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 𝛿min = 30° (see Figure-2). The refractive index of the prism material for blue light is √3. Which of the following statement(s) is(are) correct? (A) The blue light is polarized in the plane of incidence. (B) The angle of the prism is 45°. (C) The refractive index of the material of the prism for red light is √2. (D) The angle of refraction for blue light in air at the exit plane of the prism is 60°

 

Given:

1. A plane-polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. This implies that the incident light is at Brewster's angle for the blue light.

2. The angle of deviation of the emergent blue light ray is $\delta =60^{\circ }$.

3. The angle of minimum deviation for red light from the same prism is ${\delta }_{\text{min}}=30^{\circ }$.

4. The refractive index of the prism material for blue light is $\sqrt{3}$.

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Key Concepts:

1. Brewster's Angle:

When light is incident at Brewster's angle (${\theta }_B$), the reflected and refracted rays are perpendicular to each other, and the reflected light is completely polarized. Brewster's angle is given by:

$$\tan {\theta }_B=n$$

where $n$ is the refractive index of the prism material.

2. Angle of Deviation ($\delta$):

The angle of deviation for a prism is given by:

$$\delta =A+D-180^{\circ }$$

where $A$ is the angle of the prism, and $D$ is the total angle through which the light is bent.

3. Minimum Deviation (${\delta }_{\text{min}}$):

For minimum deviation, the light passes symmetrically through the prism, and the relationship between the angle of the prism ($A$), the minimum deviation (${\delta }_{\text{min}}$), and the refractive index ($n$) is:

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

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Step-by-Step Solution:

Step 1: Determine the angle of the prism ($A$) for blue light.

Since the blue light is incident at Brewster's angle, we have:

$$\tan {\theta }_B=n_{\text{blue}}=\sqrt{3}$$$${\theta }_B={\tan }^{-1}(\sqrt{3})=60^{\circ }$$

At Brewster's angle, the angle of refraction ($r$) is:

$$r=90^{\circ }-{\theta }_B=90^{\circ }-60^{\circ }=30^{\circ }$$

The angle of the prism ($A$) is related to the angle of refraction ($r$) and the angle of deviation ($\delta$). Using the formula for deviation:

$$\delta =A+D-180^{\circ }$$

For minimum deviation, $D=2r$, so:

$$\delta =A+2r-180^{\circ }$$

Substitute $\delta =60^{\circ }$ and $r=30^{\circ }$:

$$60^{\circ }=A+2(30^{\circ })-180^{\circ }$$$$60^{\circ }=A+60^{\circ }-180^{\circ }$$$$A=180^{\circ }-60^{\circ }=120^{\circ }$$

Thus, the angle of the prism is $A=120^{\circ }$.

Step 2: Verify the angle of the prism using red light.

For red light, the minimum deviation is ${\delta }_{\text{min}}=30^{\circ }$. Using the formula for minimum deviation:

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

Substitute $A=120^{\circ }$ and ${\delta }_{\text{min}}=30^{\circ }$:

$$n_{\text{red}}=\frac{\sin \left(\frac{120^{\circ }+30^{\circ }}{2}\right)}{\sin \left(\frac{120^{\circ }}{2}\right)}=\frac{\sin (75^{\circ })}{\sin (60^{\circ })}$$

Using $\sin (75^{\circ })=\frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin (60^{\circ })=\frac{\sqrt{3}}{2}$:

$$n_{\text{red}}=\frac{\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{6}+\sqrt{2}}{2\sqrt{3}}=\frac{\sqrt{2}(\sqrt{3}+1)}{2\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{6}}$$

Rationalizing:

$$n_{\text{red}}=\frac{(\sqrt{3}+1)\sqrt{6}}{6}=\frac{\sqrt{18}+\sqrt{6}}{6}=\frac{3\sqrt{2}+\sqrt{6}}{6}$$

This does not simplify to $\sqrt{2}$, so statement (C) is incorrect.

Step 3: Check the angle of refraction for blue light.

The angle of refraction for blue light at the exit plane of the prism is $r=30^{\circ }$, not $60^{\circ }$. Thus, statement (D) is incorrect.

Step 4: Verify the polarization of blue light.

Since the blue light is incident at Brewster's angle, it is polarized in the plane of incidence. Thus, statement (A) is correct.

Step 5: Verify the angle of the prism.

From Step 1, the angle of the prism is $A=120^{\circ }$, not $45^{\circ }$. Thus, statement (B) is incorrect.

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Final Answer:

Only statement (A) is correct.