In an experiment for determination of refractive index of glass of a prism by $i-\delta$, plot, it w

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 $35^{\circ}$, suffers a deviation of $40^{\circ}$ and that it emerges at angle $79^{\circ}$. In that case which of the following is closest to the maximum possible value of the refractive index?\

Professor PC Thomas Classes · Answer

(a) : Here, $i=35^{\circ}, e=79^{\circ}, \delta=40^{\circ}$ We know, $\delta=i+e-A \Rightarrow A=i+e-\delta$\ $	herefore \quad A=35^{\circ}+79^{\circ}-40^{\circ}=74^{\circ}$\ Refractive index of prism,\ $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}ight)}{\sin \left(\frac{A}{2}ight)}$\ $\mu=\frac{\sin \left(37^{\circ}+\frac{\delta_{m}}{2}ight)}{\sin 37^{\circ}}=\frac{5}{3} \sin \left(37^{\circ}+\frac{\delta_{m}}{2}ight)$\ Maximum value of $\mu$ can be $\frac{5}{3}$, so required value of $\mu$ should be less than $\frac{5}{3}$.\ Also, $\delta_{m}$ will be less than $40^{\circ}$, so\ $\mu<\frac{5}{3} \sin \left(37^{\circ}+\frac{40^{\circ}}{2}ight)=\frac{5}{3} \sin 57^{\circ}$\ $\mu<\frac{5}{3} \sin 57^{\circ}<\frac{5}{3} \sin 60^{\circ}=1.44 	herefore \mu<1.44$\ So the nearest possible value of $\mu$ for the given arrangement should be 1.5 .\