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A beam of polarized light with an average intensity of \(15 \mathrm{W} / \mathrm{m}^{2}\) is sent through a polarizer. The transmission axis makes an angle of \(25^{\circ}\) with respect to the direction of polarization. Determine the rms value of the electric field of the transmitted beam.

Short Answer

Expert verified
The rms electric field of the transmitted beam is approximately 30.8 V/m.

Step by step solution

01

Understand Malus's Law

Malus's Law states that the intensity of polarized light passing through a polarizer is given by \( I = I_0 \cos^2 \theta \), where \( I_0 \) is the initial intensity and \( \theta \) is the angle between the light's initial polarization direction and the axis of the polarizer.
02

Calculate Transmitted Intensity

Given that the initial intensity \( I_0 = 15 \, \mathrm{W/m^2} \) and the angle \( \theta = 25^\circ \), use Malus's Law to find the transmitted intensity: \[I = 15 \, \mathrm{W/m^2} \times \cos^2(25^\circ) \]Calculate \( \cos(25^\circ) \) and then square it to find \( I \).
03

Calculate Cosine Squared

First calculate \( \cos(25^\circ) \approx 0.9063 \). Then, \[\cos^2(25^\circ) \approx (0.9063)^2 = 0.8214\]
04

Calculate Transmitted Beam Intensity

Substitute the value of \( \cos^2(25^\circ) \) into the intensity formula:\[ I = 15 \, \mathrm{W/m^2} \times 0.8214 \approx 12.32 \, \mathrm{W/m^2} \].
05

Relate Intensity to Electric Field

The intensity of light can be related to the rms value of the electric field \( E_{rms} \) using the formula \[ I = \frac{1}{2} c \varepsilon_0 E_{rms}^2 \]where \( c \) is the speed of light and \( \varepsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \)).
06

Solve for Electric Field

Rearrange the formula to solve for \( E_{rms} \):\[ E_{rms} = \sqrt{\frac{2I}{c\varepsilon_0}} \]Substitute \( I = 12.32 \, \mathrm{W/m^2} \), \( c = 3 \times 10^8 \, \mathrm{m/s} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \) into the equation. Compute \( E_{rms} \).
07

Compute the Electric Field

Calculate\[ E_{rms} = \sqrt{\frac{2 \times 12.32}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \]After performing the calculation, you will get: \[ E_{rms} \approx 30.8 \, \mathrm{V/m} \]

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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 concept in optics, especially when studying polarized light. It describes how light intensity changes as it passes through a polarizer. According to Malus's Law, the intensity of light (\( I \)) after passing through a polarizer can be determined using the initial intensity (\( I_0 \)) of the polarized light and the angle (\( \theta \)) between the light's initial polarization direction and the transmission axis of the polarizer. The law is mathematically expressed as:\[ I = I_0 \cos^2(\theta) \]This formula indicates that the transmitted intensity is a product of the initial intensity and the square of the cosine of the angle. When light is unpolarized, a polarizer will typically reduce the intensity by half. However, for already polarized light, the angle matters significantly:- At \( 0^\circ \), the cosine is 1, so all light passes through.- At \( 90^\circ \), the cosine is 0, meaning no light passes through.The cosine squared term can be understood by imagining that only the component of the electric field that aligns with the polarizer's axis passes through, while the perpendicular component is blocked. Understanding Malus's Law helps predict how much light will emerge from different angles, crucial in applications like LCD screens and sunglasses.
Electric Field
The electric field (\( E \)) of light represents the force that would be exerted on a charged particle in the light's path. For a beam of light, the electric field changes direction and magnitude in a wave-like pattern. The rms (root mean square) value of the electric field (\( E_{rms} \)) is a useful measure, as it represents the effective strength of this fluctuating field.To relate the electric field to light intensity, we use the following relation:\[ I = \frac{1}{2} c \varepsilon_0 E_{rms}^2 \]Where:- \( c \), the speed of light, is approximately \( 3 \times 10^8 \; \mathrm{m/s} \).- \( \varepsilon_0 \), the permittivity of free space, is approximately \( 8.85 \times 10^{-12} \; \mathrm{C^2/N \, m^2} \).Rearranging this formula allows us to calculate the rms electric field if the intensity is known:\[ E_{rms} = \sqrt{\frac{2I}{c\varepsilon_0}} \]This equation shows the direct link between intensity and the electric field's rms value, allowing us to understand the strength of the light's electric component in physical terms.
Light Intensity
Light intensity is a measure of how much energy a light wave carries per unit area per unit time. It is typically expressed in watts per square meter (\( \mathrm{W/m^2} \)). When dealing with polarized light, the intensity indicates how bright or powerful the light is after interacting with optical devices like polarizers.The initial intensity (\( I_0 \)) is the intensity of the light before it encounters a polarizer. Once through the polarizer, the angle between the light's polarization direction and the polarizer's axis (\( \theta \)) influences the transmitted intensity (\( I \)) through Malus's Law:- A smaller angle means most of the light's intensity is preserved.- A larger angle could reduce the intensity significantly.Additionally, light intensity is directly related to the electric field's square magnitude. As intensity increases, so does the electric field strength, as evident from:\[ I = \frac{1}{2} c \varepsilon_0 E_{rms}^2 \]This relation highlights the importance of intensity in determining the light's potential impact on objects, such as how bright it will appear or how much energy it might impart to a surface.

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Most popular questions from this chapter

A distant galaxy is simultaneously rotating and receding from the earth. As the drawing shows, the galactic center is receding from the earth at a relative speed of \(u_{\mathrm{G}}=1.6 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Relative to the center, the tangential speed is \(v_{\mathrm{T}}=0.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) for locations \(A\) and \(B,\) which are equidistant from the center. When the frequencies of the light coming from regions \(A\) and \(B\) are measured on earth, they are not the same and each is different from the emitted frequency of \(6.200 \times 10^{14} \mathrm{Hz}\). Find the measured frequency for the light from (a) region \(A\) and (b) region \(B\).

Two astronauts are \(1.5 \mathrm{m}\) apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at \(343 \mathrm{m} / \mathrm{s}\) through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth. How far away from the earth is the spaceship?

Light that is polarized along the vertical direction is incident on a sheet of polarizing material. Only \(94 \%\) of the intensity of the light passes through the sheet and strikes a second sheet of polarizing material. No light passes through the second sheet. What angle does the transmission axis of the second sheet make with the vertical?

A stationary particle of charge \(q=2.6 \times 10^{-8} \mathrm{C}\) is placed in a laser beam (an electromagnetic wave) whose intensity is \(2.5 \times 10^{3} \mathrm{W} / \mathrm{m}^{2}\). Determine the magnitudes of the (a) electric and (b) magnetic forces exerted on the charge. If the charge is moving at a speed of \(3.7 \times 10^{4} \mathrm{m} / \mathrm{s}\) perpendicular to the magnetic field of the electromagnetic wave, find the magnitudes of the (c) electric and (d) magnetic forces exerted on the particle.

Suppose that a police car is moving to the right at \(27 \mathrm{m} / \mathrm{s},\) while a speeder is coming up from behind at a speed of \(39 \mathrm{m} / \mathrm{s}\), both speeds being with respect to the ground. Assume that the electromagnetic wave emitted by the police car's radar gun has a frequency of \(8.0 \times 10^{9} \mathrm{Hz}\). Find the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car and the original frequency emitted by the police car.

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