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The illumination produced by A is balanced by B on the screen when B is 60 cm apart from the screen. A smoked glass plate is placed in front of A and to balance the illumination by B is to move 15cm further away. Find the transmission coefficient of the smoked glass.
1. 0.36
2. 0.64
3. 0.49
4. 0.51

Answer
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Hint: Transmission coefficient can be calculated in two ways, we can use the amplitude or the intensity of the wave to calculate the transmission coefficient of a material. It is generally taken as the ratio of the values after the wave passes the material and before the wave passes the material.

Complete step-by-step answer:
Distance of B from the screen,
${{d}_{1}}=60cm$
Distance of B increased to balance the illumination,
${{d}_{i}}=15cm$
Final distance of B from the screen,
${{d}_{2}}={{d}_{1}}+{{d}_{i}}$
${{d}_{2}}=(60+15)cm$
${{d}_{2}}=75cm$
Now, according to the problem
$\alpha $ (Transmission coefficient)\[=\dfrac{{{I}_{1}}}{{{I}_{2}}}={{\left( \dfrac{{{d}_{1}}}{{{d}_{2}}} \right)}^{2}}\]
$\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}={{\left( \dfrac{60}{75} \right)}^{2}}$
$\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}={{\left( \dfrac{4}{5} \right)}^{2}}$
$\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{16}{25}$
$\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}=0.64$
$\alpha $ (Transmission coefficient)=0.64(Answer)
Therefore, the transmission coefficient of smoked glass is 0.64.
$\therefore $ Option 2, is the correct option.

Additional Information
The transmission coefficient gives us the amplitude, Intensity / the total power of a wave.
The transmission coefficient is a measurement of how much of an electromagnetic wave which is actually light, passes through a surface, material or an optical element.
Transmission coefficients can be calculated for both the amplitude and the intensity of the wave.

Note: We can figure out the value with the help of amplitude also, students generally make the mistake thinking that A is on one side of the material and B on the other. The ratio of the two values does not contain any units.