An unpolarized light ray passing through air (refractive index \(n_a = 1\)) is incident on a glass slab (refractive index \(n_g = \sqrt{3}\)) at an angle of 60°, as shown in the figure below. The amplitude of the in-plane (x-y) electric field component of the incident light is 4 V/m and amplitude of the out of plane (z) electric field component is 3 V/m. After passing through the glass slab, the electric field amplitude (in V/m) of the light is
Show Hint
When a problem involves reflection or transmission at an angle, always calculate Brewster's angle first (\(\tan\theta_B = n_2/n_1\)). If the angle of incidence matches, it simplifies the problem significantly, as the reflection of the p-polarized component becomes zero. For a parallel-sided slab, if the incidence is at \(\theta_B\), the p-component is transmitted with 100% amplitude.
Step 1: Understanding the Concept
The problem involves the transmission of a polarized light ray through a glass slab. We need to analyze what happens to the electric field components parallel (in-plane, p-polarized) and perpendicular (out-of-plane, s-polarized) to the plane of incidence as the light passes through both surfaces of the slab. A key aspect is to check if the incidence occurs at a special angle, like Brewster's angle. Step 2: Key Formula or Approach
1. Brewster's Angle (\(\theta_B\)): This is the angle of incidence at which light with a particular polarization is perfectly transmitted through a dielectric surface, with no reflection. It is given by \(\tan(\theta_B) = \frac{n_2}{n_1}\).
2. Fresnel's Transmission Coefficients: These coefficients describe the amplitude of the transmitted electric field relative to the incident electric field. For a light ray going from medium 1 to medium 2, the transmission coefficients for the parallel (\(t_p\)) and perpendicular (\(t_s\)) components are:
\[ t_p = \frac{2n_1 \cos\theta_1}{n_2 \cos\theta_1 + n_1 \cos\theta_2} \]
\[ t_s = \frac{2n_1 \cos\theta_1}{n_1 \cos\theta_1 + n_2 \cos\theta_2} \]
3. Snell's Law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\). Step 3: Detailed Explanation Interface 1: Air to Glass
- Given: \(n_a = n_1 = 1\), \(n_g = n_2 = \sqrt{3}\), and angle of incidence \(\theta_i = 60^\circ\).
- Let's check for Brewster's angle:
\[ \tan(\theta_B) = \frac{n_g}{n_a} = \frac{\sqrt{3}}{1} = \sqrt{3} \]
This gives \(\theta_B = 60^\circ\). The light is incident at Brewster's angle.
- We find the angle of refraction (\(\theta_r\)) using Snell's Law:
\[ n_a \sin(\theta_i) = n_g \sin(\theta_r) \]
\[ 1 \cdot \sin(60^\circ) = \sqrt{3} \cdot \sin(\theta_r) \]
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \sin(\theta_r) \implies \sin(\theta_r) = \frac{1}{2} \implies \theta_r = 30^\circ \]
Interface 2: Glass to Air
- The glass slab has parallel faces, so the angle of incidence at the second interface is \(\theta'_i = \theta_r = 30^\circ\).
- The light goes from glass (\(n_1 = \sqrt{3}\)) to air (\(n_2 = 1\)).
- Let's check for Brewster's angle at this interface:
\[ \tan(\theta'_B) = \frac{n_a}{n_g} = \frac{1}{\sqrt{3}} \]
This gives \(\theta'_B = 30^\circ\). So, the incidence at the second interface is also at Brewster's angle. Calculating Transmitted Amplitudes
A special property of a parallel slab is that if light is incident at Brewster's angle, the overall transmission coefficient for the parallel component of the electric field amplitude is 1 (assuming no interference effects). Let's verify this.
- Let \(t_{12,p}\) be the transmission coefficient from air to glass for the p-component, and \(t_{21,p}\) be from glass to air.
- For interface 1 (\(\theta_1=60^\circ, \theta_2=30^\circ\)):
\[ t_{12,p} = \frac{2(1)\cos(60^\circ)}{\sqrt{3}\cos(60^\circ) + 1\cos(30^\circ)} = \frac{2(1/2)}{\sqrt{3}(1/2) + (\sqrt{3}/2)} = \frac{1}{\sqrt{3}} \]
- For interface 2 (\(\theta_1=30^\circ, \theta_2=60^\circ\)):
\[ t_{21,p} = \frac{2(\sqrt{3})\cos(30^\circ)}{1\cos(30^\circ) + \sqrt{3}\cos(60^\circ)} = \frac{2\sqrt{3}(\sqrt{3}/2)}{(\sqrt{3}/2) + \sqrt{3}(1/2)} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
- The total transmission for the p-component amplitude is \(T_p = t_{12,p} \times t_{21,p} = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1\).
- The initial amplitude of the in-plane (p-component) is \(E_{p,i} = 4\) V/m.
- The final amplitude of the transmitted p-component is \(E_{p,f} = T_p \times E_{p,i} = 1 \times 4 = 4\) V/m.
The amplitude of the transmitted in-plane (parallel) component is exactly 4 V/m, which matches option (B). Given the options, it is highly probable that the question is asking for the amplitude of the in-plane component of the transmitted light, despite the ambiguous phrasing "the electric field amplitude".
Thus, the answer is 4 V/m.
