If the polarizing angle of a piece of glass for green light is $54.74^{\circ}$, then the angle of mi

Question

If the polarizing angle of a piece of glass for green light is $54.74^{\circ}$, then the angle of minimum deviation for an equilateral prism made of same glass is [Given: $	an 54.74=1.414$ ]\

Professor PC Thomas Classes · Answer

d) By principle of polarization, $\mu=	an 	heta_{p}$\ or $\mu=	an 54.74^{\circ}$ orn $\mu=1.414$\ For an equilateral prism, $\angle \mathrm{A}=60^{\circ}$\ $	herefore \mu=\frac{\sin \left(\frac{A+\delta}{2}ight)}{\sin (A / 2)}=\frac{\sin \left(\frac{60^{\circ}+\delta}{2}ight)}{\sin \left(60^{\circ} / 2ight)}$\ or $\frac{1.414 	imes 1}{2}=\sin \left(\frac{60^{\circ}+\delta}{2}ight)[\because 1.414=\sqrt{2}]$\ or $\frac{\sqrt{2}}{2}=\sin \left(\frac{60^{\circ}+\delta}{2}ight)$\ or $\frac{1}{\sqrt{2}}=\sin \left(\frac{60^{\circ}+\delta}{2}ight)$\ or $\sin 45^{\circ}=\sin \left(\frac{60^{\circ}+\delta}{2}ight)$ or $45^{\circ}=\left(\frac{60^{\circ}+\delta}{2}ight)$\ or $\delta=30^{\circ}$\