1. Key idea: photoelectric current depends on intensity

   For a given light frequency above threshold, the number of photoelectrons emitted per second is directly proportional to the intensity of incident light.

   $N \propto I$

2. Intensity variation with distance

   For a small bright source, intensity varies inversely as the square of the distance:

   $I \propto \frac{1}{r^2}$

3. Compare intensities at the two positions

   • Initially, source is at: $r_1 = 1\,\text{m}$
   • Then moved to: $r_2 = \frac{1}{2}\,\text{m}$

   Therefore,

   $\frac{I_2}{I_1} = \frac{1/r_2^2}{1/r_1^2} = \frac{r_1^2}{r_2^2}$

   Substituting values:

   $\frac{I_2}{I_1} = \frac{1^2}{\left(\frac{1}{2}\right)^2} = \frac{1}{1/4} = 4$

4. Effect on number of emitted electrons

   Since number of emitted electrons is proportional to intensity,

   $\frac{N_2}{N_1} = \frac{I_2}{I_1} = 4$

   So the number of electrons emitted increases by a factor of 4.

5. Option check

   • A: increases by a factor of $4$ ✅
   • B: decreases by a factor of $4$ ❌
   • C: increases by a factor of $2$ ❌
   • D: decreases by a factor of $2$ ❌

Derived answer: Option A.