Question

A lens that has convex surface at one side and concave surface at another side behaves as :

a) convergent lens when the radius of curvature of convex surface is greater than that of concave surface.
b) divergent lens when the radius of curvature of concave surface is greater than that of convex surface.
c) convergent lens when the radius of curvature of convex surface is less than that of concave.
d) divergent lens when the radius of curvature of concave surface is less than that of convex surface.

• A. a, b are true
• B. a, d are true
• C. c,d are true
• D. b,c are true

Answer 1

The correct option is C c,d are true

Radii of curvature $R_1$, $R_2$ of its surface.$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

In the formula, assuming the surrounding medium is air as nothing is specified, $\mu -1>0$
When radius of curvature of convex surface is less than that of the concave surface as in figure 1,
$\frac{1}{R_1}-\frac{1}{R_2}>0$
So the focal length will be positive and the lens will act as a converging lens.
When radius of curvature of concave surface is less than that of the convex surface as in figure 2,
$\frac{1}{R_1}-\frac{1}{R_2}<0$
So the focal length will be negative and the lens will act as a diverging lens.