Sunday, August 11, 2019

H C Verma solutions, Photometry, Questions for Short answer, Chapter-22, Concepts of Physics, Part-I

Photometry

Questions for Short Answer

1. What is the luminous flux of a source emitting radio waves?      


ANSWER:   The luminous flux is a quantity directly representing the total brightness producing capacity of a source. Since the radio waves do not produce brightness and its luminosity is zero, the luminous flux of the source will be zero.              

     



2. The luminous flux of a 1 W sodium vapor lamp is more than that of a 10 kW source of ultraviolet radiation. Comment.     


ANSWER:   The luminous flux represents the total brightness producing capacity of a source. A sodium vapor lamp produces virtually monochromatic light averaging 589 nm wavelength. Since the luminosity of the wavelength around 555 nm is maximum, the luminous flux of a 1 W sodium vapor lamp is good. The luminosity of the wavelength of ultraviolet radiation is zero, hence the luminous flux of a 10 kW source of ultraviolet light will be zero. So, the luminous flux of a 1 W sodium vapor lamp is more.            

     



  3. Light is incident normally on a small plane surface. If the surface is rotated by an angle of 30° about the incident light, does the illuminance of the surface increase, decrease or remain the same? Does your answer change if the light did not fall normally on the surface?     


ANSWER:   The light is incident normally to the small plane surface. So the normal to the plane makes an angle θ = zero with the incident light. The illuminance is proportional to the cosine of this angle. So cosθ = 1. Illuminance is maximum for this position. Given that the surface is rotated about the incident light. This means the incident light is the axis about which the plane surface is rotated. By rotating the surface by 30° the angle θ does not change, it still remains zero. Hence the illuminance remains the same. 

     For the second case when the light is not incident normally but the normal to the plane makes an angle θ with the incident light:- when the plane is rotated by 30° about the incident light again θ does not change even though the normal to the plane takes another position in the space. Thus illuminance again remains the same.       

     



  4. A bulb is hanging over a table. At which portion of the table is the illuminance maximum? If a plane mirror is placed above the bulb facing the table, will the illuminance on the table increase?     


ANSWER:   The illuminance will be maximum on the table at the portion just below the bulb because the normal to the table makes an angle zero with the incident light here.

        When a plane mirror is placed above the bulb facing the table, the light in the upward direction is reflected back towards the table. Thus the luminous flux on the table increases. It increases the luminous intensity (I) which results in increased illuminance.               

     



  5. The sun is less bright in the morning and evening as compared to at noon although its distance from the observer is almost the same. Why?      


ANSWER:   In the morning and evening the normal to the place (viewing point) makes a larger angle with the incident light. Thus the value of cosθ is very less and the illuminance at the place is low. At noon the value of θ is smaller thus the value of cosθ is more and the illuminance is more.

            Also, the sun rays have to travel a longer distance through the atmosphere to reach us in the morning and the evening. It results in the absorption and scattering of shorter wavelengths light and more of longer wavelengths like red and orange reach us. The luminosity of these wavelengths is minimum. These are the reasons why the sun is less bright in the morning and evening in comparison at noon.           

     



  6. Why is the luminous efficiency small for a filament bulb as compared to a mercury vapor lamp?     


ANSWER:   In a filament bulb, most of the electrical energy given to it is converted into heat in the infrared range which has zero luminous flux. Only a small part of the emitted light is in the visible range. So the ratio of the total luminous flux to the total radiant flux i.e. luminous efficiency is small.

          The mercury vapor lamp wastes very little energy in the infrared range. It emits light in the green and blue range and very little in the ultraviolet range. So the total luminous flux is more which results in large luminous efficiency compared to a filament bulb.             

     



  7. The yellow color has a greater luminous efficiency as compared to the other colors. Can we increase the illuminating power of a white light source by putting a yellow plastic paper around this source?     


ANSWER:   No. The illuminating power of a white light source is contributed by all wavelengths including the yellow light. When a yellow plastic paper is put around the source, all the wavelengths are blocked except the yellow light. So, the contribution of other wavelengths is zero now and only yellow light contributes to the illuminating power. So the illuminating power can not be increased by putting a yellow plastic around the source.              

     


----------------------------------------------------------------------


Click here for all links → kktutor.blogspot.com 





===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


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

Click here for → Questions for Short Answer 

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

Click here for → Questions for Short Answer

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".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

-------------------------------------------------------------------------------

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


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


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"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

Posted by at
Labels: , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)