Question

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20cm. The concave surface has a radius of curvature 60 cm. the convex side is silvered and placed on a horizontal surface.
Where should a pin be placed on the optic axis such that its image is formed at the same place?

• A. 15 cm
• B. 30 cm
• C. 40 cm
• D. 45 cm

Answer 1

The correct option is A 15 cm

For lens, $\frac{1}{fy}=(n-1)[\frac{1}{R_1}-\frac{1}{R_2}]=(1.5-1)[\frac{1}{120}-\frac{1}{60}]=\frac{1}{60}$
$\Rightarrow fg=60cm$
and $fm=\frac{R_1}{2}=\frac{20}{2}=10cm\frac{1}{F}=\frac{1}{60}+\frac{1}{10}+\frac{1}{60}=\frac{8}{60}\Rightarrow F=\frac{60}{8}=7.5cm$
For the image to be formed at the place of the object, $X=R=2F=7.5\times 2=15cm$