The back surface of a glass slab (refractive | vedclass

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(A) The laser beam undergoes partial reflection at the top surface and partial refraction. The refracted ray travels through the glass,reflects off th

Vedclass · Accepted Answer

$\frac{2 t \cos 	heta}{\sqrt{n^{2}-\sin ^{2} 	heta}}$

Vedclass · Answer

(A) The laser beam undergoes partial reflection at the top surface and partial refraction. The refracted ray travels through the glass,reflects off the back mirror,and exits the glass slab. Let the angle of incidence be $	heta$ and the angle of refraction be $r$. According to Snell's law,$\sin 	heta = n \sin r$,so $\sin r = \frac{\sin 	heta}{n}$. The horizontal distance between the point of first reflection (at the top surface) and the point where the refracted ray exits the slab is $x = 2t 	an r$. On the screen,the two rays (one reflected from the top,one from the bottom) form spots. The vertical separation $h_1 - h_2$ between these spots is related to the horizontal separation $d_1 - d_2$ by $	an 	heta = \frac{d_1 - d_2}{h_1 - h_2}$. Thus,the spacing is $h_1 - h_2 = \frac{2t 	an r}{	an 	heta}$. Substituting $	an r = \frac{\sin r}{\cos r} = \frac{\sin 	heta / n}{\sqrt{1 - (\sin 	heta / n)^2}} = \frac{\sin 	heta}{\sqrt{n^2 - \sin^2 	heta}}$,we get: $h_1 - h_2 = \frac{2t}{	an 	heta} \cdot \frac{\sin 	heta}{\sqrt{n^2 - \sin^2 	heta}} = \frac{2t \cos 	heta}{\sqrt{n^2 - \sin^2 	heta}}$.

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