/*! 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 3 Circularly polarized light of in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Circularly polarized light of intensity \(I_{0}\) (intensity means energy fux per unit area per unit time; this is proportional to a photomultiplier's output current, for light at a given frequency) is incident on a single polaroid. Show that the output intensity (intensity of the light emerging from the rear of the polaroid) is \(\frac{1}{2} I_{0}\)

Short Answer

Expert verified
The output intensity is \(\frac{1}{2} I_0\).

Step by step solution

01

Understand Circularly Polarized Light

Circularly polarized light consists of two linear components of equal amplitude that are perpendicular to each other and have a phase difference of 90 degrees. Its intensity is uniformly distributed.
02

Intensity of Light Through a Polaroid

When light passes through a polaroid, only the component of the light wave that is aligned with the axis of the polaroid is transmitted. The intensity of linearly polarized light passing through a polaroid is given by Malus's Law: \[ I = I_0 \cos^2\theta \]where \( I_0 \) is the initial intensity, and \( \theta \) is the angle between the light's polarization direction and the polaroid's axis. However, circularly polarized light has equal components in all directions, so \( \theta \) has no effect.
03

Apply the 50% Intensity Reduction Rule

For circularly polarized light, when it passes through a polaroid, only one component (of equal amplitude) that is aligned with the axis of the polaroid is transmitted. Since the light is circularly polarized, it effectively acts like two perpendicular linearly polarized components, each contributing half the intensity. Therefore, the transmitted intensity is half of the incident intensity: \[ I_{output} = \frac{1}{2} I_0 \]

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.

Intensity
In the context of physics and optics, intensity refers to the power of light striking a surface per unit area. It's a measure of how much energy the light carries at a specific point and how this energy is distributed over the area.
In simpler terms, think of intensity like the brightness of the light, an important factor when you're figuring out how much light energy is making it through a particular medium, like a window or a polaroid.

When studying circularly polarized light, the intensity is used to describe how this type of light behaves when it interacts with other surfaces or objects. Circularly polarized light, with its uniform intensity distribution, makes it easier to understand how light energy flows consistently in every direction around its path. That's why, when circularly polarized light hits a polaroid, the process is straightforward: its intensity reduces because only specific parts of this multi-directional light get through, but we'll detail that in the next sections.
Malus's Law
Malus's Law is a fundamental principle in optics, describing how light intensity changes as polarized light passes through a polarizing filter. In mathematical terms, Malus’s Law is expressed as \( I = I_0 \cos^2\theta \).
This equation tells us the new intensity \( I \) of light after it transits through a polaroid. \( I_0 \) represents the initial intensity of the light before hitting the filter. \( \theta \) is the angle between the light’s polarization direction and the polaroid’s axis.

However, here's an interesting twist: when dealing with circularly polarized light, which has equal components in all directions, the value of \( \theta \) becomes irrelevant. This is because the light is not favoring any specific direction. Imagine two identical linear components forming the circular polarization; they each provide equal intensity. The essence of Malus's Law simplifies to the idea that only half of the total intensity gets through an ideal polaroid because both components are equally viable but only one component successfully aligns with the polaroid's axis.
Polaroid
A polaroid is a special type of optical filter that allows only light waves aligned with its transmission axis to pass through. It's like a gatekeeper that blocks certain light vibrations while letting others go. This unique characteristic makes polaroids essential tools for studying the properties of light polarization.

Polaroids are particularly interesting when examining circularly polarized light. This type of light is composed of two perpendicular linear components that are continuously rotating, producing a helical light wave. Due to its uniform intensity and equal component structure, circularly polarized light behaves predictably when it meets a polaroid.
Upon encountering a polaroid, this light only allows the component aligned with the polaroid’s axis to pass. Since each component of circularly polarized light contributes equally to the total intensity, the polaroid effectively cuts the intensity in half of the incoming circularly polarized light, confirming why the resulting output intensity is \( \frac{1}{2} I_0 \).

In practical terms, polaroids are used in various applications, from 3D movies, which require polarized glasses, to scientific experiments needing precision control over light’s behavior.

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

Suppose that a beam of linearly polarized light is incident on a half-wave plate which is rotating about the beam axis with angular velocity \(\omega_{0}\). Show that the output light is linearly polarized, with the polarization direction rotating at \(2 \omega_{0}\) -

Circularly polarized light of intensity \(I_{0}\) is incident on a sandwich of three polaroids. The first and third polaroids are crossed, i.e., their easy axes are at 90 deg to one another. The middle polaroid makes an angle \(\theta\) with the axis of the first polaroid. Show that the output intensity is \(\frac{1}{2} I_{0} \cos ^{2} \theta \sin ^{2} \theta\)

Glare elimination. Suppose you wish to shine your flashlight through a glass window to illuminate something on the far side of the window. How can you get rid of the bothersome glare from light specularly reflected from the glass? Suppose instead that you are trying to look at something through rain, at night, using a flashlight beam for illumination. Will the trick used for eliminating window glare get rid of the light reflected from the raindrops? Suppose that instead of visible light you are using \(10-\mathrm{cm}\) microwaves emitted and received by the same antenna system, i.e., radar. How can you arrange the phase relations in two antennas oriented along \(x\) and \(y\) so as to eliminate glare from raindrops?

Moonlight and earthlight. When the moon appears half-full, the illuminated portion is scattering sunlight through about 90 deg to your eye. We know that for 90 deg scattering the blue sky is almost completely linearly polarized. Do you predict that half-moon-light is polarized? Do the experiment. Now think about how the earth looks from the moon at "half-earth." Is the earthlight polarized? (You can look for twenty-four hours while the earth turns.)

Navigation by the Vikings. At high latitudes (say above the Arctic Circle) the magnetic compass is unreliable. The sun is also difficult to use for navigation; it may be below the horizon even at noon. Airline navigators then sometimes use a "twilight compass" that locates the sun's position below the horizon by means of the variation with direction of the polarization of the blue sky. The compass contains a piece of polaroid. Some natural crystals have properties similar to polaroid-one such substance is tourmaline; another is cordierite. When linearly polarized light is viewed through a cordierite crystal, the crystal is clear (with a yellowish tinge) when the polarization is along the axis of easy transmission, and the crystal is dark blue when the polarization is 90 deg to this axis. Such substances are called "dichroic." The Viking saitors of the ninth century navigated their ships without benefit of either magnetic compass or polaroid. At night they used the stars. In the day they used the sun, when it was not obscured by clouds. According to ancient Scandinavian sagas, the Viking navigators could always locate the sun, even when it was behind the clouds, by using magical "sun stones." It was long a mystery what these "sun stones" were. The mystery has probably been solved by a Danish archeologist, who knew

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.