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Question: Answered & Verified by Expert
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 $=1 \cdot 5$ ]
PhysicsRay OpticsMHT CETMHT CET 2020 (13 Oct Shift 1)
Options:
  • A $-40 \mathrm{~cm}$
  • B $-20 \mathrm{~cm}$
  • C $40 \mathrm{~cm}$
  • D $20 \mathrm{~cm}$
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.
$$
\therefore R=-40 \mathrm{~cm}
$$

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