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Unpolarized light of intensity \(I_{0}\) is incident on a series of five polarizers, each rotated \(10.0^{\circ}\) from the preceding one. What fraction of the incident light will pass through the series?

Short Answer

Expert verified
Answer: The fraction of incident light intensity that passes through the series of five polarizers is approximately 0.028, or 2.8%.

Step by step solution

01

Intensity after the first polarizer

The given light is unpolarized, and a polarizer only allows light with a certain polarization plane to pass through. When unpolarized light passes through a polarizer, its intensity is halved. Thus, the intensity of the light after passing through the first polarizer is \(I_1 = \frac{1}{2} I_0\).
02

Malus's law

Malus's law states that the intensity of light after passing through a polarizer is given by: \(I = I_0 \cos^2\theta\) where \(I_0\) is the initial intensity of the light, \(\theta\) is the angle between the polarization planes of the polarizer and the light, and \(I\) is the final intensity of the light.
03

Intensity after the second polarizer

Now that we know the intensity after the first polarizer, we can apply Malus's law to find the intensity after the second polarizer. The angle between the polarization planes of the first and second polarizer is 10°. So, using Malus's law, we get: \(I_2 = I_1 \cos^2 10^{\circ} = \frac{1}{2} I_0 \cos^2 10^{\circ}\).
04

Intensity after the other polarizers

We will follow the same process for the remaining three polarizers using the intensity after the previous polarizer and the angle between the polarization planes. For each polarizer, the angle is 10° from the previous one. After third polarizer: \(I_3 = I_2 \cos^2 10^{\circ} = \frac{1}{2} I_0 \cos^2 10^{\circ} \cos^2 10^{\circ} = \frac{1}{2} I_0 \cos^4 10^{\circ}\). After fourth polarizer: \(I_4 = I_3 \cos^2 10^{\circ} = \frac{1}{2} I_0 \cos^6 10^{\circ}\). After fifth polarizer: \(I_5 = I_4 \cos^2 10^{\circ} = \frac{1}{2} I_0 \cos^8 10^{\circ}\).
05

Finding the fraction of incident light intensity

Now, we have the intensity of light after passing through all five polarizers. To find the fraction of incident light that passes through the series, we divide the final intensity by the initial intensity, which is \(I_0\): \(Fraction = \frac{I_5}{I_0} = \frac{\frac{1}{2} I_0 \cos^8 10^{\circ}}{I_0} = \frac{1}{2} \cos^8 10^{\circ} \approx 0.028\). The fraction of the incident light that passes through the series of five polarizers is approximately 0.028, or 2.8%.

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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 fundamental when understanding how the intensity of light changes as it passes through a polarizing filter. Imagine light as a wave, oscillating in all directions. A polarizer is like a fence with vertical slats; it only allows waves that move parallel to the slats to pass. Mathematically, Malus's law is expressed with the elegant equation: \[I = I_0 \times \text{cos}^2(\theta)\]where \(I\) is the intensity of the polarized light, \(I_0\) the initial intensity, and \(\theta\) the angle between the light's initial polarization direction and the polarizing filter's axis. If the light is unpolarized, like sunlight on a clear day, a single filter cuts its intensity in half—a simple concept to visualize. Any subsequent polarization depends on both the reduced intensity and the angle between the polarizers.
In practice, if you were wearing polarized sunglasses and look at the sky at a 90-degree angle to the sun, the sky would appear darker. That's Malus's law in action, minimizing glare and easing the strain on your eyes. Fully grasp Malus's law, and you'll understand why photographers use polarizing filters to enhance contrast and colors in their pictures.
Polarizer Intensity
When considering the intensity of light after it passes through a polarizer, it's crucial to understand that polarization filters light based on its plane of vibration. In our exercise, light's path through a series of polarizers is a journey of diminishing brightness. Each polarizer allows only the light oscillating in a specific direction to pass through, reducing its intensity.
To calculate the intensity after each polarizer, we apply Malus's Law recursively. The intensity after the first polarizer is set—but what about subsequent ones? They depend on the angles. If you place polarizers at 90-degree angles to each other, no light passes through. Conversely, at 0 degrees, the intensity remains unchanged. An angle in between, like the \(10^{\text{o}}\) in our example, results in a gradual decrease in intensity, reflected beautifully in the compounded effect of \[\text{cos}^2(10^{\text{o}})\] terms.
Understanding how to measure the diminishing intensity can be applied outside the classroom too. For instance, if you're trying to reduce glare on a water surface to see the aquatic life beneath, adjusting the angle of your polarized sunglasses uses this same principle to optimize clarity.
Unpolarized Light
Unpolarized light is the maverick of waves: it vibrates in numerous planes perpendicular to the direction it travels. This is the typical state of light emitted by the sun, light bulbs, and flames. Before human intervention, there is no order to this light; it's a chaotic dance of waves. By introducing a polarizer, we essentially invite the light to a formal ball. The polarizer—analogous to a dance instructor—teaches the light to only oscillate in one direction (plane).
Our exercise demonstrates the initial halving of the intensity of unpolarized light when it first encounters a polarizer. It's not unlike bringing a sunny field into the dappled shade. Subsequent polarizers further refine the light's behavior, like repeated dance lessons enhancing the elegance of the ballroom dancers' moves. Each polarizer the light encounters is an opportunity to impose further polarization—and reduce its intensity.
Whether it's creating crisp laser beams, reducing glare on car windshields, or improving the visual quality on LCD screens, the role of unpolarized light's transformation through polarization is an ever-present tool in technology and daily life. Grasping the journey from unpolarized to polarized light illuminates, quite literally, countless phenomena in both the natural and constructed worlds.

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Most popular questions from this chapter

Electromagnetic waves from a small, isotropic source are not plane waves, which have constant maximum amplitudes. a) How does the maximum amplitude of the electric field of radiation from a small, isotropic source vary with distance from the source? b) Compare this with the electrostatic field of a point charge.

The voltage across a cylindrical conductor of radius \(r\), length \(L\), and resistance \(R\) varies with time. The timevarying voltage causes a time-varying current, \(i\), to flow in the cylinder. Show that the displacement current equals \(\epsilon_{0} \rho d i / d t,\) where \(\rho\) is the resistivity of the conductor.

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Quantum theory says that electromagnetic waves actually consist of discrete packets-photons-each with energy \(E=\hbar \omega,\) where \(\hbar=1.054573 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's reduced constant and \(\omega\) is the angular frequency of the wave. a) Find the momentum of a photon. b) Find the angular momentum of a photon. Photons are circularly polarized; that is, they are described by a superposition of two plane-polarized waves with equal field amplitudes, equal frequencies, and perpendicular polarizations, one-quarter of a cycle \(\left(90^{\circ}\right.\) or \(\pi / 2\) rad \()\) out of phase, so the electric and magnetic field vectors at any fixed point rotate in a circle with the angular frequency of the waves. It can be shown that a circularly polarized wave of energy \(U\) and angular frequency \(\omega\) has an angular momentum of magnitude \(L=U / \omega .\) (The direction of the angular momentum is given by the thumb of the right hand, when the fingers are curled in the direction in which the field vectors circulate. c) The ratio of the angular momentum of a particle to \(\hbar\) is its spin quantum number. Determine the spin quantum number of the photon.

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