CBSE Class 12 · 2018 · Geometrical OpticsShort Answer

A symmetric biconvex lens of radius of curvature

R

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 tip 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: $μ_{liquid} = 2 - \frac{y}{x}$ .

Solution

The image coincides with the needle when it is at the focus of the equivalent mirror, so the measured distance equals the focal length of the lens-combination acting with the plane mirror.
Without liquid, only the glass lens acts:

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

.
With liquid, the plano-concave liquid lens

f_{2}

combines with the glass lens

f_{1} = y

\frac{1}{x} = \frac{1}{f _{1}} + \frac{1}{f _{2}}, \frac{1}{f _{2}} = (μ_{ℓ} - 1) (\frac{- 1}{R}) = \frac{1 - μ _{ℓ}}{y}

\frac{1}{x} = \frac{1}{y} + \frac{1 - μ _{ℓ}}{y} = \frac{2 - μ _{ℓ}}{y} \Rightarrow μ_{liquid} = 2 - \frac{y}{x}

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CBSE Class 12 · 2018 · Geometrical OpticsShort Answer

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