A plane 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 δ = 60° (see Figure-1). The angle of minimum deviation for red light from the same prism is δ_min = 30° (see Figure-2). The refractive index of the prism material for blue light is √3

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Which of the following statement(s) is(are) correct?

Correct Statements:

• The blue light is polarized in the plane of incidence.
• The refractive index of the material of the prism for red light is $\sqrt{2}$.
• The angle of refraction for blue light in air at the exit plane of the prism is 60°.

Step-by-step Explanation:

1. Polarization of Blue Light and Brewster's Law:

From the problem description and Figure-1, when a plane-polarized blue light ray is incident on the first surface of the prism, there is no reflection. This occurs when light is incident at Brewster's angle ($i_B$) and the electric field vectors lie entirely within the plane of incidence (p-polarized light). Therefore, the blue light is polarized in the plane of incidence.

According to Brewster's Law, the angle of incidence $i_1$ for blue light is given by:

$\tan i_1={\mu }_b=\sqrt{3}$

$\Rightarrow i_1=60°$

The angle of refraction inside the prism at the first surface $r_1$ is:

$r_1=90°-i_1=90°-60°=30°$

2. Angle of the Prism ($A$) and Emergent Angle for Blue Light (${\mathrm{i;}}_2$):

The deviation of light through a prism is given by:

$\delta =i_1+i_2-A$

Given $\delta =60°$ for blue light:

$60°=60°+i_2-A$

$\Rightarrow i_2=A$

Using Snell's law at the emergent surface for blue light:

${\mu }_b\cdot \sin r_2=1\cdot \sin i_2$

Since $A=r_1+r_2=30°+r_2$, we have $r_2=A-30°$. Substituting $i_2=A$:

$\sqrt{3}\cdot \sin (A-30°)=\sin A$

$\sqrt{3}(\sin A\cos 30°-\cos A\sin 30°)=\sin A$

$\sqrt{3}\left(\sin A\cdot \frac{\sqrt{3}}{2}-\cos A\cdot \frac{1}{2}\right)=\sin A$

$\frac{3}{2}\sin A-\frac{\sqrt{3}}{2}\cos A=\sin A$

$\frac{1}{2}\sin A=\frac{\sqrt{3}}{2}\cos A$

$\tan A=\sqrt{3}\Rightarrow A=60°$

Thus, the angle of the prism is $A=60°$, and the angle of refraction (emergence angle) for blue light in air at the exit plane of the prism is $i_2=60°$.

3. Refractive Index for Red Light:

From Figure-2, the angle of minimum deviation for red light is ${\delta }_{\text{min}}=30°$.

The relation between refractive index ${\mu }_r$, angle of prism $A$, and minimum deviation ${\delta }_{\text{min}}$ is:

${\mu }_r=\frac{\sin \left(\frac{A+{\delta }_{\text{min}}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Substituting $A=60°$ and ${\delta }_{\text{min}}=30°$:

${\mu }_r=\frac{\sin \left(\frac{60°+30°}{2}\right)}{\sin \left(\frac{60°}{2}\right)}=\frac{\sin 45°}{\sin 30°}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}$

Therefore, the refractive index of the material of the prism for red light is $\sqrt{2}$.