Question

In Fig. the liquids $L_1,L_2$ and $L_3$ have refractive indices 1.55,1.50 and 1.20 respectively. Therefore, the arrangement corresponds to

• A. biconvex lens
• B. biconcave lens
• C. concavo-convex lens
• D. convexo-concavo lens

Answer 1

Answer: (C)

For the first interface,
$\frac{{\mu }_2}{v_1}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$
$=\frac{1.5-1.55}{R_1}$
$=-\left(\frac{0.05}{R_1}\right)$
When $u=-\infty ,R_1$ is +ve.
$\Rightarrow v,$ is $-ve$.
It means the interface behaves as a concave lens. For the second interface,
$\frac{{\mu }_3}{v}-\frac{{\mu }_2}{u}=\frac{{\mu }_3-{\mu }_2}{R_2}$
$=\frac{1.2-1.5}{R_2}=\frac{-0.3}{R_2}$
When $u=-\infty ,R_2=-ve$,
$\Rightarrow v$ is $+ve$
It means the second interface acts like a convex lens. Hence the combination acts as a concavo-convex lens.