The two surfaces of a biconvex lens are of radius of curvature ' $\displaystyle R$ ' each. Obtain the condition under which its focal length ' $\displaystyle f$ ' be equal to ' $\displaystyle R$ '. If one of the two surfaces of this lens is made plane, what will be the new focal length of the lens?

$\displaystyle \frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ For $\displaystyle f=R$, $\displaystyle R_{1}=R$, $\displaystyle R_{2}=-R$: $\displaystyle \frac{1}{R}=(n-1)\left(\frac{1}{R}+\frac{1}{R}\right)$ $\displaystyle n=\frac{3}{2}$ For convex lens: $\displaystyle \frac{1}{f}=(n-1)\left(\frac{2}{R}\right)$ For plano-convex lens $\displaystyle R_{1}=R$, $\displaystyle R_{2}=\infty$: $\displaystyle \frac{1}{f_{p}}=(n-1)\left(\frac{1}{R}\right)$ $\displaystyle f_{p}=2 f$