In a microscope the objective is having focal length f_o=2 ~cm and eye-piece is having focal length f_e=4 ~cm . The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is \_\_\_\_ .

Correct answer: 100. For microscope at normal adjustment: M = L/f_o × D/f_e where L = 32 cm, f_o = 2 cm, f_e = 4 cm, D = 25 cm M = 32/2 × 25/4 = 16 × 6.25 = 100

Q72 / 75 · JEE Main 2026 (21 January Shift 1) · Physics

In a microscope the objective is having focal length $f_{o}=2 \mathrm{~cm}$ and eye-piece is having focal length $f_{e}=4 \mathrm{~cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is $\_\_\_\_$.

Correct answer: 100

Solution

For microscope at normal adjustment:
$M = \frac{L}{f_o} \times \frac{D}{f_e}$
where $L = 32$ cm, $f_o = 2$ cm, $f_e = 4$ cm, $D = 25$ cm
$M = \frac{32}{2} \times \frac{25}{4} = 16 \times 6.25 = 100$

Hence, the correct answer is 100.

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