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?
Answer & explanation
Correct answer: option 3
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