Question

Angle of minimum deviation for a prism of refractive index $1.5$, is equal to the angle of the prism. Then the angle of the prism is (given $\cos {41}^o=0.75$)

• A. ${62}^o$
• B. ${41}^o$
• C. ${82}^o$
• D. ${31}^o$

Answer 1

Answer: (C)

${82}^o$

Given,

$\mu =1.5$ refractive index

${\delta }_m=A$

The refractive index of the prism,

$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$

$\mu =\frac{sin\left(\frac{A+A}{2}\right)}{sin\left(\frac{A}{2}\right)}=\frac{sin\left(A\right)}{sin\left(\frac{A}{2}\right)}$

$\mu =\frac{2sin\left(\frac{A}{2}\right)cos\left(\frac{A}{2}\right)}{sin\left(\frac{A}{2}\right)}$ ($sin2x=2sinx.cosx$)

$\mu =2cos\left(\frac{A}{2}\right)$

$1.5=2cos\left(\frac{A}{2}\right)$

$cos\left(\frac{A}{2}\right)=0.75$

$\frac{A}{2}={41}^0$

The angle of prism, $A={82}^0$

The correct option is C.