A) 200
B) 100
C) 400
D) 150
Correct Answer: A
Solution :
Given : \[{{f}_{0}}=1.2cm;{{f}_{e}}=3.0cm\] \[{{u}_{0}}=1.25cm;{{M}_{\infty }}=?\] From\[\frac{1}{{{f}_{0}}}=\frac{1}{{{v}_{0}}}-\frac{1}{{{u}_{0}}}\]\[\Rightarrow \]\[\frac{1}{1.2}=\frac{1}{{{v}_{0}}}-\frac{1}{(-1.25)}\] \[\Rightarrow \]\[\frac{1}{{{v}_{0}}}=\frac{1}{1.2}-\frac{1}{1.25}\]\[\Rightarrow \]\[{{v}_{0}}=30cm\] Magnification at infinity,\[{{M}_{\infty }}=-\frac{{{v}_{0}}}{{{u}_{0}}}\times \frac{D}{{{f}_{e}}}\] \[=\frac{30}{1.25}\times \frac{25}{3}\](\[\because D=25cm\]least distance of distinct vision) = 200 Hence the magnifying power of the compound microscope is 200
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