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HC Verma Class XII Physics Chapter 21- Speed Of Light

Exercise : Solution of Questions on page Number : 447

Answer: 1

No, it is not advisable to define the length 1 m as the distance travelled by sound in 1/332 s because the speed of sound is affected by various factors such as temperature, humidity and nature of medium. So, it cannot be said that the distance travelled by sound in 1/332 s will be 1 m, less than 1 m or more than 1 m.

Answer: 2

We have speed of light = 299792458 m/s

To have a accuracy of 10% the light has to travel 1/10th of a second between the observers so,

Distance travelled by the light in 0.1 s = 0.1 × 299792458 = 29979 km.

The difficulty in separation of that distance will be the curvature of earth. As the earth’s surface is curved, light from one of the experimenters won’t reach the other.

Answer: 3

If the wheel is placed away from the focal plane the light returning light rays will fall in an extended area of the wheel, this will let image to appear even when the light ray is blocked by one of the teeth of the wheel.

Exercise : Solution of Questions on page Number : 448

Answer: 1

Distance between the mirrors (D) = 12.0 km = 12 × 103 m
Number of teeth in the wheel (n) = 180

Now we apply Fizeau’s apparatus
Speed of light, c = 3 × 108 m/s
We know, c = 2Dnωπ
⇒ ω = cπ2Dnred/s

= πc2Dn × 180πdeg/s
ω = 3 × 10824 × 103 = 1.25 × 104 deg/sec

Hence, the required angular speed of the wheel for which the image is not seen is 1.25 × 104 deg/sec

Answer: 2

Distance between the rotating and the fixed mirror (R) = 16 m
Distance between the lens and the rotating mirror (b) = 6 m
Distance between the source and the lens (a) = 2 m

Mirror is rotated at a speed of 356 revolutions per second
⇒ ω = 356 rev/s = 356 × 2 π rad/sec

Shift in the image (s) = 0.7 m = 0.7 × 103 m
In Foucault experiment, speed of light is given by
c = 4R2 was(R + b)
= 4×(16)2 × 356 × 2π × 2(0.7) × 10 − 3(16 + 6)
= 2.975 × 108 m/s
Therefore, the required speed of light is 2.975 × 108 m/s.

Answer: 3

Distance travelled by light between two reflections from the rotating mirror (D) = 4.8 km = 4.8 × 103 m
Number of faces of the mirror, N = 8
Angular speed of the mirror, ω = ?

In Michelson experiment, the speed of light (c) is given by
c = ωDN2π
ω = angular speed
N = number of faces in the polygon mirror

∴ ω = 2πcDN rad/s
= cDN rev/s
= 3 × 108(4.8 × 103 × 8)
= 7.8 × 103 rev/s
Hence, the required angular speed is 7.8 × 103 rev/s.

Answer: 4

There is no difficulty if the distance travelled by light is decreased. In this method, light has to travel a large distance of 8.6 km. So, the intensity of the light decreases considerably and the final image becomes dim. If somehow this distance is decreased, the final image is dark due to the increased light intensity.

Answer: 5

The advantage of using a polygonal mirror with larger number of faces in the Michelson method is it gives the value that is very near to the accurate value and minimises the error.

Answer: 1

(a) increase

If the gas is gradually pumped out, a vacuum will be created inside the closed cylindrical tube, and experimentally, light travels at the fastest speed in vacuum as compared to any other medium.

Answer: 2

(a) in vacuum but not in air

Different wavelengths travel at different speeds through different media. In vacuum, the speeds of both the red light and yellow light are same but are different in air due to some optical density of air. Both wavelengths act in a different way in the air.

Answer: 3

(b) the image will be shifted a little later than the object

Light rays emitting from a source have to cover some optical distance to form an image of the source on the other side of the lens. So, when a light source is shifted by some distance on the principal axis, then the light rays emitting from the new position of the source take some time to form a shifted image of the object on the other side of the lens. However, this delay is very small because the speed of light has a very larger value.

Answer: 1

(a), (b), (c) and (d)

The speed of light is a fundamental constant, and with respect to any inertial frame, it is independent of the motion of the light source.

Answer: 2

(c) Foucault method

Foucault gave the first laboratory method to find the velocity of light. He obtained a value of 2.98×108m/s from his measurements.

Answer: 3

(c) Foucault method

Foucault method can be used to measure the speed of light in water. One of the advantage of this method is that one can put some transparent medium (or water) between two mirrors to measure the speed of light in that medium (or water). Foucault observed that the velocity of light in water is less than that in the air.​