In the given figure,at the water-air interfac | vedclass

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(B) Let the refractive index of glass be $\mu_g$ and water be $\mu_w = 4/3$. The angle of incidence at the glass-water interface is $i$. By Snell's la

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$\frac{1}{\sin i}$

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(B) Let the refractive index of glass be $\mu_g$ and water be $\mu_w = 4/3$. The angle of incidence at the glass-water interface is $i$. By Snell's law at the glass-water interface:$\mu_g \sin i = \mu_w \sin r$At the water-air interface,the light ray is incident at the critical angle,meaning the angle of refraction in air is $90^{\circ}$. Let the angle of incidence at the water-air interface be $r$. By Snell's law at the water-air interface:$\mu_w \sin r = \mu_{air} \sin 90^{\circ}$Since $\mu_{air} = 1$,we have $\mu_w \sin r = 1$.Substituting this into the first equation:$\mu_g \sin i = 1$Therefore,$\mu_g = \frac{1}{\sin i}$.

In the given figure,at the water-air interface,a light ray is incident at the critical angle. Find the value of $\mu_g$.

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