Question

In an experiment for determination of refractive index of glass of a prism by $i-\delta$ plot, it was found that a ray incident at angle $30^{\circ }$, suffers a deviation of $14^{\circ }$ and that it emerges at angle $44^{\circ }$. In that case, the refractive index, $\mu <x$. Find the minimum value of $x$.
(Take $\tan 37^{\circ }=0.75$ )

Answer 1

Answer: 1.2

From the given data,
$A=i+e-\delta =30+44-14=60^{\circ }$
Now, $\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}<\frac{\sin \left(\frac{A+\delta }{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\therefore \mu <\frac{\sin \left(\frac{60+14}{2}\right)}{\sin \left(\frac{60}{2}\right)}=\frac{\sin \left(37^{\circ }\right)}{\sin \left(30^{\circ }\right)}$
Given: $\tan 37^{\circ }=\frac{3}{4}$
$\therefore \sin \left(37^{\circ }\right)=\sqrt{\frac{{\tan }^2\left(37^{\circ }\right)}{1+{\tan }^2\left(37^{\circ }\right)}}$
$=\sqrt{\frac{(9/16)}{1+(9/16)}}=\frac{3}{5}$
$\therefore \mu <\frac{\sin \left(37^{\circ }\right)}{\sin \left(30^{\circ }\right)}=\frac{3}{5}\times 2=1.2$
$\therefore \mu <1.2$