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 = 41.5You need to login to perform this action.
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