Question

A thin equi-convex lens (radius of curvature of either face being 33 cm) is placed on a horizontal plane mirror and a pin held 20 cm vertically above the lens coincides in position with its own image. The space between the lower surface of the lens and the mirror is filled with a liquid and then, to coincide with the image as before, the pin has to be raised to a distance of 25 cm from the lens. Find the refractive index of the liquid.

Solution

Given, a thin equi-convex lens.
Radius of curvature, R = 33 cm
Distance of the pin from the lens = 20 cm

In the first case, the image will coincide with the pin if the rays from the pin, after refraction through the lens, fall normally on the mirror and retrace their path, as shown in Fig.(a).

This means that the focal length of the convex lens is 20 cm.
i.e., f = 20 cm

In the second case, the focal length F of the combination of the convex lens and the plano-concave liquid lens is 25 cm.
i.e., F = 25 cm

Let f₂ be the focal length of the liquid lens, then

\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}

\Rightarrow

\frac{1}{f_{2}} = \frac{1}{F} - \frac{1}{f_{1}} = \frac{1}{25} - \frac{1}{20}

f_{2} = - 100 cm

For the liquid lens,
Radius of curvature of the common surface, R₁ = –33 cm,
R₂ = ∞.

If μ is refractive index of the liquid.
Using Lens makers formula,

\frac{1}{f_{2}} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

- \frac{1}{100} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

\Rightarrow

μ - 1 = \frac{33}{100} = 0.33

i.e.,

μ = 1.33 .

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