Malus's Law and Light Intensity

1. From Electric Field to Intensity: The Basis of Malus's Law

Let's recap a key idea from last time: a polarizing filter has a transmission axis. When polarized light hits this filter (which we now call an analyzer to distinguish it from the first filter), only the component of the light's electric field that is parallel to the analyzer's transmission axis gets through.

Imagine a beam of vertically polarized light with an electric field amplitude $E_0$ incident on an analyzer whose transmission axis is tilted at an angle $\theta$ to the vertical.

From simple trigonometry, the component of the incident electric field amplitude that is parallel to the analyzer's axis is $E_0\cos \theta$. This is the amplitude of the light wave that emerges from the analyzer.

But how does this relate to the intensity (the brightness) of the light? In wave physics, a fundamental principle is that the intensity of a wave is proportional to the square of its amplitude.

So, if the incident intensity is $I_0$ (proportional to $E_0^2$) and the transmitted intensity is $I$ (proportional to $(E_0\cos \theta {)}^2$), we can establish a direct relationship.

The following video provides an excellent, step-by-step derivation of this relationship. It first explains the amplitude component and then separately introduces the concept of intensity before combining them to derive the law.

This derivation leads us to the celebrated Malus's Law:

Where:

• $I$ is the intensity of the light transmitted by the analyzer.
• $I_0$ is the intensity of the polarized light incident on the analyzer.
• $\theta$ is the angle between the polarization direction of the incident light and the transmission axis of the analyzer.

For a concise written summary of this derivation, you can refer to the "Derivation of Malus Law" section in the Vedantu article.

2. Working with Malus's Law: Key Scenarios

Let's master the two most important scenarios you'll encounter.

Scenario 1: Unpolarized Light Passing Through a Polarizer

What happens when unpolarized light of intensity $I_{un}$ passes through its first polarizer? Unpolarized light has electric fields oscillating in all random directions. Malus's Law, as written, requires a single incident polarization angle $\theta$.

To solve this, we average the effect over all possible angles. The average value of ${\cos }^2\theta$ over a full cycle (0 to 360 degrees) is $\frac{1}{2}$.

Therefore, when unpolarized light passes through a polarizer, its intensity is always reduced by half, and it becomes polarized along the transmission axis of that polarizer.

This is a critical first step in almost every problem involving unpolarized light. The orientation of this first polarizer does not matter; it will always transmit 50% of the unpolarized light's intensity.

The following video explains this point very clearly.

Scenario 2: Analyzing Light with a Second Polarizer

Once the light is polarized (with intensity $I_1$), we can use Malus's Law directly to find the intensity $I_2$ after it passes through a second polarizer (the analyzer).

Here, $\theta$ is the angle between the transmission axis of the first polarizer and the second polarizer.

Let's consider two special cases:

• Parallel Polarizers ($\theta =0^{\circ }$): ${\cos }^2(0^{\circ })=1$, so $I_2=I_1$. All the polarized light passes through.
• Crossed Polarizers ($\theta =90^{\circ }$): ${\cos }^2(90^{\circ })=0$, so $I_2=0$. No light passes through. This is called extinction.

3. NEET-Style Problem Solving

Your goal is to prepare for the NEET exam, so let's apply these concepts to the types of problems you'll face. The process for a system of polarizers is sequential.

1. Start with the initial light (unpolarized or polarized).
2. Calculate the intensity after the first polarizer.
3. Use that result as the input for the second polarizer.
4. Repeat for all subsequent polarizers.

This resource contains solved examples from previous JEE exams, which are very similar in style and difficulty to NEET questions. Let's walk through them.

Worked Example 1: Two Polarizers

• Problem: Unpolarized light of intensity $I_{initial}$ passes through two polarizers with transmission axes at 30° to each other. Find the intensity of emergent light.

• Step 1: First Polarizer

  • The incident light is unpolarized. The first polarizer halves the intensity.
  • $I_1=I_{initial}/2$

• Step 2: Second Polarizer

  • The light incident on the second polarizer has intensity $I_1$ and is polarized. The angle between the two polarizers is $\theta =30^{\circ }$.
  • Apply Malus's Law: $I_2=I_1{\cos }^2(30^{\circ })$
  • Substitute $I_1$: $I_2=(I_{initial}/2)\times (\frac{\sqrt{3}}{2}{)}^2=(I_{initial}/2)\times \frac{3}{4}$
  • Result: $I_2=\frac{3}{8}{I}_{initial}$

Worked Example 2: Three Polarizers (A Classic Trick)

• Problem: Three polarizers are placed in series. The first and third have their axes perpendicular (crossed), while the middle one is at 45° to the first. Find the emergent intensity if the initial intensity of unpolarized light is $I_{initial}$.

• Step 1: First Polarizer (P1)

  • Unpolarized light. Intensity is halved.
  • $I_1=I_{initial}/2$. The light is now polarized along the axis of P1.

• Step 2: Second Polarizer (P2)

  • The angle between P1's axis and P2's axis is ${\theta }_1=45^{\circ }$.
  • $I_2=I_1{\cos }^2(45^{\circ })=(I_{initial}/2)\times (\frac{1}{\sqrt{2}}{)}^2=(I_{initial}/2)\times \frac{1}{2}=I_{initial}/4$.
  • The light is now polarized along the axis of P2.

• Step 3: Third Polarizer (P3)

  • P1 and P3 are perpendicular. P2 is at 45° to P1. So, the angle between P2's axis and P3's axis is also ${\theta }_2=90^{\circ }-45^{\circ }=45^{\circ }$.
  • $I_3=I_2{\cos }^2(45^{\circ })=(I_{initial}/4)\times (\frac{1}{\sqrt{2}}{)}^2=(I_{initial}/4)\times \frac{1}{2}$
  • Result: $I_3=I_{initial}/8$

Notice the fascinating result: if you remove the middle polarizer, the first and third are crossed, and the final intensity would be zero! The middle polarizer effectively "rotates" the plane of polarization, allowing some light to pass through the final filter.

Conclusion

You've now mastered the core mathematical tool for analyzing polarized light. By breaking down problems sequentially, you can confidently tackle questions involving multiple polarizers. Remember the two key rules, and you'll be well-prepared for this topic on the NEET.

Key Takeaways:

• Malus's Law for polarized light is $I=I_0{\cos }^2\theta$.
• $\theta$ is always the angle between the polarization of the incident light and the transmission axis of the current filter.
• The first step for unpolarized light is always to halve the intensity: $I_1=I_{un}/2$.
• For systems with multiple polarizers, calculate the intensity change at each step, using the output of one as the input for the next.

Preview of the Next Lesson:
In our previous lesson, we mentioned that polarization also occurs naturally. In the next lesson, we will focus on polarization by reflection. We will derive and apply Brewster's Law, which gives the specific angle of incidence at which reflected light from a surface becomes perfectly polarized. This will add another important quantitative tool to your wave optics arsenal.