Question

the curvature radii of a concavo-convex glass lens are $20cm$ and $60cm$ . 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. $\frac{90}{13}\text{cm}$
• B. $\frac{80}{13}\text{cm}$
• C. $\frac{20}{3}\text{cm}$
• D. $\frac{45}{8}\text{cm}$

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 ${\mathrm{f}}_3=\frac{\mathrm{R}}{2}$
$<br/>\begin{array}{l}<br/>{\mathrm{f}}_3=-10\mathrm{cm}\\ <br/>\frac{1}{-{\mathrm{f}}_{\text{eq }}}=\frac{2}{180}+\frac{2}{60}-\frac{1}{-10}\\ <br/>\frac{1}{-{\mathrm{f}}_{\text{eq }}}=\frac{26}{180}\Rightarrow {\mathrm{f}}_{\text{eq }}=\frac{-90}{13}\mathrm{cm}<br/>\end{array}<br/>$