A ray of light moving along the vector vec{v} | vedclass

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(B) The incident ray vector is $\vec{v}_i = -\hat{i} - 2\hat{j}$. The unit vector of the incident ray is $\hat{r}_i = \frac{-\hat{i} - 2\hat{j}}{\sqrt

Vedclass · Accepted Answer

$\frac{-4\hat{i} - 3\hat{j}}{5}$

Vedclass · Answer

(B) The incident ray vector is $\vec{v}_i = -\hat{i} - 2\hat{j}$. The unit vector of the incident ray is $\hat{r}_i = \frac{-\hat{i} - 2\hat{j}}{\sqrt{1^2 + 2^2}} = \frac{-\hat{i} - 2\hat{j}}{\sqrt{5}}$.The normal to the interface ($x-z$ plane) is along the $y$-axis. Let the normal vector be $\hat{n} = -\hat{j}$ (pointing into the second medium).The angle of incidence $i$ is given by $\cos i = |\hat{r}_i \cdot \hat{n}| = |(-\frac{1}{\sqrt{5}}\hat{i} - \frac{2}{\sqrt{5}}\hat{j}) \cdot (-\hat{j})| = \frac{2}{\sqrt{5}}$.Thus,$\sin i = \sqrt{1 - \cos^2 i} = \sqrt{1 - \frac{4}{5}} = \frac{1}{\sqrt{5}}$.Using Snell's Law,$\mu_1 \sin i = \mu_2 \sin r$:$2 	imes \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{2} \sin r \implies \sin r = \frac{4}{5}$.Then,$\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$.The refracted ray vector $\vec{v}_r$ lies in the same plane as the incident ray and the normal. Since the incident ray has no $z$-component,the refracted ray will also have no $z$-component. Let $\vec{v}_r = a\hat{i} + b\hat{j}$.Since the $x$-component of the vector is conserved during refraction at a plane interface,the $x$-component of the unit vector $\hat{r}_r$ must satisfy $\mu_1 \sin i = \mu_2 \sin r$ (which is already used) and the tangential component is preserved: $\mu_1 \sin i = \mu_2 \sin r$. Specifically,the $x$-component of the incident vector is $-1/\sqrt{5}$. The $x$-component of the refracted unit vector is $\sin r \cdot (	ext{direction}) = \frac{4}{5} \cdot (-1) = -4/5$.The $y$-component is $-\cos r = -3/5$.Thus,the unit vector is $\hat{r}_r = -\frac{4}{5}\hat{i} - \frac{3}{5}\hat{j} = \frac{-4\hat{i} - 3\hat{j}}{5}$.

$A$ ray of light moving along the vector $\vec{v} = -\hat{i} - 2\hat{j}$ undergoes refraction at an interface of two media,which is the $x-z$ plane. The refractive index for $y > 0$ is $\mu_1 = 2$,while for $y < 0$,it is $\mu_2 = \sqrt{5}/2$. Find the unit vector along which the refracted ray moves.

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