JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Geometrical Optics

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In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eyepiece lens (in cm) is ____.

Answer: 6.25

Solution

For a compound microscope with the final image at the near point, the magnification is

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

, where

L

is the tube length,

f_{o}

the objective focal length,

f_{e}

the eyepiece focal length and

D = 25

cm. Putting in

M = 100

L = 20

cm and

f_{o} = 1

cm,

100 = \frac{20}{1} (1 + \frac{25}{f _{e}})

, so

1 + \frac{25}{f _{e}} = 5

, giving

\frac{25}{f _{e}} = 4

and

f_{e} = 6.25

cm.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Numerical answer

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