Question

Calculate the angle of minimum deviation of ray of light passing through an equilateral prism of refractive index $1.732$

Answer 1

Given, A = ${60}^0$

$\mu =1.732$

Since, angle of minimum deviation is given by,

$\mu =\frac{sin\frac{A+\delta m}{2}}{sin\frac{A}{2}}$

$1.732\times \frac{1}{2}=sin(30+\frac{\delta m}{2})$$

$si{n}^{-1}\left(0.8666\right)=30+\frac{\delta m}{2}$

${60}^0=30\frac{\delta m}{2}$

$\delta m={60}^0$

Now,

$\delta m=i+i^{\prime }-A$

${60}^0=i+i^{\prime }-{60}^0$ ( $\delta ={60}^0$ minimum deviation )

= > i = ${60}^0$ .

so, the angle of incidence must be ${60}^0$.