JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 20)
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 eye-piece lens (in cm) is __________.
Answer
6.25
Explanation
L = 20, f0 = 1cm, M = 100
$$M = {{{v_0}} \over {{u_0}}}\left( {1 + {D \over {{f_e}}}} \right)$$
$$ \therefore $$ $$M = {L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$$ [v0 $$ \approx $$ L, u0 $$ \approx $$ f0]
$$ \Rightarrow $$ $${{20} \over 1}\left( {1 + {{25} \over {{f_e}}}} \right)$$ = 100
on solving we get
fe = 6.25 cm
$$M = {{{v_0}} \over {{u_0}}}\left( {1 + {D \over {{f_e}}}} \right)$$
$$ \therefore $$ $$M = {L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$$ [v0 $$ \approx $$ L, u0 $$ \approx $$ f0]
$$ \Rightarrow $$ $${{20} \over 1}\left( {1 + {{25} \over {{f_e}}}} \right)$$ = 100
on solving we get
fe = 6.25 cm
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