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Ch 33: The Nature and Propagation of Light
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 33: The Nature and Propagation of LightProblem 27
Chapter 32, Problem 27

Unpolarized light with intensity I0 is incident on two polarizing filters. The axis of the first filter makes an angle of 60.0° with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?

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Start by understanding that unpolarized light has an equal intensity in all directions. When it passes through the first polarizing filter, the intensity is reduced by half. Therefore, the intensity after the first filter is I1 = I0 / 2.
Next, apply Malus's Law to determine the intensity of light after it passes through the second polarizing filter. Malus's Law states that the intensity I after passing through a polarizer is given by I = I1 * cos^2(θ), where θ is the angle between the light's polarization direction and the axis of the polarizer.
In this problem, the angle θ between the light's polarization direction after the first filter (which is at 60.0° to the vertical) and the axis of the second filter (which is horizontal) is 90.0° - 60.0° = 30.0°.
Substitute the values into Malus's Law: I = (I0 / 2) * cos^2(30.0°).
Calculate cos(30.0°), which is √3/2, and then square it to find cos^2(30.0°). Substitute this value back into the equation to find the final intensity of the light after it passes through the second filter.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Unpolarized Light

Unpolarized light consists of waves vibrating in multiple planes perpendicular to the direction of propagation. When such light encounters a polarizing filter, it becomes polarized, meaning it vibrates in a single plane. The intensity of unpolarized light is reduced by half after passing through the first polarizer, as only the component aligned with the filter's axis is transmitted.
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Malus's Law

Malus's Law describes how the intensity of polarized light changes as it passes through a second polarizing filter. The transmitted intensity I is given by I = I0 * cos²(θ), where I0 is the initial intensity and θ is the angle between the light's polarization direction and the filter's axis. This law is crucial for calculating the intensity after the second filter in the given problem.
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Polarization Angle

The polarization angle is the angle between the polarization direction of light and the axis of a polarizing filter. In the problem, the first filter is at 60° to the vertical, and the second is horizontal, making the effective angle between the polarized light from the first filter and the second filter's axis 30°. This angle is used in Malus's Law to determine the final intensity.
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Related Practice
Textbook Question

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle of 38.1° with the normal in the glass, while violet light makes a 36.7° angle. What are the speeds of red and violet light in the glass?

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Textbook Question

The indexes of refraction for violet light (λ = 400 nm) and red light (λ= 700 nm) in diamond are 2.46 and 2.41, respectively. A ray of light traveling through air strikes the diamond surface at an angle of 53.5° to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

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Textbook Question

Light of original intensity I0 passes through two ideal polarizing filters having their polarizing axes oriented as shown in Fig. E33.28. You want to adjust the angle f so that the intensity at point P is equal to I0/10. If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should Φ be?

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Textbook Question

At the very end of Wagner's series of operas Ring of the Nibelung, Brünnhilde takes the golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the bottom of the river. Assuming that the ring is small enough compared to the depth of the river to be treated as a point and that the Rhine is 10.0 m deep where the ring goes in, what is the area of the largest circle at the surface of the water over which light from the ring could escape from the water?

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Textbook Question

Light of original intensity I0 passes through two ideal polarizing filters having their polarizing axes oriented as shown in Fig. E33.28. You want to adjust the angle f so that the intensity at point P is equal to I0/10. If the original light is unpolarized, what should Φ be?

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Textbook Question

A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the green part of the visible spectrum is I, find the intensity (in terms of I) of scattered light in the middle of the red part of the spectrum.

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