Question
In Fig. $$ABC$$ is the cross section of a right - angled prism and $$ACDE$$ is the cross section of a glass slab. The value of $$\theta $$ so that incident normally on the face $$AB$$ does not cross the face $$AC$$ is (given $${\sin ^{ - 1}}\left( {\frac{3}{5}} \right) = {37^ \circ }$$ ).
In Fig. $$ABC$$ is the cross section of a right - angled prism and $$ACDE$$ is the cross section of a glass slab. The value of $$\theta $$ so that incident normally on the face $$AB$$ does not cross the face $$AC$$ is (given $${\sin ^{ - 1}}\left( {\frac{3}{5}} \right) = {37^ \circ }$$ ).
A.
$$\theta \leqslant {37^ \circ }$$
B.
$$\theta < {37^ \circ }$$
C.
$$\theta \leqslant {53^ \circ }$$
D.
$$\theta < {53^ \circ }$$
Answer :
$$\theta < {37^ \circ }$$
Solution :
$$\eqalign{ & A = {90^ \circ } - \theta \cr & \Rightarrow {r_2} = A = {90^ \circ } - \theta > {\theta _c} \cr & \cos \theta > \sin {\theta _c} = \frac{{\frac{6}{5}}}{{\frac{2}{3}}} = \frac{4}{5}\,\,\left( {{\theta _c}\,{\text{is critical angle}}} \right) \cr & \theta < {\cos ^{ - 1}}\frac{4}{5} = {37^ \circ } \cr} $$
$$\eqalign{ & A = {90^ \circ } - \theta \cr & \Rightarrow {r_2} = A = {90^ \circ } - \theta > {\theta _c} \cr & \cos \theta > \sin {\theta _c} = \frac{{\frac{6}{5}}}{{\frac{2}{3}}} = \frac{4}{5}\,\,\left( {{\theta _c}\,{\text{is critical angle}}} \right) \cr & \theta < {\cos ^{ - 1}}\frac{4}{5} = {37^ \circ } \cr} $$
