MHT CET · Physics · Ray Optics
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 \(=1 \cdot 5\) ]
- A \(-40 \mathrm{~cm}\)
- B \(-20 \mathrm{~cm}\)
- C \(40 \mathrm{~cm}\)
- D \(20 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(-40 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
n=1.5, R_{1}=20 \mathrm{~cm}, R_{2}=-20 \mathrm{~cm} \\
\therefore f=20 \mathrm{~cm}
\end{array}
\)
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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