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(C) Let the distance of the wire from the corner be $L = 12 \ cm$. The observer looks through the corner,so the angle of incidence at the interface is

Vedclass · Accepted Answer

$1.73$

Vedclass · Answer

(C) Let the distance of the wire from the corner be $L = 12 \ cm$. The observer looks through the corner,so the angle of incidence at the interface is $\alpha = 45^{\circ}$.By Snell's Law at the interface: $\mu \sin \alpha = 1 \sin 	heta$,where $	heta$ is the angle of refraction.$\frac{4}{3} \sin 45^{\circ} = \sin 	heta \Rightarrow \sin 	heta = \frac{4}{3} \cdot \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Using the formula for the apparent position of an object viewed through a refracting surface,the distance of the image from the corner is $x = \frac{L \cos^2 	heta}{\mu \cos^2 \alpha}$.Since $\cos^2 \alpha = \cos^2 45^{\circ} = 0.5$ and $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{8}{9} = \frac{1}{9}$,we have:$x = \frac{12 \cdot (1/9)}{(4/3) \cdot (1/2)} = \frac{12/9}{2/3} = \frac{4}{3} \cdot \frac{3}{2} = 2 \ cm$.The angular separation between the two images is $2(	heta - \alpha)$. The linear separation $d$ between the two images is given by $d = 2x \sin(	heta - \alpha)$.$d = 2(2) \sin(	heta - 45^{\circ}) = 4(\sin 	heta \cos 45^{\circ} - \cos 	heta \sin 45^{\circ})$.$d = 4 \left( \frac{2\sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}} - \frac{1}{3} \cdot \frac{1}{\sqrt{2}} ight) = 4 \left( \frac{2}{3} - \frac{1}{3\sqrt{2}} ight) = \frac{8}{3} - \frac{4}{3\sqrt{2}} = \frac{8}{3} - \frac{2\sqrt{2}}{3} = \frac{8 - 2.828}{3} = \frac{5.172}{3} \approx 1.724 \ cm$.Rounding to the nearest option,the separation is $1.73 \ cm$.

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