A) \[(n-1)f\]
B) \[(n+1)f\]
C) \[(n+1)f/n\]
D) \[(n-1)f/n\]
Correct Answer: D
Solution :
For mirror \[m=\frac{-v}{u}=\frac{h'}{h}=n\] For real image from concave mirror u and v both are negative So \[v=-nu\] \[\Rightarrow \] From mirror equation \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\Rightarrow \] \[\frac{1}{-nu}+\frac{1}{u}=\frac{1}{f}\] \[\frac{1}{u}\left( 1-\frac{1}{n} \right)=\frac{1}{f}\] \[f\left( \frac{n-1}{n} \right)=u\]
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