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Chapter 8: Problem 4

Linearly polarized light with polarization direction at angle \(\theta\) from \(\hat{x}\) is incident on a polaroid with easy axis along \(\hat{x} .\) The first polaroid is followed by a second polaroid with its easy axis along the direction of polarization of the original incident light. Show that if the input intensity is \(I_{0}\), the output intensity is \(I_{0} \cos ^{4} \theta\).

Short Answer

Expert verified
The output intensity is \( I_0 \cos^4 \theta \).

Step by step solution

01

Determine the effect of the first polaroid

The first polaroid has its easy axis along \( \hat{x} \). When light polarized at an angle \( \theta \) to the \( \hat{x} \)-axis passes through this polaroid, the intensity of the transmitted light can be calculated using Malus's Law. According to Malus's Law, the intensity \( I_1 \) after passing through the first polaroid is given by \[ I_1 = I_0 \cos^2 \theta \].
02

Determine the effect of the second polaroid

The second polaroid is oriented along the direction of the original incident light's polarization, which is at an angle \( \theta \) to the \( \hat{x} \)-axis. This means that the light, now polarized along \( \hat{x} \), will pass through the second polaroid without any additional attenuation because its polarization direction coincides with the second polaroid's easy axis. Therefore, intensity after the second polaroid, \( I_2 \), is given by \[ I_2 = I_1 \cos^2 \theta \].
03

Express the final output intensity

Substitute \( I_1 = I_0 \cos^2 \theta \) from Step 1 into the equation derived in Step 2 for \( I_2 \). This gives: \[ I_2 = (I_0 \cos^2 \theta ) \cos^2 \theta = I_0 \cos^4 \theta \].
04

Conclusion: Output Intensity

The final intensity of the light after passing through both polaroids is \( I_0 \cos^4 \theta \), as derived from the calculations above.

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Key Concepts

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

Malus's Law
Let's talk about Malus's Law, which is an essential principle in understanding light polarization. Malus's Law describes how the intensity of light changes when it passes through a polarizing filter. Imagine light as a wave that can vibrate in many directions. A polarizer is like a gate that lets only waves vibrating in a specific direction pass through. This filtering effect is described mathematically by Malus's Law as:
  • The intensity of light after passing through a polarizer is proportional to the square of the cosine of the angle ( \( \theta \)) between the light's original direction of polarization and the axis of the polarizer.
  • The formula is: \( I = I_0 \cos^2 \theta \),
  • where \( I_0 \) is the initial intensity and \( I \) is the intensity after passing through the polarizer.
This principle highlights how the orientation of a polarizer affects the passage of light. If the light's polarization is perfectly aligned with the polarizer's axis, it passes through with maximum intensity. If it is perpendicular, no light gets through.
Polarization Angle
The Polarization Angle plays a crucial role in determining the behavior of polarized light passing through filters. Polarization is all about direction. When light is linearly polarized, its vibrations occur along a single plane. This plane, or direction, is called the polarization angle, often measured from a reference axis like \( \hat{x} \).

In our scenario, when light encounters the first polarizer, it's at an angle \( \theta \) relative to the \( \hat{x} \)-axis. This angle affects how much light can pass through due to Malus's Law, as discussed earlier. Now consider the second polarizer in the original problem. Its easy axis is aligned with the original light's polarization direction. This means the light doesn't further lose intensity when passing through this second filter because the polarization direction matches. Understanding and calculating with the polarization angle allows us to predict how light interacts with polarizing materials effectively.
Intensity of Light
The intensity of light is a measure of the power it carries per unit area. When discussing polarization, understanding intensity changes becomes vital. Initially, light travels with a certain intensity \( I_0 \), which represents the total radiant energy flowing through a unit area in a given time.

After passing through a polarizer, intensity decreases based on the polarizer's orientation regarding the light's polarization. As mentioned, Malus鈥檚 Law tells us that the intensity of polarized light after passing through a polarizing filter is \( I = I_0 \cos^2 \theta \).

In exercises like ours, where two polarizers are involved, each with its cosine squared factor, the final intensity after passing through both is \( I_0 \cos^4 \theta \). This outcome emphasizes how much a polarizer can alter light's brightness by using geometry and orientation - two critical concepts in polarization.

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Most popular questions from this chapter

Polarization "projection operator." If a piece of linear polaroid with easy axis along \(\hat{x}\) is placed in a beam of light containing a mixture of all sorts of polarization, the polaroid absorbs all light that does not have linear polarization along \hat{x} . ~ I t ~ has an "output" at the rear of the polarizer consisting of light linearly polarized along \hat?. We shall call this piece of polaroid a "projection operator." It "projects out" the x polarization without loss (neglecting small reflections) and delivers it at its output end. Note that this " \(x\) projection operator" can be run either forward or backward; i.e., either face of the polaroid may be used as the input end. Now consider a piece of circular polarizer consisting of a piece of linear polarizer (input end) glued to a quarter-wave plate with optic axis at 45 deg to the easy axis of the polaroid. This polarizer puts out (for example) right-handed light. But it absorbs half of any righthanded light incident. If it is run backward, it passes incident right-handed light and absorbs left-handed light. But when it thus passes right-handed light incident on the quarter-wave-plate face, it delivers it out the polaroid face as linearly polarized light. Therefore it is not what we are calling a polarization projection operator. Here is the problem: Invent circular polarization projection operators, one for left- handed and one for right-handed light. The right-handed projection operator should transmit incident right-handed light with no loss (neglecting small reflections) and should deliver it as right-handed light. It should absorb left-handed light. Question: Is your circular polarization projection operator reversible? Can you use either face for the input end?

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 \(2 \omega_{0}\) -

Slinky polarization. Find a slinky and a partner. You and your partner hold opposite ends of the slinky. (a) Let each shake the slinky in a clockwise circular rotation (from his own point of view). If this doesn't convince you that linear polarization is a superposition of opposite circular polarizations, nothing else will. (b) With each person using a book as a straightedge to guide his hand, let one partner shake linear polarization at \(45 \mathrm{deg}\) to the horizontal, and let the other shake linear polarization at 90 deg to the first. (The 45 deg angle is so as to prevent gravity from giving a big asymmetry.) One counts out loud, " \(1,2,3,4,1,2,3,4, \ldots "\) four beats to a cycle, or perhaps four to a half-cycle), with "one" coming at a reproducible phase of the motion of his hand. The other shakes in phase, or 180 deg out of phase, or 90 deg out of phase. It takes some concentration not to be distracted by what you see. ( \(c\) ) With the far end fixed to something (your partner can now go home), shake out a circularly polarized wave packet of one or two t?ns. Verify that it conserves angular momentum upon reflection. Verify that if the angular momentum is along the propagation direction, the shape is that of a left- handed screw, and that the handedness reverses upon reflection.

Optical activity. Suppose you send linearly polarized light through a length \(L\) of Karo corn syrup and find that, for \(L=5 \mathrm{~cm}\), red light is rotated by 45 deg. Now reflect the light that has passed through the syrup from a mirror and send it back through the syrup, so that the total length is \(10 \mathrm{~cm}\). (If you do the experiment, make the angle of reflection not quite \(180 \mathrm{deg}\); then look at the "image light bulb" through both the "real syrup" and the "image syrup." As a control experiment, you can look through the "image syrup" alone by moving your head.) Question: After the two traversals, is the linear polarization at 0 or 90 deg to the original direction?

Suppose you have linearly polarized incident light with polarization along \(\hat{x}\) You desire linearly polarized light with polarization at 30 deg to \(\hat{x}\), i.e., along $$ \hat{e}=\hat{x} \cos 30^{\circ}+9 \sin 30^{\circ} $$ How can you obtain this transmitted field \((a)\) at the cost of some loss of intensity; \((b)\) without loss of intensity and without using any polaroids?
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