In the figure shown,the angle made by the lig | 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 with the normal remains

Vedclass · Accepted Answer

$30$

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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 with the normal remains constant at every interface.Let $n_1 = 1$ be the refractive index of the first medium and $i = 45^{\circ}$ be the angle of incidence.Let $n_3 = \sqrt{2}$ be the refractive index of the medium where we need to find the angle of refraction $r$.Applying Snell's law:$n_1 \sin i = n_3 \sin r$Substituting the given values:$1 	imes \sin 45^{\circ} = \sqrt{2} 	imes \sin r$Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$:$\frac{1}{\sqrt{2}} = \sqrt{2} \sin r$$\sin r = \frac{1}{\sqrt{2} 	imes \sqrt{2}} = \frac{1}{2}$Therefore,$r = \arcsin(0.5) = 30^{\circ}$.

In the figure shown,the angle made by the light ray with the normal in the medium of refractive index $\sqrt{2}$ is.....$^o$

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