JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 1)
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 $${n \over {40}}$$ cm. The value of n is _____.
The distance between an object and the objective lens, at which the strain on the eye is minimum is $${n \over {40}}$$ cm. The value of n is _____.
Answer
50
Explanation
_5th_September_Morning_Slot_en_1_1.png)
Given, L = |v0| + |ue| = 10 cm
(ve = $$\infty $$)
$${1 \over {{v_e}}} - {1 \over {{u_e}}} = {1 \over {{f_e}}}$$
$$ \Rightarrow $$ $${1 \over \infty } - {1 \over {{u_e}}} = {1 \over 5}$$
$$ \Rightarrow $$ ue = -5
$$ \Rightarrow $$ |ue| = 5 cm
$$ \therefore $$ |v0| + 5 = 10 cm
$$ \Rightarrow $$ |v0| = 5 cm
For objective, v0 = 5 cm, f0 = 1 cm
$${1 \over {{v_0}}} - {1 \over {{u_0}}} = {1 \over {{f_0}}}$$
$$ \Rightarrow $$ $${1 \over 5} - {1 \over {{u_0}}} = {1 \over 1}$$
$$ \Rightarrow $$ u0 = $$ - {5 \over 4}$$
$$ \Rightarrow $$ |u0| = $${5 \over 4}$$ = $${{50} \over {40}}$$ = $${n \over {40}}$$
$$ \therefore $$ n = 50
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