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

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: $\tan 37^{\circ}=\frac{3}{4}$
$\therefore \sin \left(37^{\circ}\right)=\sqrt{\frac{\tan ^{2}\left(37^{\circ}\right)}{1+\tan ^{2}\left(37^{\circ}\right)}}$
$=\sqrt{\frac{(9 / 16)}{1+(9 / 16)}}=\frac{3}{5}$
$\therefore \mu < \frac{\sin \left(37^{\circ}\right)}{\sin \left(30^{\circ}\right)}=\frac{3}{5} \times 2=1.2$
$\therefore \mu < 1.2$