Question

The convex surface of a thin concavo-convex lens of glass of
refractive index 1.5 has a radius of curvature 20 cm. The
concave surface has a radius of curvature 60 cm. The convex
side is silvered and placed on a horizontal surface
(a) Where should a pin be placed on the optic axis such that
its image is formed at the same place?
(b) If the concave part is filled with water of refractive index
4/3, find the distance through which the pin should be
moved, so that the image of the pin again coincides with
the pin.

Answer 1

(a) lmage of object will coincide with it if ray of light after
refraction from the concave surface fall normally on
concave mirror so formed by silvering the convex
surface. Or image after refraction from concave surface
should form at centre of curvature of concave mirror or at
a distance of 20 cm on same side of the combination. Let
x be the distance of pin from the given optical system.
Applying
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
With proper signs
$\frac{1.5}{-20}-\frac{1}{-x}=\frac{1.5-1}{-60}$
or $\frac{1}{x}=\frac{3}{40}-\frac{1}{120}=\frac{8}{120}$
$\therefore x=\frac{120}{8}=15cm$
(b) Now, before striking with the concave surface, the ray is
first refracted from a plane surface. So, let x be the
distance of pin, then the plane surface will form its image
at a distance $\frac{4}{3}x(h_{app.}=\mu h)$ from it.
Now, using $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$ with proper signs,
we have $\frac{1.5}{-20}-\frac{4/3}{-\frac{4x}{3}}=\frac{1.5-4/3}{-60}$
or $\frac{1}{x}=\frac{3}{40}-\frac{1}{360}$
or x = 13.84 cm
$\therefore \Delta x=x_1-x_2=$ 15 cm - 13.84 cm
$=1.16cm$ (downwards)