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(B) The formula for lateral shift $d$ is given by: $d = t \frac{\sin(i-r)}{\cos r}$,where $t$ is the thickness,$i$ is the angle of incidence,and $r$ i

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The formula for lateral shift $d$ is given by: $d = t \frac{\sin(i-r)}{\cos r}$,where $t$ is the thickness,$i$ is the angle of incidence,and $r$ is the angle of refraction.Using Snell's Law: $\sin i = \mu \sin r \Rightarrow \sin 60^{\circ} = \sqrt{3} \sin r$.$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r \Rightarrow \sin r = \frac{1}{2} \Rightarrow r = 30^{\circ}$.Now,substitute the values into the lateral shift formula:$4\sqrt{3} = t \frac{\sin(60^{\circ}-30^{\circ})}{\cos 30^{\circ}}$.$4\sqrt{3} = t \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = t 	an 30^{\circ}$.$4\sqrt{3} = t \left(\frac{1}{\sqrt{3}}ight)$.$t = 4\sqrt{3} 	imes \sqrt{3} = 4 	imes 3 = 12 \, cm$.

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