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Question 149 of 199

Q.The refractive index of the medium, for the polarising angle 60° is :

(a) 1.732
(b) 1.414
(c) 1.5
(d) 1.468
Puducherry TnboardTamil Nadu HSC (DGE) Board 2017MCQ· 1mImportance★★★★★
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Brewster's law gives n=tan⁡θpn=\tan\theta_p, and tan⁡60°=3≈1.732\tan 60° = \sqrt{3} \approx 1.732.

When unpolarised light strikes the boundary of a transparent medium at a particular angle of incidence, called the polarising angle (or Brewster's angle) θp\theta_p, the reflected ray is found to be completely plane-polarised, with its electric vector oscillating perpendicular to the plane of incidence. Brewster discovered empirically (and it can be derived from the condition that the reflected and refracted rays are then exactly perpendicular to each other) that the refractive index of the medium is related to this angle by

n=tan⁡θp.n = \tan\theta_p.

This follows from combining Snell's law, n=sin⁡θpsin⁡θrn = \dfrac{\sin\theta_p}{\sin\theta_r}, with the geometric condition θp+θr=90°\theta_p + \theta_r = 90° (reflected and refracted rays perpendicular), which gives sin⁡θr=sin⁡(90°−θp)=cos⁡θp\sin\theta_r = \sin(90°-\theta_p)=\cos\theta_p, so n=sin⁡θpcos⁡θp=tan⁡θpn=\dfrac{\sin\theta_p}{\cos\theta_p}=\tan\theta_p.

Substituting θp=60°\theta_p = 60°:

n=tan⁡60°=3≈1.732.n = \tan 60° = \sqrt{3} \approx 1.732.

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