1. Key idea: photoelectric current depends on total light power falling on the cell

   The photoelectric current is proportional to the number of photoelectrons emitted per second, which in turn depends on the total light energy per second incident on the photoelectric cell (assuming frequency is above threshold and stopping conditions are unchanged).

2. What does the lens do here?

   The lens forms an image of an extended source on the photoelectric cell.

   For an extended source, the illumination at the image formed by a lens depends on the f-number or equivalently on the factor

   $$\left(\frac{D}{f}\right)^2,$$

   where $D$ is the lens diameter and $f$ is the focal length.

   Since the two lenses have the same diameter, reducing focal length from $40\,\text{cm}$ to $20\,\text{cm}$ would increase image illumination by a factor

   $$\left(\frac{f_1}{f_2}\right)^2 = \left(\frac{40}{20}\right)^2 = 4.$$

3. But the image size also changes

   For an extended object, the linear magnification is proportional to image distance, and in this situation the image size scales with focal length. So when focal length is halved, the linear dimensions of the image are halved.

   Hence the image area becomes

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

   times the original.

4. Total light falling on the cell

   Total light power on the image is

   $$\text{(illumination)} \times \text{(image area)}.$$

   Therefore,

   $$P' = 4 \times \frac{1}{4} P = P.$$

   So the total light energy incident on the photoelectric cell remains unchanged.

5. Hence photoelectric current

   Since the total incident light power is unchanged,

   $$I' = I.$$

6. Option check

   • A: $\frac{I}{2}$ — incorrect
   • B: $4I$ — incorrect
   • C: $2I$ — incorrect
   • D: $I$ — correct

Final Answer: $I' = I$ So the correct option is D.