A ray of light travelling inside a rectangula | vedclass

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(D) Given: Refractive index of glass $\mu_g = \sqrt{2}$,refractive index of air $\mu_a = 1$,and angle of incidence $i = 45^{\circ}$.Using Snell's Law

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Will emerge into the air with an angle of refraction equal to $90^{\circ}$

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(D) Given: Refractive index of glass $\mu_g = \sqrt{2}$,refractive index of air $\mu_a = 1$,and angle of incidence $i = 45^{\circ}$.Using Snell's Law at the glass-air interface: $\mu_g \sin(i) = \mu_a \sin(r)$.Substituting the values: $\sqrt{2} \sin(45^{\circ}) = 1 \cdot \sin(r)$.Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,we get: $\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \sin(r)$.This simplifies to: $1 = \sin(r)$,which implies $r = 90^{\circ}$.Since the angle of refraction is $90^{\circ}$,the ray grazes the surface of the glass block.

$A$ ray of light travelling inside a rectangular glass block of refractive index $\sqrt{2}$ is incident on the glass-air surface at an angle of incidence of $45^{\circ}$. The refractive index of air is $1$. Under these conditions,the ray:

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