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Physics 3 Marks Question

Photometry · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 4 questions.

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Question 13 Marks
A source emits light of wavelengths $555\ nm$ and $600\ nm$. The radiant flux of the $555\ nm$ part is $40W$ and of the $600\ nm$ part is $30W$. The relative luminosity at $600\ nm$ is $0\ '6$. Find:
  1. The total radiant flux.
  2. The total luminous flux.
  3. The luminous efficiency.
Answer
The radiant flux of $555\ nm$ part is $40W$ and of the $600\ nm$ part is $30W$
  1. Total radiant flux $= 40W + 30W = 70W$
  2. Luminous flux $= (L.$Fllux$)_{555\ nm} + (L.$Flux$)_{600\ nm}$
=$ 1 \times 40 \times 685 + 0.6 \times 30 \times 685 = 39730$ lumen
  1. Luminous efficiency $=\frac{\text{Total luminous flux}}{\text{Total radiant flux}}=\frac{39730}{70}=567.6\text{ lumen/W}$
So, the luminous efficiency is $568 \text{ lumen/W}$
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Question 23 Marks
The luminous flux of a 1W sodium vapour lamp is more than that of a 10kW source of ultraviolet radiation. Comment.
Answer
The luminous flux of a 1W sodium vapour lamp is more than that of a 10kW source of ultraviolet radiation. This is because the luminosity of a 1W sodium vapour lamp is 589nm or 589.6nm, whereas the luminosity of UV radiation is zero.
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Question 33 Marks
The relative luminosity of wavelength 600nm is 0.6. Find the radiant flux of 600nm needed to produce the same brightness sensation as produced by 120W of radiant flux at 555nm.
Answer
Relative luminousity $=\frac{\text{Luminous flux of a source of given wavelength}}{\text{Luminous flux of a source of 555nm of same power }}$Let the radiant flux needed be 'P' watt.
$\text{Ao},0.6=\frac{\text{Luminous flux of a source P watt}}{\text{685P}}$
$\therefore$ Luminous flux of the source = (685 P) × 0.6 = 120 × 685
$\Rightarrow\text{P}=\frac{120}{0.6}=200\text{W}$
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Question 43 Marks
A point source emitting light uniformly in all directions is placed 60cm above a table-top. The illuminance at a point on the table-top, directly below the source, is 15 lux. Find the illuminance at a point on the table-top 80cm away from the first point.
Answer

given that $\text{E}_\text{a}=15\text{ lux}=\frac{\text{l}_0}{60^2}$
$\Rightarrow\text{l}_0=15\times(0.6)^2=5.4\text{ candela}$
 So, $\text{E}_\text{B}=\frac{\text{l}_0\cos\theta}{(\text{OB})^2}=\frac{5.4\times\Big(\frac{3}{5}\Big)}{1^2}=3.24\text{ lux}$
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