In an experiment for determination of refractive index of glass of a prism by \(i - \delta\) , plot it was found thata ray incident at angle \(35 ^o \), suffers a deviation of \(40 ^o \) and that it emerges at angle \(7 9 ^o \). In that case which of the following is closest to the maximum possible value of the refractive index?

\(35^o+79^o-\mathrm{A}=40^o\)    \(\therefore \quad \mathrm{A}=74^o\)

But \(\mu  = \frac{{\sin \left( {\frac{{{\text{A}} + {\delta _{\text{m}}}}}{2}} \right)}}{{\sin {\text{A}}/2}}\) \( = \frac{{\sin \left( {\frac{{74 + {\delta _{\text{m}}}}}{2}} \right)}}{{\sin \frac{{74}}{2}}}\)

\( = \frac{5}{3}\sin \left( {{{37}^o} + \frac{{{\delta _{\text{m}}}}}{2}} \right)\)

\(\mu_{\max }\) can be \(\frac{5}{3}\). That is \(\mu_{\max }\) is less than \(\frac{5}{3}=1.67\)

But \(\delta_{\mathrm{m}}\) will be less than \(40^o\) so

\(\mu  < \frac{5}{3}\sin {57^o}\)  \( < \frac{5}{3}\sin {60^o}\) \( \Rightarrow \mu  = 1.45\)