If we need a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eye-piece, should be close to:

To find the focal length of the eye-piece we first recall that the total magnification $$M$$ of a compound microscope is the product of the magnifications produced by the objective and by the eye-piece:

$$M \;=\; m_o \, m_e$$

For a compound microscope adjusted so that the final image is formed at the least distance of distinct vision $$D$$ (normally $$25\ \text{cm}$$) the two individual magnifications are given by the standard relations

Objective: $$m_o \;=\;\dfrac{L}{f_o}$$ (because the real image produced by the objective is approximately at the focal plane of the eye-piece and the image distance is almost equal to the tube length $$L$$)

Eye-piece (simple magnifier): $$m_e \;=\;1+\dfrac{D}{f_e}$$

Combining the two, the formula for the total magnification becomes

$$M \;=\;\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$$

Now we substitute the numerical values supplied in the problem:

Tube length    $$L \;=\;150\ \text{mm}\;=\;15\ \text{cm}$$

Focal length of objective    $$f_o \;=\;5\ \text{mm}\;=\;0.5\ \text{cm}$$

Desired magnification    $$M \;=\;375$$

Least distance of distinct vision    $$D \;=\;25\ \text{cm}$$

Substituting these numbers into the magnification formula we have

$$375 \;=\;\dfrac{15}{0.5}\left(1+\dfrac{25}{f_e}\right)$$

First evaluate the fraction $$\dfrac{15}{0.5}$$:

$$\dfrac{15}{0.5} \;=\;30$$

So the equation simplifies to

$$375 \;=\;30\left(1+\dfrac{25}{f_e}\right)$$

Divide both sides by 30:

$$\dfrac{375}{30} \;=\;1+\dfrac{25}{f_e}$$

$$12.5 \;=\;1+\dfrac{25}{f_e}$$

Subtract 1 from both sides to isolate the fractional term:

$$12.5-1 \;=\;\dfrac{25}{f_e}$$

$$11.5 \;=\;\dfrac{25}{f_e}$$

Cross-multiply to solve for $$f_e$$:

$$f_e \times 11.5 \;=\;25$$

$$f_e \;=\;\dfrac{25}{11.5}$$

Carrying out the division gives

$$f_e \;\approx\;2.17\ \text{cm}$$

To express this in millimetres (since the options are quoted in mm) we multiply by 10:

$$2.17\ \text{cm} \times 10 \;=\;21.7\ \text{mm}$$

This value is most nearly equal to $$22\ \text{mm}$$ in the list of given options.

Hence, the correct answer is Option A.