CBSE Class 10 Physics Light Reflection And Refraction Worksheet Set B

Question: Define the following terms in relation to concave spherical mirror:
a. Pole
b. Centre of curvature
c. Radius of curvature
d. Principal axis
e. Principal focus
f. Aperture
g. Focal length (each one mark)
Answer: a. The mid point of mirror is known as pole.
b. The centre of curvature of a spherical mirror is the centre of that sphere of which mirror is a part,
c. The distance between pole and centre of curvature is called radius of curvature of the mirror.
d. The straight line joining the pole and centre of curvature is called principal axis.
e. The point on the principal axis through which parallel rays to the principal axis passes or appear to pass after reflection.
f. The diameter of the mirror or size of the mirror is called aperture.
g. The distance between focus and pole of a mirror is the focal length of the mirror.

Question: Write down four important characteristics of image formed by a plane mirror.
Answer: Image is virtual, erect, laterally inverted and of same size as object.

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:
f = 1/P1 m=1/ 1.5 m = + 10/15 m =+2/3m

f =+ 66.7 cm
+f means the lens is convex (converging) lens.

Question: Define the term magnification. Write its formula also.
Answer: Magnification m of a mirror gives the relative extent to which the size of image of an object is magnified w.r.t. size of object. It is the ratio of size of image (h2) to size of an object (h₁).
m = Size of image/Size of object =h₂/h₁=-v/u

Question: Write the relation between object distance and image distance from a lens and focal length of a lens.
Answer: 1/f=1/v-1/u

Question: State the types of mirrors used for (i) headlights and (ii) rear view mirrors, in motorcycles. Give reason to justify your answer in each case. 
Answer: (i) Concave mirrors are used in headlights of cars to get powerful beams of light.
(ii) Convex mirrors are used as rear-view mirrors of vehicle to get a wider field of view and and erect image of traffc behind.

Question: What is meant by power of a lens? What does its sign (+ve or –ve) indicate? State its S.I. unit related to focal length of a lens?
Answer: Positive sign (+) of power indicates that lens is convex and negative sign (–) of power indicates that lens is concave.
If focal length (f) is expressed in metres, then, power is expressed in dioptres. The SI unit of power is dioptre. Thus, 1 dioptre is the power of lens whose focal length is 1 metre. 1 D = 1m–1

Question: A spherical mirror produces an image of magnification –1 on a screen placed at a distance of 50 cm from the mirror.
(a) Write the type of mirror.
(b) Find the distance of the image from the object.
(c) What is the focal length of the mirror?
Answer: (a) Concave mirror
(b) Magnification, m = -v/u or −1= -v/(-50)
∴ Distance of the image from the object is,
v = – 50 cm
(c) As the image is formed at centre of curvature i.e., v = R.
∴ focal length of the mirror, f = -50/2= -25cm

Question: How should a ray of light be incident on a rectangular glass slab so that it comes out from the opposite side of the slab without being displaced? 
Answer: A ray of light should be incident perpendicular to the surface of the rectangular glass slab so that it comes out from the opposite side of the slab without being displaced.

Question: A spherical mirror produces an image of magnification –1 on a screen placed at a distance of 40 cm from the mirror.
(i) Write type of mirror.
(ii) What is the nature of the image formed?
(iii) How far is the object located from the mirror?
Answer: (i) This is a concave mirror.
(ii) The image is real and inverted and of same size.
(iii) As m = – 1
∴ m = -v/u ⇒ −1 = -v/u ⇒ u = v
Hence, object is located at centre of curvature i.e., at distance of 40 cm from the pole of the mirror.

Question: A girl was playing with a thin beam of light from her laser torch by directing it from different directions on a convex lens held vertically. She was surprised to see that in a particular direction the beam of light continues to move along the same direction after passing through the lens. State the reason for this observation.
Answer: The girl playing with a thin beam of light from her laser torch must incident the laser beam of light through optical centre of the convex lens.
A ray of light though the optical centre of a lens passes without suffering any deviation.

Question: The image formed by a spherical mirror is real, inverted and its magnification is –2. If the image is at a distance of 30 cm from the mirror, where is the object placed?
Find the focal length of the mirror. List two characteristics of the image formed if the object is moved 10 cm towards the mirror.
Answer: Since the image formed is real and inverted, the mirror is concave.
Magnification,m = -v/u ⇒ -2 = -v/u ⇒ v = 2u
Now, if v = – 30 cm then u = – 15 cm
As focal length of the mirror is
f = uv/u + v = -15 x -30/-15 x -30 = f =450/-45 = -10cm
If the object is shified 10 cm the mirror, then the object is between principal focus and the optical centre and the image formed will be virtual and erect.

Question: The refractive index of water is 1.33 and the speed of light in air is 3 × 108 m s–1. Calculate the speed of light in water.
Answer: Refractive index of water, nw = 1.33
Speed of light in air, c = 3 × 108 m/s
As we know that refractive index of medium,
nm = speed of light in air/speed of light in medium = c/vm
Here, our medium is water.
So, nw = c/vw
where, vw is the speed of light in water.
∴ 1. 33 = 3 x 108/vw , m/s
∴ 1. 33 = 3 x 108/vw = 2.25 x 108 m/s
∴ Speed of light in water is 2.25 × 108 m/s.

Question: In a household electric circuit, different appliances are connected in parallel to one another. Give two reasons An electrician puts a fuse of rating 5A in that part of domestic electrical circuit in which an electrical heater of rating 1.5 kW, 220V is operating. What is likely to happen in this case and why? What change, if any, needs to be made?
Answer: The two reasons for connecting the appliances in parallel are as follows.
1. Each appliance will be at same potential (voltage).
2. If one of the appliances fails the other will still keep working.
Given
Voltage = 220V;
Power = 1.5kW = 1500W
(1kW = 1000W)
Now
P = VI
⇒l=p/v=1500/200=6.8

Question: State the laws of refraction.
Answer: Snell’s laws of refraction 
a. Incident ray, refracted ray and normal to the point of incidence lie in the same plane.
b. sini/sinr=1/2n

Question: (a) Define power of a lens and write its SI unit. (b) A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the lens, if image size is equal to the object size? Also, find the power of the lens.
Answer: a. Power of a lens is the ability of a lens to bend a ray of light incident upon it. It is equal to the reciprocal of the focal length of the lens.

Question: Discuss the position and nature of the image formed by a concave mirror when the object is moved from infinity towards the pole of the mirror.
Answer: When an object is moved from 3 toward the pole of a concave mirror then its image moves from focus to infinity.

Question: A ray of light is incident obliquely on a glass slab. Draw a ray diagram showing the path of the light ray.
Clearly, mark angle of incidence, angle of refraction, angle of emergence and lateral displacement of the ray. Give a formula to find refractive index of glass slab in terms of angle of incidence and angle of refraction.
Answer: When a ray passes from optical rarer to denser medium, it bends towards the normal and vice versa.
∠i = angle of incidence
∠r = angle of refraction
∠e = angle of emergence
d = lateral displacement. 
a_gn=sinli/ 

Question: (a) A ray of light falls normally on a face of a glass slab. What are the values of angle of incidence and angle of refraction of this ray?
(b) Light enters from air to a medium X. Its speed in medium X becomes 1.5×10⁸ ms-¹. Find the refractive index of medium X.
Answer: a. When a ray of light falls normally on a glass slab then i = 0, and r = 0
b. v_m= 1.5 10⁸ m/s
c = 3×10⁸ m/s
We know Refractive index
n=c/vm
or n=3×10⁸/1.5×10⁸=2
n=2