An equiconvex lens is made of glass of refractive index 1.55. If the focal length of the lens is 15 · 0 cm, calculate the radius of curvature of its…

$\displaystyle \frac{1}{f}=\left(n_{21}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$ $\displaystyle R_{1}=R$ and $\displaystyle R_{2}=-R$ $\displaystyle \frac{1}{15}=(1.55-1)\left[\frac{2}{R}\right]$ $\displaystyle R=16.5 \mathrm{~cm}$