Physics - Geometrical Optics Question with Solution | TestHub

At point A by Snell's law $$\mu =\frac{\sin 45^{\circ }}{\sin r}\Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}$$ At point $\mathrm{B}$, for total internal reflection, $$\sin i_1=\frac{1}{\mu }$$
From figure, $i_1=90^{\circ }-r$ $$\begin{array}{rl}& \therefore \left(\sin 90^{\circ }-r\right)=\frac{1}{\mu }\\ & \Rightarrow \cos r=\frac{1}{\mu }\end{array}$$ Now $\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}$ $$=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$$ From eqs (ii) and (iii) $$\frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$$ Squaring both sides and then solving, we get $$\mu =\sqrt{\frac{3}{2}}$$