An equi-convex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle with its tip on the principal axis of the lens is moved along the axis until its inverted real image coincides with the needle itself. The distance of needle from lens is measured to be a. On removing the liquid layer and repeating the experiment, the distance is found to b. Given that two values of distances measured represent the real length values in the two cases, obtain a formula for refractive index of the liquid.

Solution

Given, an equi-convex lens and a pane mirror.

Here, combined focal length of glass lens and liquid lens, F = a, and
Focal length of convex lens, f₁ = b.
If f₂ is focal length of liquid lens, then

\frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{F} \frac{1}{f_{2}} = \frac{1}{F} - \frac{1}{f_{1}} = \frac{1}{a} - \frac{1}{b}

The liquid lens is plano-concave lens for which,
Radius of curvature,

R_{1} = - r, R_{2} = \infty

Using the lens makers formula,

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

i.e.,

\frac{1}{a} - \frac{1}{b} = (μ - 1) (\frac{1}{- r} - \frac{1}{\infty})

∴

(μ - 1) = \frac{r}{b} - \frac{r}{a}

\Rightarrow

μ = 1 + \frac{r}{b} - \frac{r}{a} .

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