A) 1.50
B) 1.55
C) 1.60
D) 1.65
Correct Answer: B
Solution :
In the first case we have\[\frac{1}{28}=\frac{2}{{{f}_{1}}}+\frac{1}{{{f}_{m}}}=\frac{2}{{{f}_{1}}}\] In the second case we have \[\frac{1}{10}=\frac{2}{{{f}_{1}}}+\frac{1}{{{f}_{m}}}\]where \[{{f}_{m}}\]is the focal length of silvered surface. By subtraction:\[\frac{1}{{{f}_{m}}}=\frac{1}{10}=\frac{1}{28}\] or, \[R=2{{f}_{m}}=\frac{280}{9}\,cm.\]\[\therefore \]\[{{f}_{1}}=28\times 2=56\,cm\] \[\frac{1}{{{f}_{1}}}=(\mu -1).\frac{1}{R}\]or \[\mu -1=\frac{R}{{{f}_{1}}}\]or \[\mu =1.55\] Hence, the correction option is [b].You need to login to perform this action.
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