A symmetric biconvex lens of the 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 bey. Obtain the expression for the refractive index of the liquid in terms of x and y.

Solution

Let,

f = focal length liquid +lens

f₁ = focal length of lens

f₂ = focal length of liquid mirror

$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \frac{1}{f_{2}} = \frac{1}{f} - \frac{1}{f_{1}} \frac{1}{f_{2}} = \frac{1}{x} - \frac{1}{y} f_{2} = \frac{xy}{y - x} Now, \frac{1}{f_{1}} = (μ - 1) (\frac{2}{R}) (convex lens) \frac{1}{y} = (1.5 - 1) \frac{2}{R} \frac{1}{y} = \frac{0.5 x 2}{R} R = y Now for liquid \frac{1}{f_{2}} = (μ' - 1) (\frac{1}{R}) - \frac{(y - x)}{xy} = (μ' - 1) (\frac{1}{R} - \frac{1}{\infty}) - \frac{(y - x)}{xy} = \frac{(μ' - 1)}{y} ∴ \frac{x - y}{x} = μ' - 1 1 + 1 - \frac{y}{x} = μ' 2 - \frac{y}{x} = μ'$

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Ray Optics And Optical Instruments

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