2 Marks Questions

Question 12 Marks

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

Answer

To get equally intense lines on the photographic plate, the radiant flux (energy) should be same. So, 10W × 12sec = 12W × t$\Rightarrow \text{t}=\frac{10\text{W}\times12\text{sec}}{12\text{W}}=10\text{sec.}$

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Question 22 Marks

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

Answer

The luminous flux of the given source of 1W is 450 lumen/ watt$\therefore$ Relative luminosity $=\frac{\text{Luminous flux of the source of given wavelength}}{\text{Luminous flux of 555nm source of same power}}=\frac{450}{685}=66\%$
So, the relative luminosity at the wavelength emitted is 66%.$\big[\therefore$ Since, luminous flux of 555nm source of 1W = 685 lumen$\big]$

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Question 32 Marks

A source emits 31.4W 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 = 31.4W, Solid angle $=4\pi$Luminous efficiency = 60 lumen/W
So, Luminous flux = 60 × 31.4 lumen
And luminous intensity $=\frac{\text{Luminous Flux}}{4\pi}=\frac{60\times31.4}{4\pi}=150\text{ candela}$

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Question 42 Marks

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

Answer

In a filament bulb, heat energy is converted into light energy. Most of the electrical energy supplied to the bulb is radiated as heat and only a small percentage is radiated as visible light. Since heat waves are not visible, they don't contribute to visibility. Thus, its luminous flux is lesser than the total radiant flux; hence, its luminous efficiency is low.
In a mercury vapour lamp, a greater amount of electrical energy supplied is converted into visible radiation. Thus, its luminous flux is relatively higher than incandescent lamps.

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Question 52 Marks

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

Answer

No, we cannot increase the illuminating power of any white light source by wrapping a yellow plastic around them.White light emits power in all wavelengths of visible radiation, and the yellow plastic will block all the other wavelengths other than the yellow one. This means that it will allow only a small amount of power, emitted by the white light, in the yellow coloured wavelength and block the rest. Thus, it will decrease the illuminating power of the source.

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Question 62 Marks

As the wavelength is increased from violet to red, the luminosity:

Continuously increases.
Continuously decreases.
Increases then decreases.
Decreases then increases.

Answer

Increases then decreases.

Explantion:
The luminosity first increases up to 555nm and then decreases.

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Question 72 Marks

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

Answer

I = luminous intensity $=\frac{628}{4\pi}=50\text{ candela}$$\text{r}=1\text{m}, $
$\theta=37^\text{o}$
So, illuminance, $\text{E}=\frac{\text{l}\cos\theta}{\text{r}^2} =\frac{50\times\cos37^\text{o}}{1^2}=40\text{ lux}$

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Question 82 Marks

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 area just below the bulb that is normal to the table.
Yes, the plane mirror will increase the radiant flux on the table, which will further increase the luminance flux of the table. Therefore, illuminance will increase.

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