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Chapter 23: Problem 58

\(\bullet\) Two ideal polarizing filters are oriented so that they transmit the maximum amount of light when unpolarized light is shone on them. To what fraction of its maximum value \(I_{0}\) is the intensity of the transmitted light reduced when the second filter is rotated through (a) \(22.5^{\circ},\) (b) \(45.0^{\circ},\) and (c) \(67.5^{\circ} ?\)

Short Answer

Expert verified
(a) 0.4268, (b) 0.25, (c) 0.0732

Step by step solution

01

Understanding Polarization

Polarizing filters reduce the intensity of light based on Malus's Law. When unpolarized light passes through the first filter, the intensity is halved. The second filter further reduces the intensity by a factor of \(\cos^2(\theta)\), where \(\theta\) is the angle between the light's polarization direction and the filter's axis.
02

Calculate for 22.5 Degrees

When the second filter is rotated through \(22.5^{\circ}\), the intensity \(I\) is given by \(I = I_0 \times \frac{1}{2} \times \cos^2(22.5^{\circ})\). Using \(\cos(22.5^{\circ}) \approx 0.9239\), we find \(I = I_0 \times 0.5 \times 0.9239^2 \approx 0.4268 I_0\).
03

Calculate for 45 Degrees

For a \(45^{\circ}\) rotation, the intensity \(I\) is \(I = I_0 \times \frac{1}{2} \times \cos^2(45^{\circ})\). Since \(\cos(45^{\circ}) = \frac{\sqrt{2}}{2}\), we find \(I = I_0 \times 0.5 \times (0.7071)^2 = 0.25 I_0\).
04

Calculate for 67.5 Degrees

When the angle is \(67.5^{\circ}\), the intensity \(I\) is calculated by \(I = I_0 \times \frac{1}{2} \times \cos^2(67.5^{\circ})\). With \(\cos(67.5^{\circ}) \approx 0.3827\), the intensity becomes \(I = I_0 \times 0.5 \times 0.3827^2 \approx 0.0732 I_0\).

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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 principle that describes how the intensity of polarized light changes when it passes through a polarizing filter. The key aspects of this law can be outlined as follows:
  • The law states that the intensity of light (\( I \)) after passing through a polarizer is proportional to the square of the cosine of the angle (\( \theta \)) between the light's initial polarization direction and the axis of the polarizer.
  • The mathematical expression of Malus's Law is given as \( I = I_0 \cos^2(\theta) \), where \( I_0 \) is the initial light intensity.
  • This law implies that as the angle \( \theta \) increases from \(0^{\circ}\) to \(90^{\circ}\), the intensity of the transmitted light decreases from maximum to zero.
Understanding Malus's Law is crucial when working with polarized light, especially in applications involving polarizing filters, where managing light intensity is essential.
Polarizing Filters
Polarizing filters are devices used to control and manipulate the polarization state of light. Here's how they work and what makes them useful:
  • These filters are designed to let only light waves vibrating in one specific direction pass through, effectively polarizing any light that was not polarized initially.
  • When unpolarized light, which contains waves vibrating in multiple directions, passes through a polarizing filter, only half of the light's intensity is transmitted.
  • A practical application of polarizing filters includes reducing glare in photography and enhancing contrast in LCD displays.
By using multiple polarizing filters in sequence, one can further reduce light intensity, which is useful in various optical devices to control brightness and glare.
Wave Intensity Reduction
Wave intensity reduction refers to the process of diminishing the power of a wave, specifically light, as it passes through a medium, such as polarizing filters. Here's how this occurs in light that encounters polarizing filters:
  • When unpolarized light first passes through a polarizing filter, its intensity is reduced by \(\frac{1}{2}\), due to the filter only allowing light waves aligned in one direction.
  • If a second polarizing filter is used, the intensity is further reduced according to Malus's Law, which is determined by the angle between the two filters.
  • Specifically, the intensity formula becomes \( I = I_0 \times \frac{1}{2} \cos^2(\theta) \), where \( \theta \) is the angle between the perpendicular axes of the two filters.
This principle is crucial in understanding how light can be controlled and manipulated in scientific and practical applications, helping in the design of devices that require specific light intensities.

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Most popular questions from this chapter

\(\bullet\) A beam of light strikes a sheet of glass at an angle of \(57.0^{\circ}\) with the normal in air. You observe that red light makes an angle of \(38.1^{\circ}\) with the normal in the glass, while violet light makes a \(36.7^{\circ}\) angle. (a) What are the indexes of refraction of this glass for these colors of light? (b) What are the speeds of red and violet light in the glass?

\(\bullet\) \(\cdot\) High-energy cancer treatment. Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of \(10^{12}\) W) pulses of light that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 5.0\(\mu \mathrm{m}\) in diameter, with the pulse lasting for 4.0 \(\mathrm{ns}\) with an average power of \(2.0 \times 10^{12} \mathrm{W}\) . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in \(\mathrm{W} / \mathrm{m}^{2} )\) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.

\(\bullet$$\bullet\) A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 \(\mathrm{cm}\) and an electric field amplitude of \(5.40 \times 10^{-2} \mathrm{V} / \mathrm{m}\) at a distance of 250 \(\mathrm{m}\) from the antenna. Calculate: (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

\(\bullet$$\bullet\) A ray of light traveling in a block of glass \((n=1.52)\) is incident on the top surface at an angle of \(57.2^{\circ}\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?
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