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A thin plano-convex lens acts like a concave mirror of focal length $0.2 \mathrm{~m}$ when silvered from its plane surface. The refractive index of the material of the lens is $1.5$. The radius of curvature of the convex surface of the lens will be
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Verified Answer
The correct answer is:
$0.1 \mathrm{~m}$
Resultant focal length $(f)$ of the given combination is $0.2 \mathrm{~m}$.
Refractive index of the material, $\mu=1.5$
From the lens Maker formula, we get
$$
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
$$
where, $R_{1}$ is the radius of curvature of convex surface and $R_{2}$ is the radius of curvature of plane surface $\left(R_{2}=\infty\right)$.
Hence,
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{f}=(15-1)\left(\frac{1}{R_{1}}-\frac{1}{\infty}\right) \\
\Rightarrow \quad & \frac{1}{0.2}=(0.5)\left(\frac{1}{R_{1}}\right) \\
\Rightarrow \quad & R_{1}=0.5 \times 0.2=0.1 \mathrm{~m}
\end{array}
$$
Refractive index of the material, $\mu=1.5$
From the lens Maker formula, we get
$$
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
$$
where, $R_{1}$ is the radius of curvature of convex surface and $R_{2}$ is the radius of curvature of plane surface $\left(R_{2}=\infty\right)$.
Hence,
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{f}=(15-1)\left(\frac{1}{R_{1}}-\frac{1}{\infty}\right) \\
\Rightarrow \quad & \frac{1}{0.2}=(0.5)\left(\frac{1}{R_{1}}\right) \\
\Rightarrow \quad & R_{1}=0.5 \times 0.2=0.1 \mathrm{~m}
\end{array}
$$
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