/*! 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 (1) Two polarizers are oriented ... [FREE SOLUTION] | 91Ó°ÊÓ

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(1) Two polarizers are oriented at \(65^{\circ}\) to one another. Unpolarized light falls on them. What fraction of the light intensity is transmitted?

Short Answer

Expert verified
The fraction of light intensity transmitted is approximately 0.089.

Step by step solution

01

Understand Malus's Law

When unpolarized light passes through a polarizer, its intensity is reduced by half. Then, according to Malus's Law, if the transmitted light from the first polarizer passes through a second polarizer at an angle \( \theta \), the intensity is further reduced by \( \cos^2(\theta) \).
02

Calculate the Intensity After First Polarizer

First, calculate the intensity of light after it passes through the first polarizer. Since the light is initially unpolarized, its intensity is reduced to half:\[ I_1 = \frac{I_0}{2} \] where \( I_0 \) is the initial intensity.
03

Use Malus's Law for Second Polarizer

Next, apply Malus's Law for the second polarizer. The intensity of light transmitted through the second polarizer is given by:\[ I_2 = I_1 \times \cos^2(\theta) \] Since \( \theta = 65^{\circ} \), it becomes:\[ I_2 = \frac{I_0}{2} \times \cos^2(65^{\circ}) \]
04

Calculate the Final Fraction of Intensity

Calculate \( \cos^2(65^{\circ}) \):\[ \cos(65^{\circ}) \approx 0.4226 \]\[ \cos^2(65^{\circ}) \approx 0.4226^2 \approx 0.178 \]Substituting back, the final intensity is:\[ I_2 = \frac{I_0}{2} \times 0.178 \approx 0.089 \times I_0 \] Thus, the fraction of the original light intensity transmitted is approximately \( 0.089 \).

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Key Concepts

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

Polarization
Polarization is an essential concept in optics that describes the orientation of waves, particularly electromagnetic waves like light. When light is polarized, its electric fields oscillate in a particular direction. In contrast, unpolarized light consists of waves oscillating in various directions. Polarization is significant in many applications from sunglasses that reduce glare to photographic filters.
There are different methods to achieve polarization:
  • Polarizing Filters: These are specialized filters that allow light waves oscillating in one direction to pass while blocking others.
  • Reflection: Light can be polarized through the process of reflection wherein the reflected light tends to be oriented parallel to the reflecting surface.
  • Scattering: Light can become polarized when it scatters, as seen in the sky's blue hues.
To fully grasp polarization, it's crucial to understand Malus's Law. This law quantifies the intensity of polarized light passing through a second polarizer, showing how the angle between them affects transmitted intensity.
Unpolarized Light
Unpolarized light contains waves that move in all possible directions perpendicular to the direction of the wave's travel. Common sources of unpolarized light include sunlight, incandescent bulbs, and most artificial lighting. When this light passes through a polarizer, it becomes polarized with half of its original intensity. This is because the polarizer only allows waves vibrating in one direction to pass through.
This initial reduction to half intensity is the key feature that distinguishes unpolarized light behavior when encountering a polarizer. The process then continues if this polarized light encounters a second polarizer, as described by Malus’s Law, leading to further reduction depending on the angle between the polarizers.
Understanding unpolarized light is vital in fields like photography, where filters are used to control glare and reflections in images, as well as in scientific experiments where controlling light behavior is necessary.
Intensity Reduction
Intensity reduction is a fundamental concept when studying the behavior of light as it interacts with polarizers. Initially, when unpolarized light encounters a polarizer, its intensity is reduced by half. This is because the polarizer only allows components of the light vibrating in a certain direction to pass through.
Further reduction in intensity is calculated using Malus's Law if a second polarizer is involved. The intensity of light that emerges from the second polarizer is determined by:\[ I_2 = I_1 \times \cos^2(\theta) \]where \(I_1\) is the intensity after the first polarizer and \(\theta\) is the angle between the axes of the two polarizers.
For instance, in the exercise, if the angle is \(65^{\circ}\), the light that passes through both polarizers will have its intensity reduced to approximately \(8.9\%\) of the original. This principle highlights the crucial role of polarizers in controlling light transmission and can be applied in various optical technologies.

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Most popular questions from this chapter

A laser beam passes through a slit of width \(1.0 \mathrm{~cm}\) and is pointed at the Moon, which is approximately \(380,000 \mathrm{~km}\) from the Earth. Assume the laser emits waves of wavelength \(633 \mathrm{nm}\) (the red light of a He-Ne laser). Estimate the width of the beam when it reaches the Moon.

Four polarizers are placed in succession with their axes vertical, at \(30.0^{\circ}\) to the vertical, at \(60.0^{\circ}\) to the vertical, and at \(90.0^{\circ}\) to the vertical. \((a)\) Calculate what fraction of the incident unpolarized light is transmitted by the four polarizers. (b) Can the transmitted light be decreased by removing one of the polarizers? If so, which one? (c) Can the transmitted light intensity be extinguished by removing polarizers? If so, which one(s)?

(II) Suppose that you wish to construct a telescope that can resolve features \(7.5 \mathrm{~km}\) across on the \(\mathrm{moon}, 384,000 \mathrm{~km}\) away. You have a 2.0 -m-focal-length objective lens whose diameter is \(11.0 \mathrm{~cm} .\) What focal-length eyepiece is needed if your eye can resolve objects \(0.10 \mathrm{~mm}\) apart at a distance of \(25 \mathrm{~cm}\) ? What is the resolution limit set by the size of the objective lens (that is, by diffraction)? Use \(\lambda=560 \mathrm{nm}\)

(I) Monochromatic light falls on a slit that is \(2.60 \times 10^{-3} \mathrm{~mm}\) wide. If the angle between the first dark fringes on either side of the central maximum is \(32.0^{\circ}\) (dark fringe to dark fringe), what is the wavelength of the light used?

X-rays of wavelength \(0.0973 \mathrm{nm}\) are directed at an unknown crystal. The second diffraction maximum is recorded when the X-rays are directed at an angle of \(23.4^{\circ}\) relative to the crystal surface. What is the spacing between crystal planes?

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