JEE Advanced 2025 Paper 1 · Q11 · Ray Optics – Mirrors
A solid glass sphere of refractive index $n=\sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\tfrac{R}{2}$, as shown in the figure above. A very thin glass layer is present at the point $O$ so that the air cavity (refractive index $n=1$) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point $O$ and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $\theta$. The value of $\sin\theta$ is _____.
Reveal answer + step-by-step solution
Correct answer:0.5, 0.75
Solution
Brewster's law at the glass→air interface inside the sphere: $\tan\theta_B = n_{\text{air}}/n_{\text{glass}}=1/\sqrt{3}$, so $\theta_B=30^\circ$ and $\sin\theta_B=0.5$. The geometry of the figure (ray from $S$ inside the glass, reflecting at $O$ on the inner surface of the air cavity, fully polarised) constrains $\theta$ at the outer-glass surface to either $\sin\theta=0.5$ or $\sin\theta=0.75$ depending on which interface the official key associates with the polarisation condition; both values are accepted in the official answer key. We surface 0.5 as the primary answer.
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