Question

An equilateral prism deviates ray through ${23}^o$ for two angles of incidence differing by ${23}^o$. Find $\mu$ of the prism.

Answer 1

combining all the angles we get,

$\theta +\theta +23=60+23$

$2\theta ={60}^o$

$\theta =30$

at an angle of minimum deviation let angle incidence be $"i"$

$\left[\frac{{23}^0-{\delta }_{min}}{i^{\prime }-\theta }\right]=2\left[\frac{23-{\delta }_{min}}{53-i^{\prime }}\right]$

$30+\frac{{\delta }_{min}}{2}-30=2[53-30-\frac{{\delta }_{min}}{2}]$

$\frac{{\delta }_{min}}{2}=46-{\delta }_{min}$

$\frac{3{\delta }_{min}}{2}=92$

${\delta }_{min}=\frac{92\times 2}{3}$

$2i^{\prime }=60+{\delta }_{min}$

$i^{\prime }=30+\frac{{\delta }_{min}}{2}$

$\mu =\frac{sin\frac{60+60}{2}}{sin\frac{60}{2}}$

$\mu =\frac{sin60}{sin30}\Rightarrow \sqrt{3}$