) (I), (III) and (IV)
Answer: (C) III and IV
As we know that, in a rectangular glass slab i = e, and refractive index of glass is 1.5.
So, sini/sinx=m
sini/sinx=1.5
sin 40⁰/sin 26⁰=1.5
0.642788/0.438371=1.5
1.4663105 = 1.5
1.5 = 1.5 (LHS = RHS)
again  sin30/sin20=1.5
0. 5/0.34202=1.5
1.505=1.5
1.5=1.5 (LHS = RHS) 

Question:  List two properties of the images formed by convex mirrors. Draw ray diagram in support of your answer. 
Answer: Following are the properties of convex minor.
(1) Image is virtual
(2) Image is Erect.
(3) Image is smaller than object (diminished) 

Question: Which type of lens has a negative power? The power of a lens is + ??D. ??ind its focal length in meters.
Answer: Concave lens has negative Power
Power of lens (P) =1/Focal length (f ) in m+ 5D =1/f(m)
f (in meter) =1+ 5= + 0.2 m.

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:  (A) A lens has a power of -4 D. What is its focal length?
(B) An object is kept at a distance of 100 cm from the above lens. Calculate the image distance and the magnification.
Answer: (A) Focal length is the reciprocal of the power in metres. 

Question:  It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm.
(A) What should be the range of the object distance in the above case?
(B) Will the image be smaller or larger than the object? Draw a ray diagram to show the formation of image in this case.
(C) Where will the image of this object be, if it is placed 2?? cm in front of the mirror?
Answer: (A) Range of distance should be 0 cm to < 12 cm.
(B) The image will larger than the object. 

(0.5 mark to be deducted if no arrows marked or wrongly marked arrows)
(C) Image will be at 24 cm in front of the mirror or the image is formed at C.

Question: A child is standing in front of a magic mirror.
She finds the image of her head bigger, the middle portion of her body of the same size
and that of the legs smaller. What will be the order of combinations for the magic mirror from the top?
Answer: A child is standing in front of a magic mirror.
Her head appears large. This means concave mirror of large focal length is used at the top.
Her middle body appears of the same size in the mirror. It means plane mirror is used in the middle.
Her legs appear smaller. It means convex mirror is used below plane mirror. 

Question:  Which spherical mirror is used as the following
and why?
(A) Shaving mirror
(B) Rear view mirror in vehicles
(C) Reflectors in search lights
Answer: The spherical mirror used in:
(A) Shaving mirror: Concave mirror. This is because when an object is placed close to a concave mirror, an erect and enlarged image of the object is formed.
Explanation: It means than object is placed  between P (Pole) and F (Focus) of a concave mirror where image formed is enlarged,virtual, erect and behind the mirror.
(B) Rear view mirror in vehicles: Convex mirror. This is because this mirror forms an erect and diminished image of the object behind the vehicle giving a wide field of view. It means the driver of a vehicle can see the traffcoverlargeareabehindhisvehicle.
(C) Reflectors in search lights: Concave mirror. This is because the beam of light after reflecting from the concave mirror emerges as a parallel beam of light which illuminates the road ahead of the vehicle.

Question: The absolute refractive indices of glass and
water are 4/3 and 3/2 respectively. If the speed of light in glass is 2 x 10^₈ m/s, calculate the speed of light in (i) vacuum, (ii) water.
Answer:

Question: Which type of phenomenon (reflection or refraction) occurs when light falls on:
(A) a highly polished surface like a mirror and 
(B) a transparent medium like glass or water
(C) Name the physical quantity which remains the same when light goes from one medium to another medium
Answer: (A) When light falls on a highly polished surface like mirror, reflection occurs.
(B) Refraction of light occurs when light falls on a transparent medium like glass or water.
When there is difference in the density of the medium, a ray tends to bend towards or away from the normal at the interface of media.
(C) Frequency of light remains the same when light goes from one medium to another medium.

Question:  Draw a ray diagram showing the path of rays of light when it enters with oblique
incidence: (A) from air into water (B) from water into air
Answer: (A) From air into water: When ray of light enters with oblique incidence from air into water,then it goes from optical rarer medium to optical denser medium and velocity of light decreases which in turn bends the incident light towards the normal. Here i > r
(B) From water into air: When ray of light enters with oblique incidence from water into air, then it goes from optical denser medium to optical rarer medium and velocity of light increases which in turn bends the incident light away from the normal. Here i < r.

Question:  Draw ray diagrams showing the image formation by a convex mirror when an object is placed:
(A) at infinity
(B) at finite distance from the mirror
Answer: (A) The ray diagram showing the image formation when the object is placed at
infinity is as follows: 

(B) The ray diagram showing the image formation when the object is placed at finite
distance from the mirror is as follows: 

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 moves the lens towards the screen.
(B) The size of the image is increase.
(C) When he moves the object very close to the lens, then no image will be formed on the screen.

Question: A student wants to project the image of a candle flame on a screen ??0cm in front of a mirror by keeping the flame at a distance of 15cm from its pole.
(A) State the type of the mirror.
(B) Find the linear magnification of the image produced.
(C)Find the distance between the object and its image.
Answer: (A) Type of mirror: He should use concave mirror as it forms a real image on the same side of the mirror.
(B) Flame is placed at 1??cm distance from pole, so object distance, u = –15 cm Screen is placed 60 cm distance from its pole, so Image distance, v = –60 cm Now, magnification = image distance/object distance
= – v/u = – 60−15
= –4 cm
Hence, linear magnification of the image is –4.
(C) Distance between the object and its image = distance of screen from pole – distance of object from pole =60cm – 15cm = 45cm

Question: Lenses play a very important part in our every dry life lenses are used in making
microscopes, teleescopes fim prrojectors,spectacles, cameras etc. Athrav saw a watch
maker using a lens to see the extremely small parts of a watch clearly. Eailier he had seen a palmist using, a magnifying glass for seeing the lines of his mothers hand. He was curiout to know what type of these lenses are.
Looking at his inquistiveness, his father gave him two different kind ot lenses and asked him to identify/distinguish the lenses without touching them.
Answer the questions (A) and (B) based on the aove case study and related stuided concepts.
(A) How will you distinguish between a convex and concave lens without touching them?
(B) 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 observation.
Answer: (A) We keep the lens close to the page of a book and see the image of the writing of the book through it. It the letters of the book appear entarged, then it is a convex lens and if the letters appear diminished. then it is a concave lens.
Explanation:
(1) The object produces an enlarged image when it a within the focus of a convex lens.
(2) A concave lens always produces a diministed image for all positions of the object.
(B) One-half a convex lens covered with a black paper, forms a complete image of the object kept in front of it.
Experiment:
(1) Take a convex lens, cover it half using a black paper.
(2) Keep a burining candle in front of the convex lens on one side and adjust a paper screen on other side.
(3) Adjust the position of the screen in order to get the sharp image of a burning candle.
(4) Finally we see the formation of image.
Explanation: When the lens is covered half with a black paper, the brightness of image formed is reduced because lesser number of light rays will be able to pass through lens.

Question:  You are given kerosene, turpentine and water. In which of these does the light travel fastest?Use the information given in the table below:

It can be inferred from the relation that speed of light is inversely proportional to the refractive index of the medium. So, we can say that light will travel the slowest in the material which has the highest refractive index and travel the fastest in the material which has the lowest refractive index.
It can be observed from table that the refractive indices of kerosene, turpentine, and water are 1.44, 1.47. and 1.33 respectively. Therefore, light travels the fastest in water as water has the lowest refractive index.

Question: A divergent lens has a focal length of 20 cm. At what distance should an object be placed so that the image is formed 10 cm away from the lens. What is the magnification produced by the lens?
Answer: Divergent lens is concave lens:
f = –20 cm and v = –10 cm because a concave lens always forms a virtual image on the same side as the object. 

Thus, object should be placed at a distance of 20 cm from the lens and magnification produced by the lens is 0.5

Long Answer Type Questions: 

Question: (A) Complete the following ray diagram:

(B) Find the nature, position and size of the image formed.
(C) Use lens formula to determine the magnification in this case. 
Answer: (B) Nature of image
On the same side of the lens as the object.
Position of image = virtual and erect
Size of image = Enlarged (10 times to that of an object)
(C) Convex lens
f = +20 cm
h = –18 cm
By using lens formula.

Question:  (A) Define the following terms:
(i) Power of a lens
(ii) Principal focus of a concave mirror
(B) Write the relationship among the object
distance (u), image distance (n) and the focal length (f) of a:
(i) Spherical lens
(ii) Spherical mirror
(C) An object is placed at a distance of 10 cm from optical centre of a convex lens of focal length 1?? cm. Draw a labelled ray diagram to show the formation of image in this case. 
Answer: (A) (i) Power of a Lens: Power of a lens is defined a the ability of the lens to converge the rays of light falling on it.
The power of a lens is also defined as the reciprocal of focal length P = 1/f (m)

(ii) Principal focus of a concave mirror: The principal focus of a concave mirror is a point on the principal axis of the mirror, at which rays of light incident on the mirror in a direction parallel to the principal axis, actually meet after reflection from the mirror.
(B) Relationship between object distance (u), image distance (v) and focal length (f) of a:
(i) spherical lens
1/f=1/v-1/u

(ii) Spherical mirror
    1/f=1/v+1/u

(C) u = – 10 cm
    f = + 15 cm

Since the object distance from optical centre of a convex lens is less than the focal length, hence the object lies between focus F₁ and optical centre O. Image formed will be enlarged, virtual and erect and on the same side of the lens as the object.

Question:  List in proper sequence the steps of the experiment for determining the approximate focal length of a given concave mirror by obtaining the image of a distant object.
Answer: Steps for determining approximate focal length of concave mirror:
(1) Fix a concave mirror on the stand such that it faces a source of bright light or a distant tree.
(2) Place a screen on a stand in front of the reflecting surface of the mirror for obtaining the image of the distant object on it.
(3) Ad??ust the position of the screen by moving it back and forth until a sharp and clear image of the distant object is obtained on it.
(4) Mark the position of the mirror and the screen and measure the distance between them with a meter scale.
(5) Repeat steps (3) and (4) above thrice and find the average distance.
(6) This gives the focal length of the concave mirror.

Question:  An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm.
(A) Use lens formula to find the distance of 60 cm. from a concave lens of focal length 30 cm.
(B) List four characteristice of the image (nature, position, size, erect??inverted) formed by the lens in this case
(C) Draw ray diagram to justify your answer of part (ii). 
Answer: In the given question,
u = object distance = –60 cm,
f = focal length = –30 cm (f is –ve as it is a concave lens having virtual focus).
Let v = image distance
(A) Using the lens formula,
1/f=1/v-1/u
where u = object distance, v = image distance and f = focal length, we get
1/−30=1/v+ 1/60
1/v =-1/30  -1/60= 2-1/60= -3/60
v=- 60/3 cm =-20 cm
∴ distance of image from the lens = 20 cm in front of the lens
(B) Magnification, m =v/u=-20/-60=1/3
Characteristics of image formed are:
(1) As m is + ve, image formed is virtual.
(2) Image is formed 20 cm in front of the lens
(3) As m < 1, size of image is smaller than size of object, i.e., image is diminished.
(4) Image is erect.
(C) Ray diagram