The following figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Solution

Suppose,
Focal length of convex lens of glass = f₁ = 30 cm

Focal length of plane concave lens of liquid be = f₂and,

Combined focal length, F = 45.0 cm

As,

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

∴

\frac{1}{f_{2}} = \frac{1}{F} - \frac{1}{f_{1}} = \frac{1}{45} - \frac{1}{30} = - \frac{1}{90}

f_{2} = - 90 cm

For glass lens, let

R_{1} = R, R_{2} = - R

Using lens maker's formula,

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

we have,

∴

\frac{1}{30} = (\frac{3}{2} - 1) (\frac{1}{R} + \frac{1}{R})

= \frac{1}{2} \times \frac{2}{R} = \frac{1}{R}

Thus,

R = 30 cm

For liquid lens,

R_{1} = - R = - 30.0 cm R_{2} = \infty \frac{1}{f_{2}} = (μ_{l} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (μ_{l} - 1) (\frac{1}{- 30} - \frac{1}{\infty})

\Rightarrow

\frac{1}{- 90} = (μ_{l} - 1) \times \frac{1}{- 30}

\Rightarrow

(μ_{l} - 1) = \frac{30}{90} = \frac{1}{3}

μ_{l} = 1 + \frac{1}{3} = \frac{4}{3} .

which is the refractive index of the liquid.

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