Question

An image of the sun is formed by a lens, of the focal length of $30cm$on a metal surface of a photoelectric cell and a photoelectric current $I$is produced. The lens forming the image is replaced by another of the same diameter but of focal length $15cm$. The photoelectric current in this case is:
A. $\dfrac{I}{2}$
B. $I$
C. $2I$
D. $4I$

Answer 1

Hint: Photoelectric current depends on the amount of radiation falling on the photoelectric surface.

Step by step answer:
We know that the amount of light passing through a lens depends on the diameter of the lens itself. The more the diameter of the lens, the more radiation (or sunlight in this case) will pass through it.
In this case, the focal length of the lens is decreased but the diameter of the lens remains unchanged.
So, the amount of sunlight falling on the photoelectric material will remain the same and the photoelectric current will also remain the same.
b) is correct.

Additional information:
Photoelectric effect is the emission of negatively charged particles (we call them electrons) from the surface of a material when a radiation of suitable energy falls on it. Every material surface has a threshold limit under which photoelectric effect does not take place. This threshold limit is known as the threshold frequency of that surface. If the frequency of the radiation is greater than the threshold frequency of the material, then the emission of electrons takes place. Threshold frequency is also known as work function in terms of energy.
When a photon of certain energy (greater than the work function) falls on a surface, some of its energy is used to overcome the work function. The remaining energy is transferred to the emitted electron (called a photoelectron) as its kinetic energy.

Note: One can argue that, when the focal length of the lens is decreased, the area exposed to the radiation also decreases which increases the intensity of the radiation and hence increasing the photoelectric current. But this is not true.
When the area exposed to the radiation decreases, the angle at which the radiation is hitting the surface also increases hence normalizing the effect of decreasing area and the intensity remains unchanged.