Problem 33
Question
Unpolarized light of intensity 20.0 \(\mathrm {W/cm}^2\) is incident on two polarizing filters. The axis of the first filter is at an angle of 25.0\(^\circ\) counterclockwise from the vertical (viewed in the direction the light is traveling), and the axis of the second filter is at 62.0\(^\circ\) counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer?
Step-by-Step Solution
Verified Answer
The intensity after the second polarizer is approximately 6.37 W/cm².
1Step 1: Calculate Intensity After First Polarizer
Use Malus's Law to determine the intensity of light after it passes through the first polarizer. Since the light is initially unpolarized, the intensity after the first filter is half of the initial intensity:\[ I_1 = \frac{I_0}{2} = \frac{20.0 \, \text{W/cm}^2}{2} = 10.0 \, \text{W/cm}^2 \]
2Step 2: Determine Angle Between Polarizers
Find the angle between the transmission axes of the two polarizers. Since the angles are given counterclockwise from the vertical, the angle \(\theta\) between them is:\[ \theta = 62.0^\circ - 25.0^\circ = 37.0^\circ \]
3Step 3: Calculate Intensity After Second Polarizer
Apply Malus's Law again to find the intensity after the second polarizer, using the angle between the polarizer axes:\[ I_2 = I_1 \cos^2(\theta) = 10.0 \, \text{W/cm}^2 \cos^2(37.0^\circ) \]First, calculate \(\cos(37.0^\circ)\), which is approximately 0.798, then:\[ I_2 = 10.0 \, \text{W/cm}^2 \times 0.798^2 \approx 10.0 \, \text{W/cm}^2 \times 0.637 = 6.37 \, \text{W/cm}^2 \]
Key Concepts
Malus's LawUnpolarized LightPolarizing Filters
Malus's Law
Malus's Law is a fundamental principle in the study of optics. It describes how light intensity changes as polarized light passes through a polarizing filter. The law states that the intensity of polarized light after passing through a polarizer is given by the equation:
\[ I = I_0 \cos^2(\theta) \]
Here, \(I\) is the transmitted light intensity, \(I_0\) is the initial light intensity, and \(\theta\) is the angle between the light's polarization direction and the axis of the polarizer.
\[ I = I_0 \cos^2(\theta) \]
Here, \(I\) is the transmitted light intensity, \(I_0\) is the initial light intensity, and \(\theta\) is the angle between the light's polarization direction and the axis of the polarizer.
- This equation highlights the cosine squared dependency, showing that intensity decreases significantly when the light's direction is not aligned with the polarizer's axis.
- For an angle \(\theta = 90^\circ\), the transmitted intensity becomes zero, effectively blocking the light.
Unpolarized Light
Unpolarized light consists of waves vibrating in multiple planes. Common sources of light, like the sun or a light bulb, emit this kind of light. It lacks a specific orientation, meaning it vibrates in all possible directions perpendicular to the direction of travel.
- When unpolarized light encounters a polarizing filter, it becomes polarized.
- Only half of the original intensity is transmitted through the first polarizer. This is because the filter aligns the light waves into a single plane.
Polarizing Filters
Polarizing filters are materials designed to convert unpolarized light into polarized light. They work by allowing light waves of a specific polarization to pass through while blocking others.
- The filter's effectiveness depends on the angle at which the light enters the filter relative to its polarization axis.
- In practical use, these filters adjust the intensity and directionality of light passing through them.
Other exercises in this chapter
Problem 31
A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the
View solution Problem 32
Three polarizing filters are stacked, with the polarizing axis of the second and third filters at 23.0\(^\circ\) and 62.0\(^\circ\), respectively, to that of th
View solution Problem 34
Three polarizing filters are stacked with the polarizing axes of the second and third at 45.0\(^\circ\) and 90.0\(^\circ\), respectively, with that of the first
View solution Problem 36
A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, the light takes 8.72 ns to travel the length of the
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