question_answer

    A compound microscope has an eyepiece of focal length 10 cm and an objective of focal length 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective, so that final image is formed at the least distance of distinct vision 20 cm.

A)  12                                         

B)  11

C)  10                                         

D)  13

Correct Answer: A

Solution :

                The following ray diagram shows the formation of image by a compound microscope. Given, \[{{f}_{e}}=10\,cm,{{f}_{o}}=4\,cm,{{u}_{o}}=-5\,cm,\] \[D=20\text{ }cm\] For objective lens                 \[\frac{1}{{{f}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}\]                 \[\frac{1}{4}=\frac{1}{{{v}_{o}}}-\frac{1}{-5}\] \[\Rightarrow \]               \[\frac{1}{{{v}_{o}}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\] \[\Rightarrow \]               \[{{v}_{o}}=20\,cm\] Magnification, \[M=-\frac{{{v}_{o}}}{{{u}_{o}}}\left( 1+\frac{D}{{{f}_{e}}} \right)\]                                 \[=\frac{20}{-5}\left( 1+\frac{20}{10} \right)\]                                 \[=4(1+2)=12\]