Monday, August 26, 2019

H C Verma solutions, Photometry, Exercises, Chapter-22, Concepts of Physics, Part-I

Photometry

EXERCISES

1. A source emits 45 joules of energy in 15 s. What is the radiant flux of the source?   


ANSWER: The total energy of radiation emitted by a source per unit time is called the total radiant flux. Here total energy, E = 45 joules, Time is taken to emit this energy, t = 15 s. Hence the radiant flux of the source = E/t =45/15 = 3 W.         


   


2. A photographic plate records sufficiently intense lines when it is exposed for 12 s to a source of 10 W. How long should it be exposed to a 12 W source radiating the light of the same color to get equally intense lines?    




ANSWER: Radiant flux of the source, P = 10 W, Time of exposure, t = 12 s. Total energy used for the exposure, E = P*t = 10*12 joules =120 joules. 


        For the other source P' = 12 W, the time taken for the same exposure = E/P' = 120/12 =10 s.          



3. Using figure, find the relative luminosity of wavelength (a) 480 nm, (b) 520 nm, (c) 580 nm and 600 nm.    
The figure for Q - 3


ANSWER: From the given figure, the relative luminosity of wavelength,

480 nm ------- = 0.2

520 nm ------- = 0.78

580 nm ------- = 0.78

600 nm ------- = 0.46          



4. The relative luminosity of wavelength 600 nm is 0·6. Find the radiant flux of 600 nm needed to produce the same brightness sensation as produced by 120 W of radiant flux at 555 nm.   


ANSWER: The luminous flux of a 555 nm wavelength source of 120 W = 120*685 lumen.

Since the relative luminosity of 600 nm is 0·6, the luminous flux of 120 W of 600 nm wavelength source = 0.6*120*685 lumen. For the same brightness sensation, the 600 nm source should also produce 120*685 lumen. From the unitary method,

0.6*120*685 lumen is produced by 120 W source
→1 lumen is ''     ''      ''  120/(0.6*120*685) W    "
→120*685 lumen  ''     '' 120*120*685/(0.6*120*685) W source
i.e. 120/0.6 W =200 W source.
Hence the radiant flux needed to produce same luminous flux (brightness) = 200 W.   


 

5. The luminous flux of a monochromatic source of 1 W is 450 lumen/watt. Find the relative luminosity at the wavelength emitted.  


ANSWER:  The luminous flux of 555 nm wavelength from 1-watt source = 685 lumen.

The relative luminosity = Luminous flux of the source/luminous flux of a 555 nm source of the same power. 

=450/685 = 0.66 = 66%.        



6. A source emits light of wavelengths 555 nm and 600 nm. The radiant flux of the 555 nm part is 40 W and of the 600 nm part is 30 W. The relative luminosity at 600 nm is 0͘·6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminous efficiency.    


ANSWER: (a) The total radiant flux = radiant flux of 555 nm part + radiant flux of 600 nm part

= 40 W + 30 W = 70 W.  

(b) The total luminous flux = luminous flux of 555 nm part + luminous flux of 600 nm part

= 40*685 + 0.6*30*685 lumen

= (40+18)*685 lumen

= 58*685 lumen

= 39,730 lumen.  

(c) The luminous efficiency 

= Total luminous flux/Total radiant flux

= 39730/70 lumen/W
≈ 568 lumen/W

   

  


7. A light source emits monochromatic light of wavelength 555 nm. The source consumes 100 W of electric power and emits 35 W of radiant flux. Calculate the overall luminous efficiency.   


ANSWER: The luminous flux of a 35 W source of wavelength 555 nm = 35*685 lumens.

Power input to the source = 100 W.

The overall luminous efficiency

= Luminous flux emitted/Power input to the source

= 35*685/100 lumen/W

≈ 240 lumen/W.         

8. A source emits 31.4 W of radiant flux distributed uniformly in all directions. The luminous efficiency is 60 lumen/watt. What is the luminous intensity of the source?   


ANSWER: Radiant flux = 31.4 W

Luminous efficiency = 60 lumen/watt.

Hence total luminous flux of the source, 

F =31.4*60 lumen. 

The luminous intensity = F/4π cd

= 31.4*60/4π cd

= 150 cd.  



       

9. A point source emitting 628 lumens of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminance on a small area placed at (1.0 m, 0, 0) in such a way that the normal to the area makes an angle of 37° with the X-axis.   


ANSWER: F = 628 lumen

The luminous intensity in any direction,
I = F/4π
= 628/4π cd
= 50 cd

The distance of the area from the source, r = 1 m.

The angle made by the normal to the area with the direction of light, θ = 37°.

Hence the illuminance at the small area,

E = I.Cosθ/r²

   = 50*Cos37°/1²

   = 40 lux.         

10. The illuminance of a small area changes from 900 lumen/m² to 400 lumen/m²when it is shifted along its normal by 10 cm. Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in the original position.  


ANSWER: Let the luminous intensity of the point source = I and the distance between the source and the area in the original position = x meter. Here the angle θ = 0°. Hence cosθ =cos0° = 1.

Given illuminance at first position, E = 900 lumen/m², and the illuminance at the second position, E' = 400 lumen/m².

Since E = I.cosθ/r²

→900 = I*1/x²

→I = 900x²

and for the second position, r = x+0.10 m, so

400 = I*1/(x+0.1)²

→I = 400*(x+0.1)²

Equating the two values of I, we get

900x² = 400(x+0.1)²

→(30x)² = {20(x+0.1)}²

→30x = 20(x +0.1) =20x + 2

→10x = 2

→x = 0.2 m = 20 cm.

        



11. A point source emitting light uniformly in all directions is placed 60 cm above a tabletop. The illuminance at a point on the tabletop, directly below the source, is 15 lux. Find the illuminance at a point on the tabletop 80 cm away from the first point.   

ANSWER: Let the luminous intensity of the source = I.

Distance = 60 cm 0.60 m. Illuminance = 15 lux.

E = I.cosθ/r² 

→15 = I. cos0°/(0.60)²

→15 = I/(0.60)²

→I = 15*(0.60)² 

The distance of the second point from the source

r =√{(60)² + (80)²} cm

=√{3600 + 6400} cm
=√{10000} cm
=100 cm
=1.0 m
Diagram for Q-11

If the angle of the normal from the direction of the light = θ, then cosθ = 60/100 =0.60

So the illuminance here = I.cosθ/r²

={15*(0.60)²}*0.60/1²

=15*0.60³ lux

=3.24 lux.          





12. Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incident light by an angle of 60°, by what fraction will the illuminance change?   


ANSWER: When the area is rotated about the incident light, neither the distance of the area from the source nor the angle between the normal to the area and the incident light is changed. Even after the rotation, the area is perpendicular to the incident light. So the illuminance will not change.         

13. A student studying a book placed near the edge of a circular table of radius R. A point source of light is suspended directly above the center of the table. What should be the height of the source above the table so as to produce maximum illuminance at the position of the book?   


ANSWER: Let the luminous intensity of the point source = I and the distance of the source from the table = r. The angle between the normal to the book and the incident light = θ.

The distance between the source and the book =r/cosθ.

cosθ = r/√(r²+R²)

Illuminance at the position of the book,

E = I*cosθ/(r/cosθ)² =I*cos³θ/r²

→E = I*r³/r²(r²+R²)³⁾² =I*r/(r²+R²)³⁾²

For maximum illuminance dE/dr = 0

I*[1*(r²+R²)³⁾² - r*(3/2)√(r²+R²)*2r]/(r²+R²)³ = 0
→√(r²+R²)*[(r²+R²) -3r²] = 0
→√(r²+R²)*[R²-2r²] = 0
Since √(r²+R²) can not be zero,
[R² - 2r²] = 0
→2r² = R²
→r² =R²/2
→r = R/√2
So the height of the source above the table should be = R/√2.

14. Figure (22-E1) shows a small diffused plane source S placed over a horizontal table-top at a distance of 2.4 m with its plane parallel to the tabletop. The illuminance at point A directly below the source is 25 lux. Find the illuminance at a point B of the table at a distance of 1.8 m from A.

The figure for Q-14
   


ANSWER:  In triangle ABS, AS =2.4 m, AB = 1.8 m. Hence BS = √(2.4²+1.8²) = √9 = 3 m.

Let angle ASB = θ,

→cosθ = 2.4/3 =0.8

Illuminance at A = 25 lux. Let the intensity along SA = Iₒ.

25 = I₀*cos0°/2.4²

→I₀ = 25*5.76 = 144 cd

Since it is a small extended source, from Lambert's cosine law, the intensity along SB, I = I₀*Cosθ
→I = 144*0.8 = 115.2 cd.  
The angle between SB and normal at B is also θ. Hence the illuminance at B,
E = I*cosθ/3² =115.2*0.8/9 Lux
→E = 10.24 Lux.  

 


15. An electric lamp and a candle produce equal illuminance at a photometer screen when they are placed at 80 cm and 20 cm from the screen respectively. The lamp is now covered with a thin paper which transmits 49% of the luminous flux. By what distance should the lamp be moved to balance the intensities at the screen again?  


ANSWER: Let the intensity of the electric lamp = I,

and the intensity of the candle = I'. For the equal illuminance,

I/80² =I'/20² 

→I = 16I'.

When the lamp is covered with a thin paper, The intensity reduces to 0.49I. Suppose now the lamp is moved by a distance x for equal illuminance.

0.49I/(80-x)² = I'/20²

→0.49*16I'/(80-x)² =I'/20²

→(80-x)² =16*20²*0.49 = 3136

→80-x = 56 cm

→x = 80 - 56 = 24 cm.          




16. Two light sources intensities 8 cd and 12 cd are placed on the same side of a photometer screen at a distance of 40 cm from it. Where should an 80 cd source be placed to balance the illuminance?   


ANSWER: Total intensity of the two sources I = 8+12 =20 cd. Its distance from the screen = 40 cm. Let the 80 cd source be placed at a distance x cm from the screen for equal illuminance. Now,

80/x² = 20/40²

→x² = 80*40²/20 = 80*80

→x = 80 cm.         

 



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Links to the Chapters



CHAPTER- 20 - Dispersion and Spectra


CHAPTER- 19 - Optical Instruments

CHAPTER- 18 - Geometrical Optics



CHAPTER- 17 - Light Waves




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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CHAPTER- 6 - Friction

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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-

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Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for EXERCISES (Question number 11 to 20)


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CHAPTER- 2 - "Physics and Mathematics"

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