The X-Y plane is taken as the boundary betwee | vedclass

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(B) The incident ray vector is $\overrightarrow{A} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}$.The normal to the $X-Y$ plane is along the $Z$-ax

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) The incident ray vector is $\overrightarrow{A} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}$.The normal to the $X-Y$ plane is along the $Z$-axis,i.e.,$\hat{k}$.The angle of incidence $i$ is the angle between the incident ray and the normal. Since the ray is travelling in $M_{1}$ $(Z \geq 0)$ towards the origin,its direction vector is $-\overrightarrow{A} = -4\sqrt{3}\hat{i} + 3\sqrt{3}\hat{j} + 5\hat{k}$.The cosine of the angle $i$ with the normal $\hat{k}$ is given by $\cos i = \frac{|A_z|}{|\overrightarrow{A}|}$.Magnitude $|\overrightarrow{A}| = \sqrt{(4\sqrt{3})^2 + (-3\sqrt{3})^2 + (-5)^2} = \sqrt{48 + 27 + 25} = \sqrt{100} = 10$.$\cos i = \frac{5}{10} = \frac{1}{2} \Rightarrow i = 60^{\circ}$.Using Snell's Law: $\mu_1 \sin i = \mu_2 \sin r$.$\sqrt{2} \sin 60^{\circ} = \sqrt{3} \sin r$.$\sqrt{2} 	imes \frac{\sqrt{3}}{2} = \sqrt{3} \sin r$.$\sin r = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow r = 45^{\circ}$.The difference between the angle of incidence and the angle of refraction is $i - r = 60^{\circ} - 45^{\circ} = 15^{\circ}$.

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