Physics - Geometrical Optics Question with Solution | TestHub

Deviations: at P $(i-r)$, at Q (reflection) $(\pi -2r)$, at R $(i-r)$; total $\delta =\pi +2i-4r$.

Minimum deviation: $\frac{d\delta }{di}=0\Rightarrow \frac{dr}{di}=\frac{1}{2}$.

Snell: $\sin i=\mu \sin r$; differentiating, $\cos i=\mu \cos r\cdot \frac{1}{2}$, so ${\mu }^2(1-{\sin }^{2}r)=4{\cos }^{2}i$.

${\mu }^2-{\sin }^{2}i=4{\cos }^{2}i\Rightarrow {\mu }^2-1=3{\cos }^{2}i\Rightarrow {\cos }^{2}i=\frac{7/4-1}{3}=\frac{1}{4}$, so $i=60^{\circ }$ - (B)

($\sin r=\frac{\sin 60^{\circ }}{\sqrt{7}/2}=\sqrt{\frac{3}{7}}\Rightarrow r=41^{\circ }$, and $\delta =180^{\circ }+120^{\circ }-164^{\circ }=136^{\circ }$, so A, C, D are wrong.)