question_answer

The focal length of an equiconvex lens e is equal to the radius of curvature of either face. What is the refractive index of the lens material?

Answer:

                Here \[{{R}_{1}}=+R,\]   \[{{R}_{2}}=-R\]and \[f=R\]                 \[\because \]     \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\]                 \[\because \]     \[\frac{1}{R}=(\mu -1)\left[ \frac{1}{R}+\frac{1}{R} \right]=\frac{(\mu -1)2}{R}\]                 or            \[\mu =\mathbf{1}\mathbf{.5}\]