In a microscope the objective is having focal length $f_{0}=2 \mathrm{~cm}$ and eye-piece is having focal length $f_{\mathrm{e}}=4 \mathrm{~cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ______.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Geometrical Optics

In a microscope the objective is having focal length

f_{0} = 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 ______.

Answer: 100

Solution

In normal adjustment the compound-microscope magnification is

m = \frac{L}{f _{0}} \cdot \frac{D}{f _{e}}

, with tube length

L = 32 cm

, least distance of distinct vision

D = 25 cm

f_{0} = 2 cm

and

f_{e} = 4 cm

. Hence

m = \frac{32}{2} \times \frac{25}{4} = 16 \times 6.25 = 100

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Integer answer

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