Question

AB & CD are 2 glass slabs with refractive indices 1.41 and 1.73. The medium between them has a refractive index 2. Find the minimum angle of incidence at Q, so that the ray is totally reflected by both the slabs.

Answer 1

$$\begin{gathered} LetusconsiderrefractiveindexofglassAB={}^a\mu _{g1}=1.41=\sqrt{2.} \\ refractiveindexofglassCD={}^a\mu _{g2}=1.73=\sqrt{3}and, \\ refractiveindexofthemediuminbetweenthetwoglasses={}^a\mu _m=2. \\ whenarayoflightstrikessurfaceABatQ,fortotalinternalreflection,lettheangleofincidentatQ=\angle i{}_{c.} \\ Therefore\frac{\sin i_c}{\sin {90}^0}=\frac{\sqrt{2}}{2}or\sin i_c=\frac{1}{\sqrt{2}}or\sin i_c=\sin 45°or\angle i_c=45°. \\ ThesamerayoflightstrikesCDsurfaceatQwithangle\angle i_0. \\ thereforeSnell\text{'}slawsaysthat\frac{\sin i_0}{\sin 90°}=\frac{\sqrt{3}}{2}or\sin i_0=\sin 60°or\angle i_0={60}^{°}. \\ ThereforetheminimumangleatwhichtherayofligjtwillgetinternallytotallyreflectedfromboththesurfacesABandCDis\angle 60°. \end{gathered}$$