JEE Advanced 2026 · Paper 2 · Physics Q12

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PhysicsNumerical+4Question 30 of 54

As shown in the figure, a ray $AB$ of unpolarized light enters from water of refractive index $n_w=4/3$ into a medium of refractive index $n_p=4/\sqrt3$ after passing through a glass plate of refractive index $n_g=1.5$ and a layer of water. At a particular incident angle $i$ the reflected ray $CD$ is polarized in the direction as shown in the figure. The value of $i$ (in degrees) is:

$Horizontal layers, bottom to top: water (n_w = 4/3), a glass plate (n_g = 1.5), a water layer (n_w = 4/3) and a medium n_p = 4/√3. Ray AB rises from the bottom water at incidence i, refracts through the glass and water layers and meets the n_p boundary; part is transmitted upward, and the reflected part returns down through the layers as CD. AB carries both dots and bars (unpolarized); CD carries only dots (polarized perpendicular to the plane of the figure).$

Your answer

Answer and worked solution

Official answer: 60

The reflected ray $CD$ has only the component perpendicular to the plane of incidence. So the reflection at the water- $n_p$ boundary is at Brewster's angle:
$\tan\theta_B=\frac{n_p}{n_w}=\frac{4/\sqrt3}{4/3}=\sqrt3\implies\theta_B=60^\circ.$
The layers are parallel, so $n\sin(\text{angle})$ is the same in every layer. The bottom medium is also water, so
$n_w\sin i=n_w\sin60^\circ\implies i=60^\circ.$

Topics:Polarisation and Brewster's law Refraction at plane surfaces

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JEE Advanced 2026 · Paper 2, Physics question 12