/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 Unpolarized light with intensity... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

Short Answer

Expert verified
(a) Intensity is approximately 0.2854 \( I_{0} \). (b) The light is polarized at 41.0°.

Step by step solution

01

Apply Malus's Law for the First Polarizer

When unpolarized light passes through an ideal polarizing filter, it reduces its intensity by half. Therefore, the intensity of light after passing through the first polarizer is \( I_{1} = \frac{I_{0}}{2} \).
02

Calculate Intensity after the Second Polarizer using Malus’s Law

The intensity of light after it passes through a second polarizer, which is at an angle \( \theta \) to the first, is given by Malus's Law: \( I_{2} = I_{1} \cos^{2}(\theta) \). Here, \( I_{1} = \frac{I_{0}}{2} \) and \( \theta = 41.0^{\circ} \). Therefore, \( I_{2} = \frac{I_{0}}{2} \cos^{2}(41.0^{\circ}) \).
03

Calculate \( \cos^{2}(41.0^{\circ}) \)

Calculate \( \cos(41.0^{\circ}) \) and then square the result: \( \cos(41.0^{\circ}) \approx 0.7547 \), so \( \cos^{2}(41.0^{\circ}) \approx 0.5707 \).
04

Determine Final Intensity \( I_{2} \)

Substitute \( \cos^{2}(41.0^{\circ}) \approx 0.5707 \) into the formula for \( I_{2} \): \( I_{2} = \frac{I_{0}}{2} \times 0.5707 = 0.2854 \times I_{0} \). Hence, the intensity of light emerging from the second polarizer is approximately \( 0.2854 \times I_{0} \).
05

Describe the State of Polarization

After passing through the first polarizer, the light becomes polarized along the axis of the first polarizer. The light then remains polarized but at an angle of \(41.0^{\circ}\) with respect to the original polarization direction after passing through the second polarizer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 optics that describes how the intensity of light changes after passing through a polarizing filter. This law states that when polarized light encounters another polarizer at an angle, the intensity changes according to the cosine square of that angle. Specifically, if the initial intensity of polarized light is \( I_1 \), and it passes through a second polarizer set at an angle \( \theta \), the new intensity \( I_2 \) is given by \[I_2 = I_1 \cos^2(\theta).\]This means that as the angle between the light's polarization direction and the polarizer's axis increases, the transmitted light intensity decreases.
Conversely, when the angle is 90 degrees, no light passes through, because the cosine of 90 degrees is zero, making \( \cos^2(90^{\circ}) = 0 \).
Unpolarized Light
Unpolarized light is light that vibrates in multiple directions as it travels. Natural light sources like sunlight and many artificial sources produce unpolarized light. Unlike polarized light, which oscillates in just one direction, unpolarized light spreads its electromagnetic waves evenly across different planes.
When unpolarized light is passed through a polarizing filter, it becomes polarized because the filter blocks all but one direction of light vibration. As a result, the intensity of unpolarized light reduces to half when it first passes through an ideal polarizer. So, if the initial intensity of unpolarized light is \( I_0 \), after passing through the first polarizer, it becomes \( \frac{I_0}{2} \). This is the first step in solving problems related to polarized light transmission.
Polarizing Filter
A polarizing filter is a device that allows only light waves vibrating in a particular direction to pass through it, converting unpolarized light into polarized light. Polarizers are commonly used in photography to reduce glare or in sunglasses to eliminate reflections from surfaces like water or glass.
When light encounters the surface of a polarizing filter, only a specific orientation of wave vibrations is allowed to continue. This results in polarized light exiting the filter. In practice, polarizing filters can be used in pairs to control and manipulate light intensity further. When stacked at different angles, such filters can significantly reduce or change the intensity of the outgoing light, as dictated by Malus's Law. A crucial point is that when an unpolarized light goes through the first polarizer, half of it is absorbed, and the remaining half is aligned along the filter's axis.
Intensity of Light
The intensity of light is a measure of the energy carried by light waves per unit area. In the context of polarized light, this intensity changes as light passes through polarizers. Initially, unpolarized light has an intensity level noted as \( I_0 \).
As it passes through the first polarizing filter, its intensity reduces to \( \frac{I_0}{2} \). When this now polarized light encounters another polarizer with its axis at an angle \( \theta \), its intensity changes again, following Malus's Law to become \( I_2 = \frac{I_0}{2} \cos^2(\theta) \).
This process shows how light intensity can be controlled and adjusted using polarizers, allowing practical applications like reducing glare or improving visibility through reflective surfaces.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Radiation falling on a perfectly reflecting surface produces an average pressure \(p .\) If radiation of the same intensity falls on a perfectly absorbing surface and is spread over twice the area, what is the pressure at that surface in terms of \(p ?\)

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahighenergy (in the range of \(10^{12} \mathrm{~W}\) ) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk \(5.0 \mu \mathrm{m}\) in diameter, with the pulse lasting for \(4.0 \mathrm{~ns}\) with an average power of \(2.0 \times 10^{12} \mathrm{~W}\). We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

A plane sinusoidal electromagnetic wave in air has a wavelength of \(3.84 \mathrm{~cm}\) and an \(\vec{E}\) -field amplitude of \(1.35 \mathrm{~V} / \mathrm{m}\). (a) What is the frequency of the wave? (b) What is the \(\vec{B}\) -field amplitude? (c) What is the intensity? (d) What average force does this radiation exert perpendicular to its direction of propagation on a totally absorbing surface with area \(0.240 \mathrm{~m}^{2} ?\)

Some insect eyes have two types of cells that are sensitive to the plane of polarization of light. In a simple model, one cell type (type \(\mathrm{H}\) ) is sensitive to horizontally polarized light only, and the other cell type (type \(\mathrm{V}\) ) is sensitive to vertically polarized light only. To study the responses of these cells, researchers fix the insect in a normal, upright position so that one eye is illuminated by a light source. Then several experiments are carried out. To vary the angle as well as the intensity of polarized light, ordinary unpolarized light is passed through one polarizer with its transmission axis vertical, and then a second polarizer is placed between the first polarizer and the insect. When the light leaving the second polarizer has half the intensity of the original unpolarized light, which statement is true about the two types of cells? A. Only type H detects this light. B. Only type \(V\) detects this light. C. Both types detect this light, but type \(\mathrm{H}\) detects more light. D. Both types detect this light, but type \(\mathrm{V}\) detects more light.

A radio broadcasts at a wavelength of \(3.06 \mathrm{~m}\). Near the broadcast tower the electric-field amplitude is \(2700 \mathrm{~N} / \mathrm{C}\). Calculate (a) the frequency, (b) the wave number, and (c) the magnetic-field amplitude of the electromagnetic wave.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.