Question

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

• A. ${45}^o$
• B. ${54.74}^o$
• C. ${60}^o$
• D. ${30}^o$

Answer 1

Answer: (D)

${30}^o$

By principle of polarisation, $\mu$ = tan${\theta }_p$
or $\mu$ = tan 54.74${}^o$ or $\mu$ = 1.414
For an equilateral prism, $\angle A$= 60${}^o$
$\therefore \mu$ = $\frac{sin\left(\frac{A+\delta }{2}\right)}{sin\left(A/2\right)}$ = $\frac{sin\left(\frac{{60}^0+\delta }{2}\right)}{sin\left({60}^0/2\right)}$
or, $\frac{1.414\times 1}{2}$ = $sin\left(\frac{{60}^0+\delta }{2}\right)$ [$\therefore 1.414=\sqrt{2}]$
or, $\frac{\sqrt{2}}{2}$ = $sin\left(\frac{{60}^0+\delta }{2}\right)$ or $\frac{1}{\sqrt{2}}$ = $sin\left(\frac{{60}^0+\delta }{2}\right)$
or, sin45${}^0$ = $sin\left(\frac{{60}^0+\delta }{2}\right)$ or 45${}^0$ = $\left(\frac{{60}^0+\delta }{2}\right)$

$\delta ={90}^0-{60}^0={30}^0$