Question

A ray of light passes through an equilateral prism $(\mu =1.5)$. The angle of minimum deviation is

• A. ${45}^o$
• B. ${37}^o{12}^{\prime }$
• C. ${20}^o$
• D. ${30}^o$

Answer 1

Answer: (B)

${37}^o{12}^{\prime }$

$\mu =\frac{sin(A+{\delta }_m)/2}{sinA/2}$

$\Rightarrow \frac{sin(A+{\delta }_m)}{2}=\mu sinA/2=1.5\times sin{30}^o=0.75$

$\frac{A+{\delta }_m}{2}=si{n}^{-1}\left(0.75\right)={48}^o{36}^{\prime }\therefore {\delta }_m={37}^o{12}^{\prime }$