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(A) Applying Snell's Law at the glass-water interface:$\mu_g \sin i = \mu_w \sin r$Given $\mu_w = \frac{4}{3}$,so $\mu_g \sin i = \frac{4}{3} \sin r$

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

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(A) Applying Snell's Law at the glass-water interface:$\mu_g \sin i = \mu_w \sin r$Given $\mu_w = \frac{4}{3}$,so $\mu_g \sin i = \frac{4}{3} \sin r$  --- $(1)$Applying Snell's Law at the water-air interface,where the ray grazes the surface (angle of refraction is $90^\circ$):$\mu_w \sin r = \mu_{air} \sin 90^\circ$$\frac{4}{3} \sin r = 1 	imes 1 = 1$$\sin r = \frac{3}{4}$Substituting $\sin r = \frac{3}{4}$ into equation $(1)$:$\mu_g \sin i = \frac{4}{3} 	imes \frac{3}{4} = 1$$\mu_g = \frac{1}{\sin i}$

$A$ ray of light travels in the fashion as shown in the figure. After passing through water,the ray grazes along the water-air interface. The value of $\mu_g$ in terms of $i$ is

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