In the figure shown here,the angle made by th | vedclass

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(A) According to Snell's Law,for a series of parallel interfaces,the product of the refractive index and the sine of the angle of incidence remains co

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$30$

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(A) According to Snell's Law,for a series of parallel interfaces,the product of the refractive index and the sine of the angle of incidence remains constant at each interface.Let $n_1 = 1$ be the refractive index of the first medium and $i_1 = 60^o$ be the angle of incidence.Let $n_3 = \sqrt{3}$ be the refractive index of the third medium and $r_3$ be the angle of refraction in this medium.Applying Snell's Law between the first and the third medium:$n_1 \sin(i_1) = n_3 \sin(r_3)$Substituting the given values:$1 \cdot \sin(60^o) = \sqrt{3} \cdot \sin(r_3)$Since $\sin(60^o) = \frac{\sqrt{3}}{2}$:$\frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r_3)$Dividing both sides by $\sqrt{3}$:$\sin(r_3) = \frac{1}{2}$Therefore,$r_3 = \arcsin(0.5) = 30^o$.The angle made by the light ray with the normal in the medium of refractive index $\sqrt{3}$ is $30^o$.

In the figure shown here,the angle made by the light ray with the normal in the medium of refractive index $\sqrt{3}$ is.......$^o$

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