Light Reflection and Refraction Class 10 Science Important Questions

Please refer to Light Reflection and Refraction Class 10 Science Important Questions given below. These solved questions for Light Reflection and Refraction have been prepared based on the latest CBSE, NCERT and KVS syllabus and books issued for the current academic year. We have provided important examination questions for Class 10 Science all chapters.

Class 10 ScienceLight Reflection and RefractionImportant Questions

Question. Name some phenomenon associated with light during image formation by mirrors.

Question. Define reflection of light.
Answer : The phenomenon of coming back of light in the same medium after striking a plane and polished surface is called reflection of light.

Question. State laws of reflection.
Answer : Incident ray, reflected ray and normal at the point of incidence all lie in the same plane. The angle of incidence is equal to the angle of reflection.

Question. What are the properties of image formed by a plane mirror?
Answer : Image is virtual and erect.
 Size of the image is equal to that of object
 Image is laterally inverted.
 The image formed by a plane mirror is always at the same distance as the object
is in front of it.

Question. What are spherical mirrors?
Answer : Mirrors whose reflecting surface are part of a sphere are called spherical mirrors.

Question. Define magnification of mirror.
Answer : The ratio of height of the image to the height of the object is called magnification. It
is represented by ‘m’.
m = Height of image (h′)/Height of object (h) = –v/u
Magnification of real image is negative and of virtual image is positive.

Question. Define refraction of light.
Answer : The change in direction of light, when it travels from one medium to another medium is called refraction of light.

Question. Define centre of curvature, principal axis, optical centre, aperture, focus and focal length for a lens.
Answer : (a) Centre of curvature: It is the centre of the spheres of which the each surface of the lens forms a part. Represented by C or 2f.
(b) Principal axis: An imaginary straight line passing through the two centres of curvatures.
(c) Optical centre: It is the central point of the lens. Represented by O.
(d) Aperture: It is the diameter of circular outline of a spherical lens.
(e) Focus: The point at which rays of light parallel to principal axis converges (convex lens) or appears to diverge (concave lens) after refraction. Represented by F.
(f) Focal length: The distance between focus and optical centre is called focal length. It is represented by f.

Question. State laws of refraction.
Answer : The ratio of sin of angle of incidence to the sin of angle of refraction for a light of given colour and for a given pair of media is constant. This is called Snell’s law.
i.e.,
sin i/sin r = Constant
The incident ray, refracted ray and the normal at the point of incidence lie on the same plane.

Question. Define pole, centre of curvature, radius of curvature, principal axis, aperture, focus and focal length of a spherical mirror.
Answer : Pole: the centre of reflecting surface. It is represented by letter P.
Centre of Curvature: The centre of the sphere of which the mirror forms the part.
Represented by “C”.

Question. What is light?
Answer : Light is a form of electromagnetic radiation that causes the sensation of sight. It doesn’t require any material medium to travel.
Radius of Curvature: The radius of the sphere of which the mirror forms the part.
Represented by “R”.
Principal axis: The straight line joining the pole (P) and the centre of curvature. It is normal to the mirror at its pole.
Aperture: The diameter of the spherical mirror is called its aperture. The reflecting surface of the mirror.
Focus: The point of the principal axis at which the rays parallel to principal axis meet (concave mirror) or appear to meet (convex mirror) after reflection. Represented by F.
Focal Length: The distance between the pole and the principal focus of a spherical mirror is called focal length. Represented by f.

Question. Give some uses of concave mirror.
(b) Used as shaving mirror.
(c) Used by dentist.
(d) Used in solar furnance.

Question. State mirror formula and write it mathematically.
Answer : The relation between focal length of mirror, distance of the object and distance of the image is known as mirror formula. It is given by
1/u + 1/ν = 1/f
u = Image distance
ν = Object distance
f = Focal length

Question. Give the relation between focal length and radius of curvature.

Question. Where should be an object placed in front of convex lens so as to use it as a magnifier?
Answer : Between the pole and the focal length.

Question. Define the term angle of incidence.
Answer: The angle between an incident ray and the normal at the point of incidence is called angle of incidence.

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

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. What is the name given to the centre of the mirror ?
Answer: The centre of reflecting surface of a spherical mirror is known as Pole.

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. What is silvering of mirror?
Answer : Silvering of mirror means coating the surface of mirror with a thin layer of silver, aluminium or some other shiny, opaque material.

Question. What are the two types of lenses?
Answer : Spherical lens: combination of two spherical refracting surfaces.

Question. What do you observe when light ray passes through rectangular slab?
Answer : (a) Angle of incidence is equal to angle of emergence.
(b) Incident ray is parallel to the emergent ray.
(c) Lateral displacement is proportional to the thickness of glass slab.
(d) Lateral displacement is proportional to the angle of incidence.

Question. Define lateral displacement.
Answer : Lateral displacement is the perpendicular distance between the incident ray and theemergent ray.

Question. Give uses of convex mirror.
Answer : (a) Used as rear view mirror in vehicles.
(b) Used to see full length image of a tall building.

Question. Define refractive index.
Answer : Refractive index is defined as the ratio of speed of light in medium 1 to the speed of light in medium 2 and is represented as n21 and is read as refractive index of medium
2 with respect to medium 1.
n21 = speed of light in medium 1/speed of light in medium 2.

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 : Focal length of concave mirror = 15 cm.
Object should be place in front of given concave mirror at a distance less than 15 cm.
Image formed is virtual and erect.
Image size is larger than object.

Question. Name the type of mirror used in the following situations.
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace.
Answer : (a) For headlight of a car— Concave mirror is used to get a powerful beam of light after reflection.

(b) Convex mirror is used for side/rear view mirror of a vehicle. Convex mirror forms an erect and diminished image of vehicles and gives wider view of rear.
(c) In solar furnace concave mirror is used as a reflector, it concentrates sun light at a point where the temperature increases sharply to 180°C – 200°C.

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 : Yes, one-half of a convex lens when covered with a black paper, the lens produces a complete or full image of an object.
To verify experimentally:
Take a convex lens, cover half part of it as shown in the figure, with a paper. Place it on a stand. Focus a distant object on a screen, the image obtained on the screen is complete.
Observation and conclusion: Image formed on the screen does not depend on the size of the lens. The brightness of the image decreases as less number of rays pass through the lens.   (Img 145)

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.

Hence, the image formed at 16.67 cm from the lens on the other side. The size of the image is 3.3 cm, i.e., reduced and inverted.

Question. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

f = –15 cm
ν = –10 cm
1/v – 1/u = 1/f
1/-10 – 1/u = 1/-15
∴ 1/u = – 1/30 ∴ u = –30 cm

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.

f = +15 cm, u = – 10 cm.
1 /f = 1/ν + 1/u
1/15 = 1/ν + 1/(– 10)
1 /ν = 1/15 + 1/10
1 /ν = 5/30
ν = + 6 cm

The image is formed 6 cm behind the mirror, virtual image is formed.

Question. The magnification produced by a plane mirror is +1. What does this mean?
+ indicates virtual image.
1 indicates that the object size and image size is same.

Question. How are the images formed when an object is moved from infinity to the convex lens?
Answer: Object at Infinity. When object is at infinity, a real image is formed at F on the other side of the lens

Object at infinity. Image at F on the other side of lens.
However if the rays are parallel to themselves but not parallel to principal axis, then
these rays after refraction will form image at focus F’ and not at principal focus F.

Object at infinity, rays parallel to themselves but not parallel to principal axis. Image is formed at F’, the focus on the other side of lens. Object beyond 2F. When the object
is beyond 2F, a real, inverted, diminished image is formed between F and 2F.

Object beyond 2F, real, inverted, diminished image between F and 2F. Object at 2F.
When the object is at 2F, a real, inverted image of the same size is formed on the other side of the lens at 2F as given in Fig.

Object at 2F, image at 2F on the other side of the lens. Image is of size same as that of the object. When the object is between F and 2F, its real, inverted, magnified image is formed on the other side of the convex lens as shown in fig.

Object between F and 2F real, inverted, magnified image is formed beyond 2F on the other side of lens.
Object at F. When object is placed principal focus, a real, inverted, very highly magnified image is formed at infinity.

Object at F, a very highly magnified, real, inverted image is formed at infinity.
Object between F and C. When an object is placed between principal focus and optical centre of the lens, virtual, erect, magnified image is formed on the same side of the lens.

Object between F and C; a virtual, erect, magnified image is formed on the same side.

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 image is at a distance of 7.5 cm from lens (in front of lens).
The magnification m = v/u = -7.5/30
= 1//4
= +0.25
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. Two friends Kapil and Rohit were studying in the same class. One day Rohit observed that Kapil was having pain in gums during lunch time. Rohit told Kapil that his father was dentist and asked him to visit his father’s clinic. Rohit’s father examined Kapil with the help of a mirror and advised him not to eat too many chocolates and soft drinks. Kapil follow ed the advise of the doctor and soon he got recovered. After that he starts taking care of his mouth, as he washes his mouth properly after every meal and also starts taking a calcium rich diet. Read the given passage and answer the following questions:
i. Identify the mirror used by the dentist.
ii. Name the phenomenon of light by which doctor is able to exam ine Kapil.
iii. What values are shown by doctor, his son and Kapil?
Answer:  i. The mirror used by the dentist is concave mirror.
ii. The phenomenon of light by which doctor is able to examine Kapil is reflection of light.
iii. The doctor gave the correct advise to Kapil on how to keep his mouth clean and gums healthy. The doctor’s son and doctor was helping in nature. Kapil followed the advise given by the doctor. So, he is an obedient boy.

Question. Draw ray diagram showing the image formation by a convex lens when an object is placed at twice the focal length of the lens.

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. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Question. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
u = –27 cm, f = –18 cm, h = 7.0 cm
Mirror formula 1/v + 1/u = 1/f
∴ 1/v + 1/(– 27) = 1/(-18)
∴ 1/v =  1/-18 + 1/27 = -3+2 /54 = -1/54
ν = –54 cm.
hi/he = v/u
hi = vxhe / u
= 54×7/27 = 14 cm

The image is real, inverted and enlarged.

Question. Find the focal length of a lens of power –2.0 D. What type of lens is this?
Answer :  P = – 2.0 D
P = 1/f
∴ f = 1/P = 1/-2.0D = -0.5cm
∴ The lens is concave lens as f = –ve.

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 :    P = +1.5 D, P = 1/f
Focal length of the lens f = 1/P = 1/+1.5D = + 0.67 m

Power of the lens is +ve, and it is converging lens i.e., convex lens.

Question. Define pole and centre of curvature of spherical mirrors.
Answer : Pole: The centre of the reflecting surface of a spherical mirror is a point called the pole. It lies on the surface of the mirror. It is represented by “P”. Centre of Curvature: The reflecting surface of a spherical mirror forms a part of a sphere this sphere has a centre and this point is called the centre of curvature of the spherical mirror.

Question. If the speed of light in water is 2.25 × 108 m/s and the speed in vacuum is 3×108m/s.
Calculate the refractive index of water.
Answer : Refractive index of water =  Speed of light in 1 medium (air) / Speed of light in 2 medium (water)
nm = c/v
nm = 3 x 108 / 2.25 x 108
nm = 1.33
∴The refractive index of water = 1.33.

Question. The refractive index of water is 1.33 and kerosene is 1.44. Calculate the refractive index of kerosene with respect to water.
Answer : Refractive index of water = nw = 1.33
Refractive index of kerosene = nk = 1.44
∴Refractive index of kerosene with respect to water is

nkw =  = 1.44/1.33
= 1.082

Question. Take down this diagram on to your answer book and complete the path of the ray.

Question. What kind of mirrors are used in big shopping stores to watch activities of customers?
Answer : Convex mirror as the image is independent of position of the object.

Question. Draw a ray diagram to determine the position of image formed of an object placed between the pole and the focus of a concave mirror.

Question. Draw ray diagrams to show the image formed by a concave lens for the object placed at (i) infinity (ii) Between f and 2f of the lens.

Answer : (i) object at infinity

Object ⎯→ at infinity
Image ⎯→ at focus
Size of image ⎯→ Point sized
Nature ⎯→ virtual and erect

(ii) object between f and 2f

Object ⎯→ between f and 2f
image ⎯→ between O and F.
Size of image ⎯→ Diminished
Nature ⎯→ virtual and erect.

Question. Draw a ray diagram to show the path of light when it travels through glass slab.

Incident ray I enters the glass slab forms an angle of incidence ‘i’. Its bends towards the normal and forms an angle of refraction ‘r’.
The emergent ray is parallel to the incident ray.

Question. (a) It is desired to obtain an erect image of an object using a concave mirror of focal length 20 cm.
(i) What should be the range of distance of the object from the mirror?
(ii) Will the image be bigger or smaller than the object?
(iii) Draw a ray diagram to show the image formation in this case.
(b) One-half of a convex lens of focal length 20 cm is covered with a black paper.
(i) Will the lens produce a complete image of the object?
(ii) Show the formation of image of an object placed at 2F1 of such covered lens with the help of a ray diagram.
(iii) How will the intensity of the image formed by half covered lens compare with non-covered lens?

Answer : (a) (i) Range of the object distance is 0 to 20 cm from the pole.
(ii) Image will be bigger than the object.
(iii) Ray diagram:

(b) (i) Yes, complete image will be formed.
(ii)

(iii) Intensity will be reduced as the light falling on the lower (covered) portion will not reach the position of image.

Question. Which type of mirrors are used to give an erect and enlarged image of an object?

Question. If a light ray IM is incident on the surface AB as shown, identify the correct emergent ray.

Answer : Q as it has to be parallel to S.
Question. An object of 2 cm high is placed at a distance of 64 cm from a white screen on placing a convex lens at a distance of 32 cm from the object it is found that a distant image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.
Answer : Since the object-screen distance is double of object-lens separation, the object is at a distance of 2f from lens and the image should be of the same size of the object.

So 2f = 32 ⇒ f = 16 cm
Height of image = Height of object = 2 cm

Question. The power of a lens is –4.0 D. What is the nature of this lens?
Answer : Negative power is associated with only concave lens.

Question. Redraw the given diagram and show the path of refracted ray.

Question. A convex lens has a focal length of 10 cm. At what distance from the lens should the object be placed so that it gives a real and inverted image 20 cm away from the lens?
What would be the size of the image formed if the object is 2 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
Answer : f = +10 cm, ν = +20 cm as image is real and inverted. Height of the object = 2 cm (say +ve)

Using 1/f = 1/v – 1/u , we get

1/u = 1/v – 1/f
= 1/+20 – 1/10 = +1-2 / 20 = -1/20

Image will be of the same size as that of object (as u = ν) and hence, the height of the image will be 2 cm.

Question. Redraw the given diagram and show the path of the refracted ray.

Question. Why does a ray of light bend when it travels from one medium into another?
Answer : Due to change in velocity in the medium and to reduce the time taken to travel the same, a ray of light bends when it travels from one medium to another.

Question. Draw the given diagram in your answer book and complete it for the path of ray of light beyond the lens.

Question. A convex lens has a focal length of 15 cm. At what distance from the lens should the object be placed so that is forms on its other side a real and inverted image 30 cm away from the lens? What would be the size of image formed if the object is 5 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
Answer :      f = 15 cm
ν = + 30 cm as image is real and inverted
using           1/f = 1/v – 1/u   we get
1/u  = 1/v – 1/f
= 1/30 – 1/15 = 1-2/30 = -1/30

So, image should be of the same size as the object. Height of image = 5 cm.

Question. Redraw the given diagram and show the path of retracted ray.

Question. A convex lens has a focal length of 12 cm. At what distance from the lens should an object of height 6 cm be placed so that on the other side of the lens its real and inverted image is formed 24 cm away from the lens? What would be the size of the image formed?
Draw a ray diagram to show the image formed in this case.
Answer : f = +12 cm
Real and inverted image so v = +24 cm
Using

1/f = 1/v – 1/u we get
1/u = 1/v – 1/f
= 1/24 – 1/12 = 1-2 / 24 = -1/24
u = –24 cm (= 2f)

So image will be of the same size as the object. Height of image = 6 cm.

Question. In a small town fair Akshay took his friend and showed him a mirror in which his image showed upper half body very fat and lower body very thin. Akshay’s friend got upset but Akshay explained him by showing his similar image in the mirror.
(a) Name two mirrors used in this fair shop.
(b) Name the mirror in which the size of image is small.
(c) What value of Akshay is reflected?

Answer : (a) Concave and convex mirror.
(b) Convex mirror gives small size image.
(c) Akshay showed compassion and empathy.

Questions On High Order Thinking Skills

Question. Write the laws of reflection.
Answer : (a) The angle of incidence is equal to angle of reflection.
(b) The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane.

Question. Give characteristics of image formed by plane mirror
Answer : Image is virtual and erect.
Size is same as of the object.
It is formed at same distance.

Question. Give uses of plane mirror.
Used in submarines
Solar cooker
Kaleidoscope

Question. Name two types of spherical mirror.
Convex mirror

Question. Give uses of concave mirror.
Used as shaving mirror.
Used in solar cooker to focus the sunlight on one point.

Question. Give uses of convex mirror.
Answer : It is used as the rear view mirror in cars.
It is used in street lights as it diverge the light over larger area.

Question. What are the two types of refractive index?
Answer : Relative refractive index– It is the ratio of speed of light in one medium to the speed of light in another medium
Absolute refractive index– It is the ratio of light in vacuum to the speed of light in another medium

Question. Why do we prefer a convex mirror as a rear view mirror in vehicles?
Answer : Convex mirrors are used as rear view mirror in cars because it produces erect and diminished image of the traffic behind the vehicle. It also gives a wider view.

Question. Name the type of mirror used in the following situations.