JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 7)
If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :
22 mm
12 mm
33 mm
2 mm
Explanation
Case 1 : Near – point adjustment
M.P = $${L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$$
$$ \Rightarrow $$ 375 = $${{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)$$
$$ \Rightarrow $$ fe = 21.7 mm $$ \approx $$ 22 mm
Case-2 : If final image is at inifinity
M.P = $${L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)$$
$$ \Rightarrow $$375 = $${{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)$$
$$ \Rightarrow $$ fe = 20 mm
M.P = $${L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$$
$$ \Rightarrow $$ 375 = $${{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)$$
$$ \Rightarrow $$ fe = 21.7 mm $$ \approx $$ 22 mm
Case-2 : If final image is at inifinity
M.P = $${L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)$$
$$ \Rightarrow $$375 = $${{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)$$
$$ \Rightarrow $$ fe = 20 mm
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