A ray of light falls on a transparent glass s | vedclass

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(A) According to Brewster's Law,when the reflected and refracted rays are mutually perpendicular,the angle of incidence is known as the Brewster's ang

Vedclass · Accepted Answer

$	an^{-1}(1.62)$

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(A) According to Brewster's Law,when the reflected and refracted rays are mutually perpendicular,the angle of incidence is known as the Brewster's angle $(i_p)$.From Snell's Law,we have $\mu = \frac{\sin i}{\sin r}$.Given that the reflected and refracted rays are perpendicular,the angle between them is $90^{\circ}$.Since the angle of reflection equals the angle of incidence $(i)$,we have $i + 90^{\circ} + r = 180^{\circ}$,which implies $r = 90^{\circ} - i$.Substituting this into Snell's Law: $\mu = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = 	an i$.Therefore,$i = 	an^{-1}(\mu)$.Given $\mu = 1.62$,the angle of incidence is $i = 	an^{-1}(1.62)$.

$A$ ray of light falls on a transparent glass slab with a refractive index (relative to air) of $1.62$. The angle of incidence for which the reflected and refracted rays are mutually perpendicular is

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