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If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to : 
 
Option: 1 12mm 
Option: 233mm
Option: 3 22mm  
Option: 4 2mm
 

Answers (1)

best_answer

The magnification is given by

 M=\frac{L}{f_0}(1+\frac{D}{f_e})\\ \Rightarrow 375=\frac{150}{5}(1+\frac{25}{f_e})\\ \Rightarrow f_e=22 \ mm

So option (3) is correct.

Posted by

Ritika Jonwal

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