question_answer

The radii of curvature of both the surfaces of a lens are equal. If one of the surfaces is made plane by grounding, how will the focal length and power change?

Answer:

                 For the original lens: \[{{R}_{1}}=+R\] and \[{{R}_{2}}=-R,\]so we can write \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{R}+\frac{1}{R} \right]=\frac{2[\mu -1]}{R}\] When one surface is made plane by grounding, we have \[{{R}_{1}}=+R\] and\[{{R}_{2}}=-\infty \]. Therefore, \[\frac{1}{f'}=(\mu -1)\left[ \frac{1}{R}+\frac{1}{\infty } \right]=\frac{\mu -1}{R}\] \[\therefore \]  \[\frac{f'}{f}=2\]               or \[f'=2f\] Thus the focal length becomes double and power becomes one-half.