CBSE Class 10 Science Light Reflection and Refraction Assignment Set H

Very Short Answer Type Questions

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

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

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

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. 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. 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.

Short Answer Type Questions

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. Give some uses of concave mirror.
Answer: (a) Used in torches, search lights and vehicle headlights.
(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. 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 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. 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: Convex mirror 
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?
Answer: Magnification, m=+ 1
+ indicates virtual image.
1 indicates that the object size and image size is same.
The negative sign indicates the virtual nature of the image.
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.
Answer: 

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.
Answer: Concave mirror
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. 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 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. 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. 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. Take down this diagram on to your answer book and complete the path of the ray.    

Answer: 

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.
Answer: 

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

Answer: 

Long Answer Type Questions

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

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 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.  

Answer: 

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. (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: 

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

 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.
Answer: