A biconvex lens $\left(\mathrm{R}_{1}=\mathrm{R}_{2}=20 \mathrm{~cm}\right)$ has focal length equal to focal…
A biconvex lens $\left(\mathrm{R}_{1}=\mathrm{R}_{2}=20 \mathrm{~cm}\right)$ has focal length equal to focal length of
concave mirror. The radius of curvature of concave mirror is
[R.I. of glass lens $=1 \cdot 5$ ]
$-40 \mathrm{~cm}$
$-20 \mathrm{~cm}$
$40 \mathrm{~cm}$
$20 \mathrm{~cm}$
Solution
$\begin{array}{l}
\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
n=1.5, R_{1}=20 \mathrm{~cm}, R_{2}=-20 \mathrm{~cm} \\
\therefore f=20 \mathrm{~cm}
\end{array}$
for a concave mirror, $R=2 f$ and it is negative.
$\therefore R=-40 \mathrm{~cm}$