Question

As shown in the figure, a ray \(AB\) of unpolarized light enters from water of refractive index \(n_w=\frac{4}{3}\) into a medium of refractive index \(n_p=\frac{4}{\sqrt{3}}\) after passing through a glass plate of refractive index \(n_g=1.5\) and a layer of water. At a particular incident angle \(i\), the reflected ray \(CD\) is polarized in the direction as shown in the figure. The value of \(i\) (in degrees) is:

Hint:

For light traveling through multiple parallel slabs, the condition $n \sin \theta = \text{constant}$ is extremely useful. It allows you to skip intermediate layers (like the glass plate in this problem) and relate the first and last angles directly.

Answer 1

Correct Answer: 60

Step 1: Understanding the Question:
The reflected ray $CD$ is completely polarized. According to Brewster's Law, this occurs when the reflected and refracted rays are perpendicular, and the angle of incidence at that specific interface is equal to Brewster's angle. We need to find the initial incident angle $i$ in water.

Step 2: Key Formula or Approach:

• Brewster's Law: $\tan \theta_B = \frac{n_2}{n_1}$.

• Snell's Law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$.

Step 3: Detailed Explanation:

• Brewster's angle at the interface:
The polarization occurs at the reflection from the water ($n_w = 4/3$) to the medium ($n_p = 4/\sqrt{3}$).
Let $\theta_B$ be the angle of incidence at this interface.
$\tan \theta_B = \frac{n_p}{n_w} = \frac{4/\sqrt{3}}{4/3} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Therefore, $\theta_B = 60^\circ$.

• Applying Snell's Law:
The ray passes through several parallel layers. For parallel interfaces, the product $n \sin \theta$ remains constant throughout.
$n_{\text{initial}} \sin i = n_{\text{interface}} \sin \theta_B$.
$(4/3) \sin i = (4/3) \sin 60^\circ$.
$\sin i = \sin 60^\circ$.
$i = 60^\circ$.

Step 4: Final Answer:
The value of $i$ is 60.