$\mu$ = ${{\sin \left( {{{{\delta _m} + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$

= ${{\sin \left( {{{A + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$

= ${{\sin A} \over {\sin \left( {{A \over 2}} \right)}}$

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

= $2\cos \left( {{A \over 2}} \right)$

The angle of minimum deviation is given as

$\delta$_min = i + e – A

for minimum deviation

$\delta$_min = A and i = e then

2A = i + i

$\Rightarrow$ i = A

i_min = 0 = A_min $\Rightarrow$ $\mu$_min = 2

i_max = 90⁰ = A_max $\Rightarrow$ $\mu$_max = $\sqrt 2$

$\therefore$ $\mu$ lies between 2 and $\sqrt 2$.