In the figure shown,frac{sin i}{sin r} is equ | vedclass

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(B) According to Snell's Law at the first interface between medium $\mu_1$ and $\mu_2$:$\mu_1 \sin i = \mu_2 \sin r_1$where $r_1$ is the angle of refr

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$\frac{\mu_3}{\mu_1}$

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(B) According to Snell's Law at the first interface between medium $\mu_1$ and $\mu_2$:$\mu_1 \sin i = \mu_2 \sin r_1$where $r_1$ is the angle of refraction in the second medium.At the second interface between medium $\mu_2$ and $\mu_3$:$\mu_2 \sin r_1 = \mu_3 \sin r$where $r$ is the angle of refraction in the third medium.From these two equations,we can equate the common term $\mu_2 \sin r_1$:$\mu_1 \sin i = \mu_3 \sin r$Therefore,the ratio is:$\frac{\sin i}{\sin r} = \frac{\mu_3}{\mu_1}$

In the figure shown,$\frac{\sin i}{\sin r}$ is equal to:

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