A photosensitive surface is illuminated by a point source of light 1 m away. When the source is shifted to 2 m, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → number of emitted electrons is a quarter of the initial number
Given: A point source of light is moved from 1 m to 2 m away from a photosensitive surface.
Since light intensity varies inversely as the square of distance,
$I \propto \frac{1}{r^2}$
Therefore,
$\frac{I_2}{I_1} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
Photoelectric current (number of emitted electrons per second) ∝ Intensity.
Thus, when the source distance doubles, the number of emitted electrons becomes one-fourth of the original.
Energy of each emitted electron depends only on frequency of light, not on intensity.
Final Answer: number of emitted electrons is a quarter of the initial number.