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Chapter 33: Problem 33

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?

Short Answer

Expert verified
The intensity after the second polarizer is approximately 6.37 W/cm虏.

Step by step solution

01

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:\[ I_1 = \frac{I_0}{2} = \frac{20.0 \, \text{W/cm}^2}{2} = 10.0 \, \text{W/cm}^2 \]
02

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:\[ \theta = 62.0^\circ - 25.0^\circ = 37.0^\circ \]
03

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:\[ I_2 = 10.0 \, \text{W/cm}^2 \times 0.798^2 \approx 10.0 \, \text{W/cm}^2 \times 0.637 = 6.37 \, \text{W/cm}^2 \]

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Key Concepts

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

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.
  • 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.
Understanding Malus's Law is crucial in applications like photography, where polarizing filters manage reflections and glare.
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.
This transformation using polarized filters is key in reducing glare and enhancing visual clarity in various optical devices.
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.
These filters are used in photography to enhance image quality by reducing reflections from surfaces like water or glass. They are also used in sunglasses to help reduce glare, making them invaluable in everyday life.

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Most popular questions from this chapter

A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angle of incidence is 62.0\(^\circ\), and the ray originates in the glass with \(n\) = 1.70. Compute the angle of refraction.

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of \(\theta_a = 36.2{^\circ}\), the ray refracted into the water makes an angle of 49.8\(^\circ\) with the normal to the interface. What is the smallest value of the incident angle \(\theta_a\) for which none of the ray refracts into the water?

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 ray of light is traveling in a glass cube that is totally immersed in water. You find that if the ray is incident on the glass-water interface at an angle to the normal larger than 48.7\(^\circ\), no light is refracted into the water. What is the refractive index of the glass?

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 light is \(\theta\), the intensity of the emerging beam is \(I\). If you now want the intensity to be \(I/2\), what should be the angle (in terms of \(\theta\)) between the polarizing angle of the filter and the original direction of polarization of the light?
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