b
As  \(\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

\(\mu  = \frac{{\sin \left( {\frac{{A + A}}{2}} \right)}}{{\sin \left( {\frac{A}{2}} \right)}}\)  \( = \frac{{\sin A}}{{\sin \left( {\frac{A}{2}} \right)}}\)    \(\left( {\because {\delta _m} = A({\text{ Given }})} \right)\)

\( = \frac{{2\sin \left( {\frac{A}{2}} \right)\cos \left( {\frac{A}{2}} \right)}}{{\sin \left( {\frac{A}{2}} \right)}}\) \( = 2\cos \left( {\frac{A}{2}} \right)\)

As \(\delta=i+e-A\)

At minimum deviation, \(\delta=\delta_{m}, i=e\)

\(\therefore {\delta _m} = 2i - A\)

\(2i = {\delta _m} + A\)

\(i = \frac{{{\delta _m} + A}}{2} = \frac{{A + A}}{2} = A\)        \(\left( {\because \,\,{\delta _m} = A({\text{ given }})} \right)\)

\({i_{\min }} = {0^o} \Rightarrow {A_{\min }} = {0^o}\)

Then, \(\mu_{\max }=2 \cos 0^o=2\)

\(\therefore \) \(i_{\max }=\frac{\pi}{2} \Rightarrow A_{\max }=\frac{\pi}{2}\)

Then, \(\mu_{\min }=2 \cos 45^o=2 \times \frac{1}{\sqrt{2}}=\sqrt{2}\)