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\(\bullet\) Light of original intensity \(I_{0}\) passes through two ideal polarizing filters having their polarizing axes oriented as shown in Figure \(23.62 .\) You want to adjust the angle \(\phi\) so that the intensity at point \(P\) is equal to \(I_{0} / 10 .\) (a) If the original light is unpolarized, what should \(\phi\) be? (b) If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should \(\phi\) be?

Short Answer

Expert verified
(a) \( \phi \approx 63.4^\circ \); (b) \( \phi \approx 71.6^\circ \).

Step by step solution

01

Understand Malus's Law

Malus's Law states that the intensity of polarized light after passing through a polarizer is given by \( I = I_{0} \cos^2 \theta \), where \( I_{0} \) is the initial intensity and \( \theta \) is the angle between the light's initial polarization direction and the axis of the polarizer.
02

Unpolarized light through a polarizing filter

When unpolarized light passes through a polarizer, its intensity reduces to half, so after the first polarizer, the intensity \( I_{1} = \frac{I_{0}}{2} \).
03

Apply Malus's Law for the second polarizer with angle \( \phi \)

The light then passes through a second polarizer rotated at angle \( \phi \). We will use Malus's Law: \( I_{P} = I_{1} \cos^2 \phi \).Substitute \( I_{1} = \frac{I_{0}}{2} \) to get \( I_{P} = \frac{I_{0}}{2} \cos^2 \phi \).
04

Solve for \( \phi \) in part (a)

We need \( I_{P} = \frac{I_{0}}{10} \). Thus, \( \frac{I_{0}}{2} \cos^2 \phi = \frac{I_{0}}{10} \). Simplify and solve for \( \cos^2 \phi = \frac{1}{5} \), which means \( \phi = \cos^{-1}(\sqrt{0.2}) \approx 63.4^\circ \).
05

Consider linear polarization in the same direction as the first filter

If the light is already polarized in the direction of the first polarizer's axis, it passes through unchanged with intensity \( I_1 = I_{0} \).
06

Apply Malus's Law for linearly polarized light through the second polarizer

As in Step 3, we use \( I_{P} = I_{0} \cos^2 \phi \). We need \( I_{P} = \frac{I_{0}}{10} \), so \( \cos^2 \phi = \frac{1}{10} \), meaning \( \phi = \cos^{-1}(\sqrt{0.1}) \approx 71.6^\circ \).

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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 a fundamental concept in the study of light polarization. It describes how the intensity of light changes when it passes through a polarizing filter. According to this law, if polarized light with an initial intensity of \(I_0\) encounters a polarizer at an angle \(\theta\), the resulting intensity \(I\) is given by the equation \(I = I_0 \cos^2 \theta\).
This means that the light intensity depends on the cosine squared of the angle \(\theta\) between the light's initial direction of polarization and the axis of the polarizer. Malus's Law helps us predict how much light will be let through by a polarizer when the light is originally polarized. If the light is not aligned with the polarizer, less light will pass through, and the intensity decreases accordingly. This principle is critical in applications involving the control and manipulation of light intensity using polarizers.
Unpolarized Light
Unpolarized light consists of waves that have oscillations in multiple directions. This is the most common form of light we encounter in daily life, such as sunlight or light from a traditional bulb. Unlike polarized light, unpolarized light does not have a specific orientation of the electric field.
When unpolarized light is incident on a polarizing filter, the filter only allows the component of light that is aligned with its axis to pass through. This effectively reduces the light's intensity by half, because half of the oscillations are eliminated; only those in the direction parallel to the filter’s polarization axis are transmitted. Hence, if unpolarized light with intensity \(I_0\) hits a polarizer, the resulting intensity \(I_1\) becomes \(I_1 = \frac{I_0}{2}\). This concept is a stepping stone for more complex calculations involving additional polarizing filters.
Polarizing Filters
Polarizing filters are key tools in controlling the intensity and direction of light. They work by allowing only light waves of a specific polarization to pass through, thereby reducing unwanted reflections, glare, or any other undesired light effects.
There are many practical applications for polarizing filters, such as in photography to reduce glare from water surfaces or glass. When light passes through multiple polarizers, their arrangement and the angle between their axes determine the final intensity of the light. In scenarios involving two polarizers, you can use Malus's Law to determine how the light's intensity changes based on the angle \(\phi\) between their axes. For instance, if the first filter cuts the unpolarized light's intensity in half, the second polarizer will further modify the intensity depending on the cosine square of the angle \(\phi\). By adjusting \(\phi\), one can achieve the desired intensity at the output.
Intensity Reduction
Intensity reduction refers to the deliberate lowering of light intensity using polarizing filters. This concept is particularly useful when a specific light intensity is required, like in the problem where the intensity at point P needs to be \(\frac{I_0}{10}\).
After unpolarized light passes through the first filter, the intensity is halved. If a second polarizer is used and angled at \(\phi = 63.4^\circ\) or \(71.6^\circ\) depending on the initial polarization, the final intensity can be fine-tuned. In the case where the original light is unpolarized, setting \(\phi\) to approximately \(63.4^\circ\) achieves the desired intensity reduction. If the light is initially polarized in the same direction as the first filter, a \(\phi\) of approximately \(71.6^\circ\) is needed. Understanding these principles is crucial when working with optical systems that require precise control over light intensity.

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Most popular questions from this chapter

\(\bullet$$\bullet\) In a physics lab, light with wavelength 490 nm travels in air from a laser to a photocell in 17.0 ns. When a slab of glass 0.840 m thick is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light 21.2 ns to travel from the laser to the photocell. What is the wavelength of the light in the glass?

\(\bullet\) Ultraviolet radiation. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 \(\mathrm{nm}\) to 400 nm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and \(320 \mathrm{nm},\) is much more dangerous, because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

\(\bullet$$\bullet\) A thin layer of ice \((n=1.309)\) floats on the surface of water \((n=1.333)\) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

\(\bullet\) The polarizing angle for light in air incident on a glass plate is \(57.6^{\circ} .\) What is the index of refraction of the glass?

\(\bullet\) A light beam travels at \(1.94 \times 10^{8} \mathrm{m} / \mathrm{s}\) in quartz. The wavelength of the light in quartz is 355 \(\mathrm{nm}\) . (a) What is the index of refraction of quartz at this wavelength? (b) If this same light travels through air, what is its wavelength there?

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