A) \[-\frac{f\,n'(n-1)}{n'-n}\]
B) \[-\frac{f(\,n'-n)}{n'(n-1)}\]
C) \[-\frac{n'(\,n-1)}{f\,(n'-n)}\]
D) \[\frac{fn'n}{n-n'}\]
Correct Answer: A
Solution :
\[\frac{1}{f}=\left( \frac{n-1}{1} \right)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]and \[\frac{1}{f'}=\left( \frac{n-n'}{n'} \right)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\therefore \] \[\frac{f'}{f}=\frac{n-1}{1}\times \frac{n'}{n-n'}\] \[\Rightarrow \]\[f=\frac{fn'(n-1)}{n'-n}\]
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