JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Geometrical Optics

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An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to ____ cm.

Answer: 25

Solution

Initial magnification with simple microscope:

m_{1} = 1 + \frac{D}{f _{o}} = 6

, where

D = 25

cm. So

f_{o} = \frac{25}{5} = 5

cm. For compound microscope with final image at infinity:

m = \frac{v _{o}}{u _{o}} \times \frac{D}{f _{e}}

, where total magnification = 12. Tube length

L = v_{o} + f_{e} = 60

cm. For objective:

v_{o} - f_{o} = \frac{v _{o} f _{o}}{u _{o}}

, and magnification

\frac{v _{o}}{∣ u _{o} ∣} \times \frac{25}{f _{e}} = 12

. With

v_{o} = 60 - f_{e}

and solving:

f_{e} = 5

cm.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Geometrical Optics

Solution

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