Question

A symmetric biconvex lens of radius of curvature and made of glass of refractive index 1.5 is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its top on the principal axis of the lens is moved along the axis until its real inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x. on removing the liquid layer and repeating the experiment the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y.

Answer 1

When a liquid is placed on top of plane mirror and convex lens is placed over it, then this whole system would become a combination of convex lens of glass and planoconcave lens of liquid.

This is shown in the figure.

 

Let focal length of convex lens = f₁

focal length of planoconcave liquid lens = f₂

combined focal length = F

in both case image coincides with needle; hence ray is normal to plane mirror. So, needle position is focal lengths of convex lens and combined system respectively. According to the question,

f₁ = y unit

F = x unit

We also know that for combination of two lenses

\(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\)

\(\frac{1}{f_2}=\frac{1}{F}-\frac{1}{f_1}\)

\(\frac{1}{f_2}=\frac{1}{x}-\frac{1}{y}\)

\(f_2=\frac{xy}{x-y}\)

for glass lens, let R1 = R, R2 = −R.

From lens maker formula

\(\frac{1}{f}=(n-1)(\frac{1}{r}-(-\frac{1}{r}))\)

\(\frac{1}{y}=(1.5-1)(\frac{1}{R}+\frac{1}{R})\)

\(\frac{1}{y}=\frac{1}{R}\)

R = y unit

For liquid planoconcave lens

R₁ = −R, R₂ = ∞.

From lens maker formula

\(\frac{1}{f_2}=(n_i-1)(-\frac{1}{r}-\frac{1}{∞})\)

\(\frac{y-x}{yx}=-(n_1-1)\times \frac{1}{y}\)

(∵ R = y)

\(1-n_i=\frac{y-x}{x}\)

\(n_1=1-\frac{y-x}{x}\)

\(n_1=\frac{x-y+x}{x}\)

\(n_1=\frac{2x-y}{x}\)