Before You Begin

Question #43598

Difficulty: 8/10 | 🟢 Free

The Question

A wide, uniform, linearly polarized light beam propagates along the $+x$ -axis. Its electric field vector is initially aligned with the $+y$ -axis. The beam passes through a thin, non-uniform phase plate located at $x=0$ . Upon exiting the plate, the rays are deflected in the $xy$ -plane such that the ray at a transverse coordinate $y$ makes an angle $\theta(y) = \frac{\pi y}{3 y_0}$ with the $x$ -axis. The magnitude of the Poynting vector along the new propagation direction remains a constant $I_0$ (in $\text{W/m}^2$ ) for all rays.

Immediately after the plate, the light falls on an ideal flat linear polarizer placed in the $yz$ -plane. The transmission axis of the polarizer is parallel to the $y$ -axis. The polarizer transmits only the component of the electric field parallel to its transmission axis, such that the magnitude of the transmitted Poynting vector is given by Malus's law.

What is the total transmitted power $P$ through a rectangular aperture on the polarizer defined by $-y_0 \le y \le y_0$ and $0 \le z \le h$ ?

Options

A.
$\frac{9\sqrt{3}}{4\pi} I_0 h y_0$
B.
$\left(1 + \frac{3\sqrt{3}}{4\pi}\right) I_0 h y_0$
C.
$\frac{3\sqrt{3}}{\pi} I_0 h y_0$
D.
$\frac{3\sqrt{3}}{2\pi} I_0 h y_0$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Malus's Law|Poynting Vector and Intensity|Non-Uniform Dielectric Filling|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between the scalar intensity often used in Malus's Law and the vector nature of the Poynting vector when calculating power flux.

Students commonly fall into the trap of directly integrating the transmitted intensity $I_0 \cos^2\theta$ over the area (Option B).

However, because the rays are incident obliquely on the polarizer, the power intercepted by a differential area $dA$ on the $yz$ -plane is governed by the flux integral $\vec{S} \cdot d\vec{A} = S \cos\theta dA$ .

This introduces an additional geometric projection factor of $\cos\theta$ , making the correct integrand proportional to $\cos^3\theta$ .

Failing to account for the projection of the area normal to the propagation direction is a classic conceptual error in wave optics and electromagnetism.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Malus's Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free