A light ray incident along a vector vec{AO} ( | vedclass

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(A) From Snell's law, $\mu_1 \sin \alpha = \mu_2 \sin \beta$.Given $\mu_1 = 1$ and $\mu_2 = 1.5$.The vector $\vec{AO} = 2\hat{i} - 3\hat{j}$ makes an

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(A) From Snell's law, $\mu_1 \sin \alpha = \mu_2 \sin \beta$.Given $\mu_1 = 1$ and $\mu_2 = 1.5$.The vector $\vec{AO} = 2\hat{i} - 3\hat{j}$ makes an angle $\alpha$ with the vertical axis (y-axis). Thus, $	an \alpha = \frac{|x|}{|y|} = \frac{2}{3}$.Therefore, $\sin \alpha = \frac{2}{\sqrt{2^2 + 3^2}} = \frac{2}{\sqrt{13}}$.Using Snell's law: $\sin \beta = \frac{\mu_1}{\mu_2} \sin \alpha = \frac{1}{1.5} 	imes \frac{2}{\sqrt{13}} = \frac{2}{1.5 \sqrt{13}} = \frac{4}{3\sqrt{13}}$.The refracted vector is $\vec{OB} = C\hat{i} - 4\hat{j}$. The angle $\beta$ is with the vertical axis, so $\sin \beta = \frac{|C|}{\sqrt{C^2 + (-4)^2}} = \frac{C}{\sqrt{C^2 + 16}}$.Equating the two expressions for $\sin \beta$: $\frac{C}{\sqrt{C^2 + 16}} = \frac{4}{3\sqrt{13}}$.Squaring both sides: $\frac{C^2}{C^2 + 16} = \frac{16}{9 	imes 13} = \frac{16}{117}$.$117C^2 = 16(C^2 + 16) \implies 117C^2 = 16C^2 + 256$.$101C^2 = 256 \implies C^2 = \frac{256}{101} \approx 2.534$.$C = \sqrt{2.534} \approx 1.59 \approx 1.6$.

$A$ light ray incident along a vector $\vec{AO}$ $(\vec{AO} = 2\hat{i} - 3\hat{j})$ emerges along a vector $\vec{OB}$ $(\vec{OB} = C\hat{i} - 4\hat{j})$ as shown in the figure. The value of $C$ is . . . . . . .

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