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  • The mixture a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension.

The mixture of a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in traveling a horizontal distance d << h, the height of the column.

Answers (1)

Explanation:-

Let us consider a portion of a ray between x and x+dx inside the liquid solution.

Let us assume that the angle of the incidence ray is \theta and

Let the ray enter the thin column at the height y.

As a result of the refraction, the ray deviates from its original path and subsequently emerges at x+dx with an angle

\theta +d\theta and at a height y+dy.

From Snell's law,

\mu \left ( y \right )\sin \theta =\mu \left ( y+dy \right )\sin\left ( \theta+d\theta \right )......(i)  

   

Let the refractive index of the liquid at position y be \mu (y)=\mu, then

\mu\left ( y+dy \right )=\mu+\left ( \frac{d\mu}{dy} \right )dy=\mu+kdy

Where k=\left ( \frac{d\mu }{dy} \right )= refractive index gradient along the vertical dimension.

Hence from (i), \mu \sin\theta=\left ( \mu+kdy \right )\sin \left ( \theta+d \theta \right )

\mu \sin\theta=\left ( \mu+kdy \right ).\left ( \sin \theta.\cos d \theta+\cos \theta. \sin d\theta \right )

\mu \sin\theta=\left ( \mu+kdy \right ).\left ( \sin \theta.1+\cos \theta.d \theta \right )\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)

For small angle \sin d \theta \approx d \theta \; and \; \cos d \theta\approx 1

\mu \sin \theta=\mu \sin \theta+kdy \; \sin \theta+\mu \cos \theta .d\theta+k\; \cos\; \theta dy.d \theta

kdy \sin \theta +\mu \cos \theta.d\theta=0\Rightarrow d \theta=-\frac{k}{\mu}\tan \theta dy

But \tan \theta=\frac{dx}{dy}\; and \: k=\left ( \frac{d\mu}{dy} \right )

d\theta=-\frac{k}{\mu}\left ( \frac{dx}{dy} \right )dy\Rightarrow d\theta=-\frac{k}{\mu}dx

Integrating both sides, \int_{0}^{\delta }d\theta =-\frac{k}{\mu}dx

\Rightarrow \delta =-\frac{kd}{\mu }=-\frac{d}{\mu }\left ( \frac{d\mu }{dy} \right )

 

Posted by

infoexpert23

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