A lens is 5 cm thick and its radius of curvature of its surfaces are 10cm and 25 cm respectively. A point object is placed at a distance 12 cm from the surface whose radius of curvature is 10 cm. How far beyond the other surface image is formed?

95 cm

We know that $\frac{μ_2}{v}-\frac{μ_1}{u}=\frac{μ_2-μ_1}{R}$

u = –12 cm, R =10 cm, $μ_1 = 1, μ_2 = 1.5$

$\frac{1.5}{v}-\frac{1}{-12}=\frac{1.5-1}{10}⇒v=-45cm$

This image will service as an object for the second surface. For the second surface = object distance

u = 5 + 45 = 50 cm

For the second surface again

u = – 50 cm, R = -25 cm, $μ_1 = 1.5, μ_2 = 1$

$\frac{μ_2}{v}-\frac{μ_1}{u}=\frac{μ_2-μ_1}{R}⇒\frac{1}{v}-\frac{1.5}{-50}=\frac{1-1.5}{-25}$

or v = –100 cm

Final image will be at a distance –95 cm from the first surface on the same side as the objects.