HC Verma i Solutions for Class 12 Science Physics Chapter 21 Speed Of Light are provided here with simple step-by-step explanations. These solutions for Speed Of Light are extremely popular among class 12 Science students for Physics Speed Of Light Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma i Book of class 12 Science Physics Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s HC Verma i Solutions. All HC Verma i Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 447:

#### Question 1:

#### Answer:

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.

#### Page No 447:

#### Question 2:

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:

We have speed of light = 299792458 m/s

To have a accuracy of 10% the light has to travel 1/10^{th} 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.

#### Page No 447:

#### Question 3:

We have speed of light = 299792458 m/s

To have a accuracy of 10% the light has to travel 1/10^{th} 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:

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.

#### Page No 448:

#### Question 1:

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.

#### Answer:

Distance between the mirrors (*D*) = 12.0 km = 12 × 10^{3} m

Number of teeth in the wheel (*n*) = 180

Now we apply Fizeau's apparatus

Speed of light, *c* = 3 × 10^{8} m/s

$\mathrm{We}\mathrm{know},c=\frac{2Dn\omega}{\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =\frac{c\pi}{2Dn}\mathrm{red}/\mathrm{s}$

$=\frac{\pi c}{2\mathrm{D}n}\times \frac{180}{\pi}\mathrm{deg}/\mathrm{s}$

$\omega =\frac{3\times {10}^{8}}{24\times {10}^{3}}\phantom{\rule{0ex}{0ex}}=1.25\times {10}^{4}\mathrm{deg}/\mathrm{sec}$

Hence, the required angular speed of the wheel for which the image is not seen is $1.25\times {10}^{4}\mathrm{deg}/\mathrm{sec}$

#### Page No 448:

#### Question 3:

Distance between the mirrors (*D*) = 12.0 km = 12 × 10^{3} m

Number of teeth in the wheel (*n*) = 180

Now we apply Fizeau's apparatus

Speed of light, *c* = 3 × 10^{8} m/s

$\mathrm{We}\mathrm{know},c=\frac{2Dn\omega}{\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =\frac{c\pi}{2Dn}\mathrm{red}/\mathrm{s}$

$=\frac{\pi c}{2\mathrm{D}n}\times \frac{180}{\pi}\mathrm{deg}/\mathrm{s}$

$\omega =\frac{3\times {10}^{8}}{24\times {10}^{3}}\phantom{\rule{0ex}{0ex}}=1.25\times {10}^{4}\mathrm{deg}/\mathrm{sec}$

Hence, the required angular speed of the wheel for which the image is not seen is $1.25\times {10}^{4}\mathrm{deg}/\mathrm{sec}$

#### Answer:

Distance travelled by light between two reflections from the rotating mirror (*D*) = 4.8 km = 4.8 × 10^{3} m

Number of faces of the mirror, *N* = 8

Angular speed of the mirror, $\omega $ = ?

In Michelson experiment, the speed of light (*c*) is given by

$c=\frac{\omega DN}{2\mathrm{\pi}}$

where*ω = *angular speed*N* = number of faces in the polygon mirror

$\therefore \omega =\frac{2\mathrm{\pi}c}{DN}rad/\mathrm{s}$

$=\frac{c}{DN}\mathrm{rev}/\mathrm{s}$

$=\frac{3\times {10}^{8}}{(4.8\times {10}^{3}\times 8)}$

= 7.8 × 10^{3} rev/s

Hence, the required angular speed is 7.8 × 10^{3} rev/s.

#### Page No 448:

#### Question 2:

Distance travelled by light between two reflections from the rotating mirror (*D*) = 4.8 km = 4.8 × 10^{3} m

Number of faces of the mirror, *N* = 8

Angular speed of the mirror, $\omega $ = ?

In Michelson experiment, the speed of light (*c*) is given by

$c=\frac{\omega DN}{2\mathrm{\pi}}$

where*ω = *angular speed*N* = number of faces in the polygon mirror

$\therefore \omega =\frac{2\mathrm{\pi}c}{DN}rad/\mathrm{s}$

$=\frac{c}{DN}\mathrm{rev}/\mathrm{s}$

$=\frac{3\times {10}^{8}}{(4.8\times {10}^{3}\times 8)}$

= 7.8 × 10^{3} rev/s

Hence, the required angular speed is 7.8 × 10^{3} rev/s.

#### Answer:

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 × 10^{3} m

In Foucault experiment, speed of light is given by

$c=4{\mathrm{R}}^{2}\frac{wa}{s(\mathrm{R}+b)}$

$=\frac{4\times (16{)}^{2}\times 356\times 2\mathrm{\pi}\times 2}{(0.7)\times {10}^{-3}(16+6)}$

= 2.975 × 10^{8} m/s

Therefore, the required speed of light is 2.975 × 10^{8} m/s.

#### Page No 448:

#### Question 4:

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 × 10^{3} m

In Foucault experiment, speed of light is given by

$c=4{\mathrm{R}}^{2}\frac{wa}{s(\mathrm{R}+b)}$

$=\frac{4\times (16{)}^{2}\times 356\times 2\mathrm{\pi}\times 2}{(0.7)\times {10}^{-3}(16+6)}$

= 2.975 × 10^{8} m/s

Therefore, the required speed of light is 2.975 × 10^{8} m/s.

#### Answer:

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.

#### Page No 448:

#### Question 5:

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:

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.

#### Page No 448:

#### Question 1:

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:

(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.

#### Page No 448:

#### Question 2:

(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:

(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.

#### Page No 448:

#### Question 3:

(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:

(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.

#### Page No 448:

#### Question 1:

(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:

(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.

#### Page No 448:

#### Question 2:

(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:

(c) Foucault method

Foucault gave the first laboratory method to find the velocity of light. He obtained a value of $2.98\times {10}^{8}\mathrm{m}/\mathrm{s}$ from his measurements.

#### Page No 448:

#### Question 3:

(c) Foucault method

Foucault gave the first laboratory method to find the velocity of light. He obtained a value of $2.98\times {10}^{8}\mathrm{m}/\mathrm{s}$ from his measurements.

#### Answer:

(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.â€‹

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