Physics - Geometrical Optics Question with Solution | TestHub

Critical angles at 1 and 2



${\theta }_{c_1}={\text{sin}}^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\right)={\text{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^0$

${\theta }_{c_2}={\text{sin}}^{-1}\left(\frac{{\mu }_3}{{\mu }_2}\right)={\text{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^0$

Therefore, minimum angle of incidence for total internal reflection to take place on both slabs should be 60⁰