In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of $θ$ so that light incident normally on the face AB does not cross the face BC is :- (given $sin^{- 1} (3/5) = 3 7^{\circ}$ )

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TestHub · Accepted Answer

Ray enters face AB normally, so it travels undeviated and hits face BC at incidence $i = 9 0^{\circ} - θ$ .

To not cross BC, TIR at prism-slab interface: $sin i > \frac{n _{2}}{n _{1}} = \frac{6/5}{3/2} = \frac{4}{5}$ .

$cos θ > \frac{4}{5} \Rightarrow θ < 3 7^{\circ}$ - option (B).

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