Question

the curvature radii of a concavo-convex glass lens are $20 \, cm$ and $60 \, cm$ . The convex surface of the lens is silvered. With the lens horizontal, the concave surface is filled with water. The focal length of the effective mirror is ( $\mu $ of glass = $1.5$ , $\mu $ of water = $\frac{4}{3}$ )

• A. $\frac{9 0}{1 3} \text{cm}$
• B. $\frac{8 0}{1 3} \text{cm}$
• C. $\frac{2 0}{3} \text{cm}$
• D. $\frac{4 5}{8} \text{cm}$

Answer 1

Answer: (A)

$R=60cm$ , $R_{2}=20cm$
Let $P_{1}$ be power of lens of water , $P_{2}$ be power of glass lens, and $P_{3}$ be power of concave mirror
$\mathrm{P}_{\mathrm{eq}}=2 \mathrm{P}_1+2 $
$\mathrm{P}_2+\mathrm{P}_3 $
$\frac{1}{-\mathrm{f}_{\mathrm{eq}}}=\frac{2}{\mathrm{f}_1}+\frac{2}{\mathrm{f}_2}-\frac{1}{\mathrm{f}_3}$
for the lens of water
$ \frac{1}{f_1}=\left(\frac{4}{3}-1\right)\left(\frac{1}{\infty}-\frac{1}{-60}\right) $
$\frac{1}{f_1}=\frac{1}{180}$
for glass lens
$\frac{1}{f_2}=\left(\frac{3}{2}-1\right)\left(\frac{1}{-60}-\frac{1}{-20}\right) $
$\frac{1}{f_2}=\frac{1}{60}$


For concave mirror $\mathrm{f}_3=\frac{\mathrm{R}}{2}$
$
\begin{array}{l}
\mathrm{f}_3=-10 \mathrm{~cm} \\
\frac{1}{-\mathrm{f}_{\text {eq }}}=\frac{2}{180}+\frac{2}{60}-\frac{1}{-10} \\
\frac{1}{-\mathrm{f}_{\text {eq }}}=\frac{26}{180} \Rightarrow \mathrm{f}_{\text {eq }}=\frac{-90}{13} \mathrm{~cm}
\end{array}
$