Question

A light beam given by \(\mathbf{E}(z, t) = E_{01} \sin(kz - \omega t)\hat{i} + E_{02} \sin(kz - \omega t + \frac{\pi}{6})\hat{j}\) passes through an ideal linear polarizer whose transmission axis is tilted by 60\(^\circ\) from x-axis (in x-y plane). If \(E_{01} = 4\) V/m and \(E_{02} = 2\) V/m, the electric field amplitude of the emerging light beam from the polarizer is ______ V/m. (up to two decimal places)

Hint:

When combining two sinusoidal functions of the same frequency but different phases, like \(A\sin(\omega t) + B\cos(\omega t)\), the resulting wave is also a sinusoid with the same frequency, and its amplitude is always \(\sqrt{A^2 + B^2}\). This is a very useful identity in wave physics and AC circuit analysis.

Answer 1

Correct Answer: 3.61

Step 1: Understanding the Concept:
An ideal linear polarizer transmits only the component of the incident electric field that is parallel to its transmission axis. The transmitted electric field is the projection of the incident electric field vector onto the transmission axis. The amplitude of the emerging beam is the maximum value of this transmitted field component.
Step 2: Key Formula or Approach:
1. Represent the transmission axis as a unit vector \(\hat{p}\). 2. The transmitted electric field is given by \(E_{trans} = \mathbf{E}_{inc} \cdot \hat{p}\). 3. The amplitude of this resulting scalar wave is found by combining the two sinusoidal terms into a single sinusoid of the form \(A \cos(\phi - \delta)\), where the amplitude is \(A\).
Step 3: Detailed Explanation:
The incident electric field is \(\mathbf{E}_{inc} = E_x \hat{i} + E_y \hat{j}\), where: \[ E_x = 4 \sin(kz - \omega t) \] \[ E_y = 2 \sin(kz - \omega t + \pi/6) \] The transmission axis of the polarizer is at 60\(^\circ\) to the x-axis, so its unit vector is: \[ \hat{p} = \cos(60^\circ)\hat{i} + \sin(60^\circ)\hat{j} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \] The transmitted electric field is \(E_{trans} = \mathbf{E}_{inc} \cdot \hat{p}\): \[ E_{trans} = E_x \cos(60^\circ) + E_y \sin(60^\circ) = \frac{1}{2} E_x + \frac{\sqrt{3}}{2} E_y \] Substitute the expressions for \(E_x\) and \(E_y\): \[ E_{trans} = \frac{1}{2}[4 \sin(kz - \omega t)] + \frac{\sqrt{3}}{2}[2 \sin(kz - \omega t + \pi/6)] \] Let \(\phi = kz - \omega t\). \[ E_{trans} = 2 \sin(\phi) + \sqrt{3} \sin(\phi + \pi/6) \] Use the angle addition formula \(\sin(A+B) = \sin A \cos B + \cos A \sin B\): \[ E_{trans} = 2 \sin(\phi) + \sqrt{3} [\sin(\phi)\cos(\pi/6) + \cos(\phi)\sin(\pi/6)] \] \[ E_{trans} = 2 \sin(\phi) + \sqrt{3} [\sin(\phi)\frac{\sqrt{3}}{2} + \cos(\phi)\frac{1}{2}] \] \[ E_{trans} = 2 \sin(\phi) + \frac{3}{2}\sin(\phi) + \frac{\sqrt{3}}{2}\cos(\phi) \] \[ E_{trans} = \left(2 + \frac{3}{2}\right)\sin(\phi) + \frac{\sqrt{3}}{2}\cos(\phi) = \frac{7}{2}\sin(\phi) + \frac{\sqrt{3}}{2}\cos(\phi) \] This is a sinusoidal function of the form \(A\sin(\phi) + B\cos(\phi)\). Its amplitude is \(\sqrt{A^2 + B^2}\). \[ \text{Amplitude} = \sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13} \] \[ \sqrt{13} \approx 3.6055 \] Step 4: Final Answer:
The electric field amplitude of the emerging light beam is 3.61 V/m.