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Unpolarized light with intensity \(I_0\) is incident on two polarizing filters. The axis of the first filter makes an angle of 60.0\(^\circ\) with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?

Short Answer

Expert verified
The final intensity is \(\frac{3I_0}{8}\).

Step by step solution

01

Understand Unpolarized Light and Malus's Law

Unpolarized light can be thought of as having its electric field vectors oriented in all possible directions perpendicular to the direction of propagation. When it encounters a polarizer, only the component of the electric field in the direction of the polarizer's axis is transmitted. Malus's Law states that the intensity of polarized light through a second polarizer is given by \( I = I_0 \cos^2(\theta) \), where \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.
02

First Polarizing Filter

The first polarizer has its axis at an angle of 60° with the vertical. Since the light is originally unpolarized, the initial intensity after passing through this filter is halved. Thus, the intensity after the first filter is \( I_1 = \frac{I_0}{2} \).
03

Second Polarizing Filter

Now, the intensity after the first filter is polarized at 60° to the vertical. The second filter’s axis is horizontal (90° to the vertical). Use Malus's Law to find the intensity after the second filter: \( I_2 = I_1 \cos^2(90° - 60°) = \frac{I_0}{2} \cos^2(30°) \).
04

Calculate Final Intensity

Substitute \( \cos(30°) = \frac{\sqrt{3}}{2} \) into the equation: \[ I_2 = \frac{I_0}{2} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}. \] Thus, the final intensity of light after passing through both filters is \( \frac{3I_0}{8} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Malus's Law
One of the fundamental principles in understanding the behavior of polarized light is Malus's Law. This law describes how the intensity of polarized light changes as it passes through a polarizing filter. Malus's Law states that after polarized light passes through a polarizer, its intensity (\( I \)) is related to the original intensity (\( I_0 \)) and the cosine of the angle (\( \theta \)) between the light's polarization direction and the polarizer's axis: \[ I = I_0 \cos^2(\theta) \]. This shows us that as the angle increases from 0 to 90 degrees, the intensity diminishes, reaching zero when the light is completely perpendicular to the axis.
  • Max intensity occurs when the light is aligned (0 degrees) with the polarizer's axis.
  • Min intensity, or zero, when it is perpendicular (90 degrees).
This law is crucial for numerous optical applications, including controlling light intensity in photography and displays.
Unpolarized Light
Unpolarized light is light that vibrates in multiple planes perpendicular to the direction of propagation. Most natural light sources, like the sun or a typical light bulb, emit unpolarized light. This means that the waves are oscillating in every possible plane around the direction they are moving.
When unpolarized light encounters a polarizing filter, it becomes polarized. This process filters out all those light waves not aligned with the filter’s axis. Hence, the intensity of the light reduces by half after passing through the first polarizer. This phenomenon is a direct consequence of selecting only the components of the light that are parallel to the filter's axis.
By understanding how unpolarized light is converted to polarized light, we can predict and calculate the behavior and reduction in light intensity as it passes through polarizers.
Intensities of Light
The concept of light intensity is central to understanding light behavior as it interacts with polarizing filters. Light intensity typically refers to the power per unit area carried by a wave. In terms of polarized light passing through polarizers, this is directly impacted by both the arrangement and orientation of these filters.
Let's break down the process with the example of light passing through two filters:
  • First, unpolarized light passes through the first polarizer, its intensity reduces by half as only one plane of the electromagnetic wave continues.
  • Next, when this now polarized light meets a second polarizer, we apply Malus's Law to determine the final intensity.
If we consider the original setup with the light encountering two filters at specific angles, using the given angles and \( \cos^2 \) calculations, we see how the final intensity after the second filtering would be \( \frac{3I_0}{8} \). Understanding these calculations allows us to manipulate light as needed in various technological applications, from camera filters to sunglasses.

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Most popular questions from this chapter

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. (a) If unpolarized light of intensity \(I_0\) is incident on the stack, find the intensity and state of polarization of light emerging from each filter. (b) If the second filter is removed, what is the intensity of the light emerging from each remaining filter?

A glass plate 2.50 mm thick, with an index of refraction of 1.40, is placed between a point source of light with wavelength 540 nm (in vacuum) and a screen. The distance from source to screen is 1.80 cm. How many wavelengths are there between the source and the screen?

A ray of light traveling \(in\) a block of glass (\(n\) = 1.52) is incident on the top surface at an angle of 57.2\(^\circ\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

A parallel beam of light in air makes an angle of 47.5\(^\circ\) with the surface of a glass plate having a refractive index of 1.66. (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface of the glass?

A thin layer of ice (\(n\) = 1.309) floats on the surface of water (\(n\) = 1.333) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

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