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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 $=1 \cdot 5$ ]
Options:
concave mirror. The radius of curvature of concave mirror is
[R.I. of glass lens $=1 \cdot 5$ ]
Solution:
1613 Upvotes
Verified Answer
The correct answer is:
$-40 \mathrm{~cm}$
$$
\begin{array}{l}
\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
n=1.5, R_{1}=20 \mathrm{~cm}, R_{2}=-20 \mathrm{~cm} \\
\therefore f=20 \mathrm{~cm}
\end{array}
$$
for a concave mirror, $R=2 f$ and it is negative.
$$
\therefore R=-40 \mathrm{~cm}
$$
\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}
$$
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