question_answer

A compound microscope is 20 cm long and the power of its eye-piece is 20 D. If the objective is bi-convex (radius of curvature of either surface 2 cm) and made of a material of = 1.5, the magnifying power of the microscope is

A)  54               

B)  60

C)  13.5             

D)  41.5

Correct Answer: D

Solution :

: For objective lens, \[\frac{1}{{{f}_{0}}}=(\mu -1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)=(1.5-1)\times \frac{2}{R}=\frac{0.5\times 2}{2}=\frac{1}{2}\] Or \[{{f}_{0}}=2\,cm\]                       ...(i) \[{{f}_{c}}=\frac{100}{D}=\frac{100}{20}=5\,cm\] For eyepiece,\[\frac{1}{{{v}_{e}}}-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}}\]Final image is at 25 cm from it. \[\therefore \]\[\frac{-1}{25}-\frac{1}{{{u}_{e}}}=\frac{1}{5}\] or\[{{u}_{e}}=-\frac{25}{6}cm=\]Image is on the object (virtual) side of lens. \[\therefore \]Tube length\[={{u}_{e}}+{{v}_{0}}\] or \[{{v}_{0}}=20-\frac{25}{6}=\frac{95}{6}cm\]                ...(iii) For objective lens, \[\frac{1}{{{v}_{0}}}-\frac{1}{{{u}_{0}}}=\frac{1}{{{f}_{0}}}\] Or \[\frac{1}{{{u}_{0}}}=\frac{1}{{{v}_{0}}}-\frac{1}{{{f}_{0}}}=\frac{6}{95}-\frac{1}{2}=\frac{12-95}{190}=\frac{-83}{190}\] Or \[{{u}_{0}}=\frac{-190}{83}cm\] \[\therefore \]Magnifying power of microscope \[=\frac{{{v}_{0}}}{{{u}_{0}}}\left( 1+\frac{D}{{{f}_{e}}} \right)\] Here final image is formed at least distance of distinct vision (25 cm) from eyepiece. The final image is virtual and magnified. \[M=\frac{{{v}_{0}}}{{{u}_{0}}}\left( 1+\frac{D}{{{f}_{e}}} \right)\] \[M=\frac{95/6}{190/83}\left( 1+\frac{25}{5} \right)=\frac{95}{6}\times \frac{83}{190}(1+5)\] \[=\frac{83\times 6}{6\times 2}=41.5\] Magnifying power = 41.5