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JEE Main 2017 · Physics · Ray and Wave Optics · hard

A ray of unpolarised light is incident at the polarising (Brewster) angle on a parallel-sided glass slab of thickness 3 cm and refractive index √3…

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A ray of unpolarised light is incident at the polarising (Brewster) angle on a parallel-sided glass slab of thickness 3 cm3\ \mathrm{cm} and refractive index 3\sqrt3 placed in air. The perpendicular distance between the emergent ray and the undeviated extension of the incident ray (the lateral shift) is
  1. A. 1.5 cm1.5\ \mathrm{cm}
  2. B. 3 cm\sqrt3\ \mathrm{cm}
  3. C. 3 cm3\ \mathrm{cm}
  4. D. 332 cm\dfrac{3\sqrt3}{2}\ \mathrm{cm}
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Answer: (B) 3 cm\sqrt3\ \mathrm{cm}

Brewster's law: tan⁡iB=μ=3\tan i_B = \mu = \sqrt3, so i=60∘i = 60^\circ. Snell's law gives sin⁡r=sin⁡60∘/3=12\sin r = \sin60^\circ/\sqrt3 = \tfrac12, so r=30∘r = 30^\circ (indeed i+r=90∘i + r = 90^\circ, the reflected and refracted rays being perpendicular). The lateral shift of a ray through a parallel slab is d=tsin⁡(i−r)cos⁡r=3sin⁡30∘cos⁡30∘=3(1/2)3/2=3 cmd = \dfrac{t\sin(i - r)}{\cos r} = \dfrac{3\sin30^\circ}{\cos30^\circ} = \dfrac{3(1/2)}{\sqrt3/2} = \sqrt3\ \mathrm{cm}. The value 33/23\sqrt3/2 comes from using sin⁡i\sin i in place of sin⁡(i−r)\sin(i - r).
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