Q144 / 200 · Manipal (MET) 2012 · Physics
A thin plano-convex lens acts like a concave mirror of focal length 0.2 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
- A0.1 m
- B0.75 m
- C0.4 m
- Correct answerD0.2 m
Solution
After silvering the plane surface, plano-convex lens behaves as a concave mirror of focal length
$\begin{aligned}
& \frac{1}{F} & =\frac{2}{f_{\text {lens }}} \\
\text { But } & F & =0.2 \mathrm{~m} \\
\therefore \quad & f_{\text {lens }} & =2 F=2 \times 0.2=0.4 \mathrm{~m}
\end{aligned}$
Now, from lens maker's formula $\frac{1}{f_{\text {lens }}}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\begin{array}{ll}\therefore & \frac{1}{0.4}=(1.5-1) \times \frac{1}{R_1} \quad\left[\because R_2=\infty\right] \\ \Rightarrow & R_1=0.5 \times 0.4=0.2 \mathrm{~m}\end{array}$
Now, from lens maker's formula $\frac{1}{f_{\text {lens }}}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\begin{array}{ll}\therefore & \frac{1}{0.4}=(1.5-1) \times \frac{1}{R_1} \quad\left[\because R_2=\infty\right] \\ \Rightarrow & R_1=0.5 \times 0.4=0.2 \mathrm{~m}\end{array}$
Hence, the correct answer is option: 'D'.