A light ray is incident from a denser medium | vedclass

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(A) According to Snell's law,$\mu_{D} \sin i = \mu_{R} \sin r'$,where $r'$ is the angle of refraction.Thus,the relative refractive index $\mu = \frac{

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$\sin^{-1}(	an\, i)$

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(A) According to Snell's law,$\mu_{D} \sin i = \mu_{R} \sin r'$,where $r'$ is the angle of refraction.Thus,the relative refractive index $\mu = \frac{\mu_D}{\mu_R} = \frac{\sin r'}{\sin i} \dots (i)$.Since the reflected ray and refracted ray are perpendicular,the sum of the angle of reflection $(i)$,the angle between them $(90^{\circ})$,and the angle of refraction $(r')$ is $180^{\circ}$.Therefore,$i + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - i$.Substituting this into equation $(i)$,we get $\mu = \frac{\sin(90^{\circ} - i)}{\sin i} = \frac{\cos i}{\sin i} = \cot i$.The critical angle $	heta_c$ is defined as $	heta_c = \sin^{-1}(\frac{1}{\mu})$.Substituting $\mu = \cot i$,we get $	heta_c = \sin^{-1}(\frac{1}{\cot i}) = \sin^{-1}(	an i)$.

$A$ light ray is incident from a denser medium to a rarer medium at an angle of incidence $i$. If the reflected and refracted rays are mutually perpendicular,then the critical angle for the interface is:

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