## BdPhO Regional (Dhaka-South) Secondary 2019/6

Discuss Physics and Physics Olympiad related problems here
SINAN EXPERT
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### BdPhO Regional (Dhaka-South) Secondary 2019/6 BdPhO Regional (Dhaka-South) Secondary 2019(6).png (10.21 KiB) Viewed 1472 times
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\$

SINAN EXPERT
Posts: 38
Joined: Sat Jan 19, 2019 3:35 pm
Contact:

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

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\$ BdPhO Regional (Dhaka-South) Secondary 2019(6S).png (13.12 KiB) Viewed 1411 times
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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