A ray of light from a denser medium strikes a | vedclass

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(A) Let the angle of incidence be $i$. According to the law of reflection,the angle of reflection $r$ is equal to the angle of incidence,so $r = i$.Gi

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$\sin^{-1}(	an r)$

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(A) Let the angle of incidence be $i$. According to the law of reflection,the angle of reflection $r$ is equal to the angle of incidence,so $r = i$.Given that the reflected and refracted rays are perpendicular to each other,the sum of the angle of reflection $r$,the angle between the reflected ray and the interface,and the angle of refraction $r'$ must be $180^{\circ}$. Since the angle between the reflected ray and the interface is $(90^{\circ} - r)$,we have $r + 90^{\circ} + r' = 180^{\circ}$,which simplifies to $r' = 90^{\circ} - r$.Applying Snell's Law at the interface: $\mu \sin i = 1 \cdot \sin r'$,where $\mu$ is the refractive index of the denser medium relative to the rarer medium.Substituting $i = r$ and $r' = 90^{\circ} - r$ into Snell's Law:$\mu \sin r = \sin(90^{\circ} - r)$$\mu \sin r = \cos r$$\mu = \frac{\cos r}{\sin r} = \cot r$The critical angle $	heta_c$ is given by $\sin 	heta_c = \frac{1}{\mu}$.Substituting $\mu = \cot r$:$\sin 	heta_c = \frac{1}{\cot r} = 	an r$$	heta_c = \sin^{-1}(	an r)$.

$A$ ray of light from a denser medium strikes a rarer medium. The angle of reflection is $r$ and that of refraction is $r'$. The reflected and refracted rays make an angle of $90^{\circ}$ with each other. The critical angle will be

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