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. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d < < h, the height of the column.
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Let us consider a portion of a ray between x and x + dx inside the liquid. Let us suppose the angle of incidence at x be $\theta$ and let it enter the thin column at height y
Because of the bending it shall emerge at x = dx with an angle $\theta+\text{d}\theta$ and at a height y + dy. From Snell's law,
$\mu(\text{y})\sin\theta=\mu(\text{y}+\text{dy})\sin(\theta+\text{d}\theta)$
$\Rightarrow\ \mu(\text{y})\sin\theta=\Big(\mu(\text{y})+\frac{\text{d}\mu}{\text{dy}}\text{dy}\Big)(\sin\theta\cos\text{d}\theta+\cos\theta\sin\text{d}\theta)$
$\big[\because\ \sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A}\big]$
Since $\text{d}\theta$ is too small.
Hence, $\cos\text{d}\theta\approx1\text{ and }\text{d}\theta=\text{d}\theta$
$\therefore\ \mu(\text{y})\sin\theta=\Big(\mu(\text{y})+\frac{\text{d}\mu}{\text{dy}}\text{dy}\Big)(\sin\theta+\cos\theta\text{d}\theta)$
$\Rightarrow\ \mu(\text{y})\sin\theta=\mu(\text{y})\sin\theta+\mu(\text{y})\cos\theta\text{d}\theta+\sin\theta\frac{\text{d}\mu}{\text{dy}}\text{dy}+\frac{\text{d}\mu}{\text{dy}}\text{dy}\cos\theta\text{d}\theta$
$\Rightarrow\ \mu(\text{y})\cos\theta\text{d}\theta=-\frac{\text{d}\mu}{\text{dy}}\text{dy}\sin\theta\ \ \bigg[\text{Neglecting}\frac{\text{d}\mu}{\text{dy}}\text{dy}\cos\theta\text{d}\theta\bigg]$
$\Rightarrow\ \text{d}\theta=\frac{-\text{d}\mu}{\mu\text{dy}}\text{dy}\tan\theta$
From the figute, we have $\tan\theta=\frac{\text{dx}}{\text{dy}}$
$\therefore\ \text{d}\theta=\frac{-1\text{d}\mu}{\mu\text{dy}}\text{dx}$
Solving this variable separable from of differential equation, we get
$\therefore\ \theta=\frac{-1\text{d}\mu}{\mu\text{dy}}\int_0^\text{d}\text{dx}=\frac{-1\text{d}\mu}{\mu\text{dy}}\text{d}$
Because of the bending it shall emerge at x = dx with an angle $\theta+\text{d}\theta$ and at a height y + dy. From Snell's law,
$\mu(\text{y})\sin\theta=\mu(\text{y}+\text{dy})\sin(\theta+\text{d}\theta)$
$\Rightarrow\ \mu(\text{y})\sin\theta=\Big(\mu(\text{y})+\frac{\text{d}\mu}{\text{dy}}\text{dy}\Big)(\sin\theta\cos\text{d}\theta+\cos\theta\sin\text{d}\theta)$
$\big[\because\ \sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A}\big]$
Since $\text{d}\theta$ is too small.
Hence, $\cos\text{d}\theta\approx1\text{ and }\text{d}\theta=\text{d}\theta$
$\therefore\ \mu(\text{y})\sin\theta=\Big(\mu(\text{y})+\frac{\text{d}\mu}{\text{dy}}\text{dy}\Big)(\sin\theta+\cos\theta\text{d}\theta)$
$\Rightarrow\ \mu(\text{y})\sin\theta=\mu(\text{y})\sin\theta+\mu(\text{y})\cos\theta\text{d}\theta+\sin\theta\frac{\text{d}\mu}{\text{dy}}\text{dy}+\frac{\text{d}\mu}{\text{dy}}\text{dy}\cos\theta\text{d}\theta$
$\Rightarrow\ \mu(\text{y})\cos\theta\text{d}\theta=-\frac{\text{d}\mu}{\text{dy}}\text{dy}\sin\theta\ \ \bigg[\text{Neglecting}\frac{\text{d}\mu}{\text{dy}}\text{dy}\cos\theta\text{d}\theta\bigg]$
$\Rightarrow\ \text{d}\theta=\frac{-\text{d}\mu}{\mu\text{dy}}\text{dy}\tan\theta$
From the figute, we have $\tan\theta=\frac{\text{dx}}{\text{dy}}$
$\therefore\ \text{d}\theta=\frac{-1\text{d}\mu}{\mu\text{dy}}\text{dx}$
Solving this variable separable from of differential equation, we get
$\therefore\ \theta=\frac{-1\text{d}\mu}{\mu\text{dy}}\int_0^\text{d}\text{dx}=\frac{-1\text{d}\mu}{\mu\text{dy}}\text{d}$
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