Question
The angle of a prism is $$A.$$ One of its refracting surfaces is silvered. Light rays falling at an angle of incidence $$2A$$ on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index $$\mu ,$$ of the prism is
A.
$$2\sin A$$
B.
$$2\cos A$$
C.
$$\frac{1}{2}\cos A$$
D.
$$\tan A$$
Answer :
$$2\cos A$$
Solution :
According to question, diagram is shown below.
$$\eqalign{ & {\text{So,}}\,\,\angle MON = {90^ \circ } - A \cr & {\text{and}}\,\,\angle r = {90^ \circ } - \left( {{{90}^ \circ } - A} \right) \cr & \Rightarrow \angle r = A \cr} $$
By Snell's law, $$\frac{{\sin i}}{{\sin r}} = \mu $$
$$\eqalign{ & \frac{{\sin \left( {2A} \right)}}{{\sin \left( A \right)}} = \mu \cr & \frac{{2\sin A\cos A}}{{\sin A}} = \mu \cr & \Rightarrow \mu = 2\cos A \cr} $$
According to question, diagram is shown below.
$$\eqalign{ & {\text{So,}}\,\,\angle MON = {90^ \circ } - A \cr & {\text{and}}\,\,\angle r = {90^ \circ } - \left( {{{90}^ \circ } - A} \right) \cr & \Rightarrow \angle r = A \cr} $$
By Snell's law, $$\frac{{\sin i}}{{\sin r}} = \mu $$
$$\eqalign{ & \frac{{\sin \left( {2A} \right)}}{{\sin \left( A \right)}} = \mu \cr & \frac{{2\sin A\cos A}}{{\sin A}} = \mu \cr & \Rightarrow \mu = 2\cos A \cr} $$
