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QJEEM20Phf744: JEE Main 2020 Physics - Class 12 JEEM20Phf744 Solution: A compound microscope consists of an objective lens of focal…

A compound microscope consists of an objective lens of focal length 1 cm and an eyepiece of focal length 5 cm with a separation of 10 cm.
The distance between an object and the objective lens, at which the strain on the eye is minimum is n40{n \over {40}} cm. The value of n is _____.

Step-by-Step Verified Solution

JEE Main 2020 (Online) 5th September Morning Slot Physics - Geometrical Optics Question 141 English Explanation

Given, L = |v0| + |ue| = 10 cm

(ve = ∞\infty )

1ve−1ue=1fe{1 \over {{v_e}}} - {1 \over {{u_e}}} = {1 \over {{f_e}}}

⇒\Rightarrow 1∞−1ue=15{1 \over \infty } - {1 \over {{u_e}}} = {1 \over 5}

⇒\Rightarrow ue = -5

⇒\Rightarrow |ue| = 5 cm

∴\therefore |v0| + 5 = 10 cm

⇒\Rightarrow |v0| = 5 cm

For objective, v0 = 5 cm, f0 = 1 cm

1v0−1u0=1f0{1 \over {{v_0}}} - {1 \over {{u_0}}} = {1 \over {{f_0}}}

⇒\Rightarrow 15−1u0=11{1 \over 5} - {1 \over {{u_0}}} = {1 \over 1}

⇒\Rightarrow u0 = −54- {5 \over 4}

⇒\Rightarrow |u0| = 54{5 \over 4} = 5040{{50} \over {40}} = n40{n \over {40}}

∴\therefore n = 50

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