The x-z plane separates two media A and B wit | vedclass

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(A) The $x-z$ plane is the interface,so the $y$-axis is normal to the interface.Let $i$ be the angle of incidence and $r$ be the angle of refraction w

Vedclass · Accepted Answer

$\mu_1a = \mu_2\alpha$

Vedclass · Answer

(A) The $x-z$ plane is the interface,so the $y$-axis is normal to the interface.Let $i$ be the angle of incidence and $r$ be the angle of refraction with respect to the normal ($y$-axis).The unit vectors are $\vec{r}_A = a\hat{i} + b\hat{j}$ and $\vec{r}_B = \alpha\hat{i} + \beta\hat{j}$. Since these are unit vectors,their magnitudes are $1$,so $a^2 + b^2 = 1$ and $\alpha^2 + \beta^2 = 1$.The components along the normal ($y$-axis) are $\cos i = \vec{r}_A \cdot \hat{j} = b$ and $\cos r = \vec{r}_B \cdot \hat{j} = \beta$.The components along the interface ($x$-axis) are $\sin i = \vec{r}_A \cdot \hat{i} = a$ and $\sin r = \vec{r}_B \cdot \hat{i} = \alpha$.According to Snell's Law,$\mu_1 \sin i = \mu_2 \sin r$.Substituting the values,we get $\mu_1 a = \mu_2 \alpha$.

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