Question

A plain polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is \(\delta\) = 60\(^{\circ}\) (figure 1). The angle of min. deviation for red light from the same prism is min \(\delta_{min}\) = 30\(^{\circ}\) (figure 2). If the refractive index of prism material for blue light is √3, then which of the following options is correct? 

PPBL = Plane Polarized Blue Light                 PPRL = Plane Polarized Red Light
 

• A. Angle r for blue light in air at the exit plane of the prism is 60\(^{\circ}\)
• B. The angle of the prism is 45\(^{\circ}\)
• C. Blue light is polarized in the plane of incidence
• D. The refractive index of the material of the prism for red light is √2

Answer 1

The Correct Option is B, C, D

The correct options are: (B),(C) and (D).
For blue light,
\(\mu=tan\,i_p\)
\(\sqrt3=tan\,i_p\)
\(i_p=60^{\circ}\)
\(\delta=i+e-A\)
\(60=60+e-A\)
\(e=A\)
\(sin\,60^{\circ}=\sqrt3\,sin\,r_1\)
\(\frac{\sqrt3}{2}=\sqrt3\,sin\,r_1\)
\(r_1=30^{\circ}\)
\(r_2=A-30^{\circ}\)
\(\sqrt3sin\,(A-30^{\circ})=sin\,A\)
\(\frac{3}{2}sinA-\frac{\sqrt3}{2}cosA=sinA\)
\(\frac{-\sqrt3}{2}cosA=-\frac{sinA}{2}\)
\(tanA=\sqrt3\)
\(A=e=60^{\circ}\)
For red light, \(\frac{sin(\frac{A+\delta_{min}}{2})}{sin\frac{A}{2}}=n_R\)
\(\frac{sin\frac{90}{2}}{sin\,30}=n_R\)
\(n=\sqrt2\)