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.