The refractive index of an equilateral triangular prism kept in air is √ 2. Calculate: (a) the angle of minimum deviation. (b) the critical angle for…

$\displaystyle \mu=\frac{\sin \left(\dfrac{\delta_{m}+A}{2}\right.}{\sin (A / 2)}$

$$\sqrt{2}=\frac{\sin \left(\dfrac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)}{\sin $\displaystyle 30$^{\circ}} \Rightarrow \sin \left(\frac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)=\sqrt{2} \times \frac{1}{2}

$$$$ \sin \left(\frac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)=\frac{1}{\sqrt{2}}=>\left(\frac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)=$\displaystyle 45$^{\circ}

$$$$ \delta_{m}=$\displaystyle 90$^{\circ}-$\displaystyle 60$^{\circ} \Rightarrow \delta_{m}=$\displaystyle 30$^{\circ}

$$$$ \text { b) } \sin \theta_{c}=\frac{1}{\mu}=\frac{1}{\sqrt{2}}

$$$$ \theta_{c}=$\displaystyle 45$^{\circ}