Rohit has been given two converging lenses of focal lengths 1.25 cm and 5 cm, respectively to design a compound microscope. If it is desired to have a magnification of 30, what would be the separation between the objective and the eyepiece when the final image is formed at the least distance of distinct vision?

6.25 cm

The correct answer is Option (3) → 6.25 cm

Focal length of objective: $f_o=1.25 \,\text{cm}$

Focal length of eyepiece: $f_e=5 \,\text{cm}$

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

Total magnification: $M=30$

Magnification of compound microscope:

$M = M_o \cdot M_e$

For objective (approximate formula): $M_o = \frac{L}{f_o}$

For eyepiece: $M_e = 1 + \frac{D}{f_e}$

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

Substitute values:

$30 = \frac{L}{1.25}\left(1+\frac{25}{5}\right)$

$30 = \frac{L}{1.25}(1+5)$

$30 = \frac{L}{1.25} \cdot 6$

$30 = \frac{6L}{1.25}$

$30 = 4.8L$

$L = \frac{30}{4.8} = 6.25 \,\text{cm}$

Separation between objective and eyepiece = 6.25 cm