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 :
Solution
<b>Case 1 :</b> Near – point adjustment
<br><br>M.P = ${L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$
<br><br>$\Rightarrow$ 375 = ${{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)$
<br><br>$\Rightarrow$ f<sub>e</sub> = 21.7 mm $\approx$ 22 mm
<br><br><b>Case-2 :</b>
If final image is at inifinity
<br><br>M.P = ${L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)$
<br><br>$\Rightarrow$375 = ${{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)$
<br><br>$\Rightarrow$ f<sub>e</sub> = 20 mm
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
This question is part of PrepWiser's free JEE Main question bank. 197 more solved questions on Optics are available — start with the harder ones if your accuracy is >70%.