BdPhO Regional (Dhaka-South) Secondary 2019/6

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SINAN EXPERT
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BdPhO Regional (Dhaka-South) Secondary 2019/6

Unread post by SINAN EXPERT » Mon Jan 28, 2019 9:17 pm

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A light ray is incident on a rectangular slab $ABCD$ at an angle $θ$ with the normal. The refractive index varies with vertical distance $y$ in the rectangular slab.
Refractive index, $n(y) = ky$, for $0 < y < h$, where $k$ is a constant.
At what angle with the normal the light ray will come out of the surface $AB$?
$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$

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Re: BdPhO Regional (Dhaka-South) Secondary 2019/6

Unread post by SINAN EXPERT » Thu Apr 25, 2019 9:24 pm

Given, $n(y)=ky,$ $where$ $0<y<h$. Suppose that, $y_1<y_2<y_3$.

Then, $n_1<n_2<n_3$, Hence, $sin θ_1>sinθ_2>sinθ_3$ (Snell's law)
$⇒$ $θ_1>θ_2>θ_3$
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By Snell's law, $\dfrac{n_1}{n_2}=\dfrac{sinθ_2}{sinθ_1}$
$∴$ $n_1 sinθ_1=n_2 sinθ_2=n_3 sinθ_3=...=n_4 sinθ_4$, Here, $n_1=n_4=n_{air}=1$

$∴$ $sinθ_1=sinθ_4$

So, the required angle is $\boxed{θ}$.
$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$

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