JEE Challenger
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PhyMedAdv 2026 • Paper 2

Incident Angle for Polarized Reflected Ray in Layered Media

As shown in the figure, a ray ABAB of unpolarized light enters from water of refractive index nw=4/3n_w = 4/3 into a medium of refractive index np=4/3n_p = 4/\sqrt{3} after passing through a glass plate of refractive index ng=1.5n_g = 1.5 and a layer of water. At a particular incident angle ii the reflected ray CDCD is polarized in the direction as shown in the figure. The value of ii (in degrees) is:

Question Diagram 1
Official Numerical Answer60

Topics & Concepts

Ray Optics and Optical InstrumentsPolarizationReflection and Refraction

Step-by-Step Solution

To find the angle of incidence ii for which the reflected ray from the interface between water (nw=4/3n_w = 4/3) and the medium (np=4/3n_p = 4/\sqrt{3}) is completely polarized, we apply Brewster's Law at that interface.

The Brewster angle θB\theta_B at the top interface satisfies: tan⁡θB=npnw=4/34/3=3\tan \theta_B = \frac{n_p}{n_w} = \frac{4/\sqrt{3}}{4/3} = \sqrt{3} Thus, θB=60∘\theta_B = 60^\circ.

Since the interfaces of the media layers are parallel, applying Snell's Law across the planar media gives: nwsin⁡i=nwsin⁡θBn_w \sin i = n_w \sin \theta_B sin⁡i=sin⁡θB  ⟹  i=60∘\sin i = \sin \theta_B \implies i = 60^\circ

Therefore, the value of ii in degrees is 6060.

Incident Angle for Polarized Reflected Ray in Layered Media | Physics PYQ Solution - JEE Challenger