Assertion (A) : Brewster angle is the angle of incidence at which there is no reflection.
Reason (R) : For parallel polarisation, Brewster angle is given by following expression :
\[\begin{aligned} \dfrac{E_{r}}{E_{i}} = {} & \dfrac{N}{D} \\ N = {} & \cos \theta_{i} + \sqrt{\dfrac{\epsilon_{r2}}{\epsilon_{r1}} - \sin^{2}\theta_{i}} \\ D = {} & \cos \theta_{i} - \sqrt{\dfrac{\epsilon_{r2}}{\epsilon_{r1}} + \sin^{2}\theta_{i}} \end{aligned}\]UGC NET - 2014,
Answer: C
(A) is true and (R) is false. At the Brewster angle the parallel (p) polarised wave is not reflected at all, so the reflected field is zero. This makes the assertion correct.
The reason is false. The ratio printed is not the standard reflection coefficient for parallel polarisation: its numerator and denominator are made with unequal signs inside the root, so it cannot be used to read off the Brewster angle. The Brewster angle is the one at which that reflection ratio is zero, and it satisfies \(\tan\theta_{B} = \sqrt{\epsilon_{r2}/\epsilon_{r1}}\). Hence option (C).
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