JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Geometrical Optics

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

(A)22 mm✓ Correct
(B)33 mm
(C)12 mm
(D)2 mm

Answer: (A)

Solution

The magnifying power of a compound microscope with the final image at the least distance of distinct vision

D = 25 cm

M = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})

. With

M = 375

, tube length

L = 150 mm

and

f_{o} = 5 mm

the prefactor is

\frac{L}{f _{o}} = 30

, so

375 = 30 (1 + \frac{25}{f _{e}})

, giving

\frac{25}{f _{e}} = 11.5

and

f_{e} \approx 2.17 cm \approx 22 mm

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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