Sunday, August 4, 2019

H C Verma solutions, Speed of Light, EXERCISES, Chapter-21, Concepts of Physics, Part-I

Speed of Light

EXERCISES


1. In an experiment to measure the speed of light by Fizeau's apparatus, the following data are used:

Distance between the mirrors = 12.0 km

The number of teeth in the wheel = 180.

Find the minimum angular speed of the wheel for which the image is not seen.     


ANSWER: In Fizeau's method the speed of light is calculated as

c = 2Dn⍵/π

Given that D = 12.0 km

n = 180

We know c = 3x10⁵ km/s
Hence, ⍵ = πc/2Dn =π*3x10⁵/(2*12*180) rad/s
= π*3x10⁵/4320)*(180/π) deg/s
= 180*3x10⁵/4320 deg/s
= 12500 deg/s
= 1.25x10⁴ deg/s

           

 


2. In an experiment with Foucault's apparatus, the various distances used are as follows:

Distance between the rotating and the fixed mirror = 16 m.

Distance between the lens and the rotating mirror = 6 m,

Distance between the source and the lens = 2 m.

When the mirror is rotated at a speed of 356 revolutions per second, the image shifts by 0.7 mm. Calculate the speed of light from this data.      


ANSWER:  From the given data

a = the distance from the lens to source = 2 m

b = the distance from rotating mirror to lens = 6 m

s = shift = 0.7 mm = 7x10⁻⁴ m

⍵ = the angular speed of the plane mirror =2π*356 rad/s

 R = the distance between the rotating and the fixed mirror = 16

The speed of the light in Foucault's method is given as

c = 4R²⍵a/s(R+b)
   = 4*16²*2π*356*2/{7x10⁻⁴(16+6)}
   = 2.975x10⁸ m/s

      


 


3. In a Michelson experiment for measuring the speed of light, the distance traveled by light between two reflections from the rotating mirror is 4.8 km. The rotating mirror has the shape of a regular octagon. At what minimum angular speed of the mirror (other than zero) the image is formed at the position where a non-rotating mirror forms it?      


ANSWER: Given that

D = the distance traveled by light between two reflections from the rotating mirror = 4.8 km = 4800 m

N = number of faces in the mirror = 8

We take c = speed of light = 3x10⁸ m/s  

The speed of light in the Michelson method is given as 

c = D⍵N/2π

→⍵ = 2πc/DN rad/s

      =2πc/DN)*(1/2π) rev/s

      = c/DN rev/s

      = 3x10⁸/(4800*8) rev/s

      = 7812.5 rev/s

      ≈ 7.8x10³ rev/s     

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Links to the Chapters





CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Vector related Problems"

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Click here for "OBJECTIVE-II"

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