CBSE Class 10 Science Chapter-10 Light Reflection and Refraction – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Science Chapter 10 – Light Reflection and Refraction prepared by expert Science teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Science Chapter-10 Light Reflection and Refraction Important Questions
CBSE Class 10 science Important Questions Chapter 10 – Light Reflection and Refraction
3 Mark Questions
1. An object is placed at a distance of 12 cm in front of a concave mirror. It forms a real image four times larger than the object. Calculate the distance of the image from the mirror
Ans. 
2. Draw a ray diagram to represent the nature, position and size of the image formed by a convex lens for the object placed at
(a) infinity
(b) Between 
and optical centre (O)
Ans. (a) Size – Point sized
Position – At focus
Nature – Real & Inverted
(b) Size – highly magnified
Position – same side of the lens where the object is
placed
Nature – virtual & erect.
3. 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.
Ans. 
Image is virtual.
4. A concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed?
Ans. 
H = 5 cm
V = -15 cm
U = ?
U = -60 cm
5. An object is kept at a distance of 15 cm from a
(a) convex mirror
(b) concave lens
(c) Plane mirror.
The focal length of the convex mirror and the concave lens are 10 cm each.
Draw the appropriate ray diagrams, showing the formation of image, in each of the three cases.
Ans. Here U=15 cm f=10 cm for convex & concave
(a) 
Between Pole and focus virtual & erect Diminished
(b) 
Between optical centre & focus diminished, virtual & erect
(c) 
6. State the mirror formula for determining the focal length of spherical mirrors write the meanings of the symbols used An object is placed at a distance of 25 cm. from a concave mirror of focal length 15 cm. Calculate the distance of the image from the mirror.
Ans. 
Where f is the focal length of the mirror
U is the object distance
V is the Image distance
U=-25 cm
F=-15 cm
V=?
7. Find the position, nature and size of the image formed by a convex lens of focal length 12 cm of an object 5 cm high placed at a distance 20 cm from it.
Ans. 
V = -20 cm
is negative so the image is real and invested
8. An object is kept at a distance of
(a) 
(b) 
from a convex lens having focal length of magnitude (a) Draw ray diagrams showing the formation of images formed in the two cases.
Ans. (a) for 
Position of the object –Between O & F
(b) 
For
Position of the object – Between F and 2F
9. A concave mirror is used to form an erect and enlarged image of a given object. Where is the image located with respect to the mirror? Draw the corresponding ray diagram.
Ans.
Image is located behind the mirror. It is highly magnified and virtual and erect.
10. How can you show that if a ray enters a rectangular glass slab obliquely and emerges from the opposite face, the emergent ray is parallel to the incident ray?
Ans. 
(Alternate angles)
At face AB 
At face CD 
From (1) & (2)
11. Light enters from air to glass having refractive index 1.50. What is the speed of light in glass? The speed of light in vacuum is
.
Ans. Speed of light in vacuum (c) =
.
Refractive index = c/v.
Speed of light in glass = 
= 
12. 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 mirror? What is the nature of image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Ans. Object must be placed in front of concave mirror between its pole and principal focus at a distance less than 15 cm. The image formed will be virtual and erect. The size of the image is larger the object. The ray diagram is as follows:
13. Name the type of mirror used in the following situations:
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Ans. (a) Headlights of a car- concave mirror to give parallel beam of light after reflection from concave mirror.
(b) Side/rear-view mirror of vehicle- convex mirror as it forms virtual erect and diminished image to give wider view field.
(c) Solar furnace- concave mirror to concentrate sunlight to produce heat in solar furnace.
14. 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 image.
Ans. f = +15 cm, u = -10 cm.
1/f = 1/v +1/u
1/v = 1/15 +1/10
1/v = 5/30
v = + 30 cm.
The image is formed 6 cm behind the mirror, it is a virtual and erect image.
15. The magnification produced by a plane mirror is +1. What does this means?
Ans. 
Magnification produced by a plane mirror is +1 which means that size of image formed is exactly equal to size of object behind the mirror.
16. 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.
Ans. Radius of curvature (R) = 30 cm
f = R/2 = 30/2 = 15 cm
u = -20 cm, h= 5 cm.
1/v +1/u = 1/f
1/v = 1/15+ 1/20 = 7/60
v = 60/7 = 8.6 cm.
image is virtual and erect and formed behind the mirror.
Size of image is 2.2 cm.
17. 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 focused image can be obtained? Find the size and the nature of the image.
Ans. u = – 27 cm, f = – 18 cm. 
1/v = 1/f- 1/u
1/v = -1/18 + 1/27 = -1/54
V = – 54 cm.
Screen must be placed at a distance of 54 cm from the mirror in front of it.
Thus, the image is of 14 cm length and is inverted image.
18. Find the focal length of a lens of power -2.0 D. What type of lens is this?
Ans. Power of lens (P) = -2.0 D
P = 1/f or f = 1/m
f = 1/-2.0 = -0.5 m.
(-ve) sign of focal length means that the lens is concave lens.
19. 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?
Ans. P = +1.5 D
f = 1/P = 1/+1.5 = 0.67 m.
As the power of lens is (+ve), the lens is converging lens.
20. You are given three lenses.
i) a concave lens of focal length 25 cm.
ii) a convex lens of focal length ¼ m and
iii) a convex lens of focal length 100 cm.
Which combination out of these three lenses will form a lens of zero power?
Ans. Combination of concave lens of focal length of 25 cm and a convex lens of focal length of ¼ m
21. A rod of length 10 cm lies along the principal axis of a concave mirror of 10 cm in such a way that the end closer to the pole is 20 cm away from it. Find the length of image?
Ans. R =2f=2O cm. Thus the nearer end B of the rod AB is at C and hence its image will be formed at B itself For end A u = -30 cm, f= -10 cm, v = -15 cm Length of image will be at 5 cm