Ch. 34 - The Wave Nature of Light: Interference and Polarization

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

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Giancoli Douglas 5th edition

Ch. 34 - The Wave Nature of Light: Interference and Polarization

Problem 49

Chapter 33, Problem 49

What is Brewster’s angle for an air-glass (n = 1.56) surface? Specify two answers.

Verified step by step guidance

Understand the concept: Brewster's angle is the angle of incidence at which light with a particular polarization is perfectly transmitted through a surface, with no reflection. It is given by the formula:

θ B = arctan (n 2 / n 1)

, where

n 1

is the refractive index of the first medium (air, in this case, with

n 1 = 1

) and

n 2

is the refractive index of the second medium (glass, with

n 2 = 1.56

Substitute the given values into the formula:

θ B = arctan (1.56 / 1)

. This simplifies to

θ B = arctan (1.56)

Calculate the arctangent of 1.56 to find the Brewster's angle in degrees. Use a scientific calculator or software to perform this step.

Recognize that Brewster's angle is typically measured relative to the normal of the surface. The second answer refers to the complementary angle, which is

90 ° - θ B

Summarize: The two answers are the Brewster's angle itself and its complementary angle. Ensure both are expressed in degrees for clarity.

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Key Concepts

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

Brewster's Angle

Brewster's angle is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. It is given by the formula θ_B = arctan(n), where n is the refractive index of the second medium relative to the first. This angle is significant in optics, particularly in applications involving polarized light.

Refractive Index

The refractive index (n) is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. For example, in the case of air-glass interface, the refractive index of glass is typically greater than that of air, affecting how light bends when entering or exiting the material.

Polarization of Light

Polarization refers to the orientation of the oscillations of light waves in a particular direction. Light can be polarized in various ways, such as through reflection, refraction, or by passing through polarizing filters. Understanding polarization is crucial for applications involving Brewster's angle, as it determines the conditions under which light is transmitted or reflected at the interface.

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Introduction to Polarization

Related Practice

Textbook Question

Light of wavelength 5.0 x 10⁻⁷ passes through two parallel slits and falls on a screen 5.0 m away. Adjacent bright bands of the interference pattern are 2.0 cm apart.

(a) Find the distance between the slits.

(b) The same two slits are next illuminated by light of a different wavelength, and the fifth minimum for this light occurs at the same point on the screen as the fourth minimum for the previous light. What is the wavelength of the second source of light?

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Textbook Question

A single optical coating reduces reflection to zero for λ = 550 nm. By what factor is the intensity reduced by the coating for λ = 430 nm and λ = 670 nm as compared to no coating? Assume normal incidence.

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Textbook Question

Show that the radius r of the mᵗʰ dark Newton’s ring, as viewed from directly above (Fig. 34–18), is given by r = √mλR where R is the radius of curvature of the curved glass surface and λ is the wavelength of light used. Assume that the thickness of the air gap is much less than R at all points and that r ≪ R . [Hint: Use the binomial expansion.]

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Textbook Question

What would Brewster’s angle be for reflections off the surface of water for light coming from beneath the surface? Compare to the angle for total internal reflection, and to Brewster’s angle from above the surface.

Textbook Question

Light of wavelength 690 nm passes through two narrow slits 0.66 mm apart. The screen is 1.75 m away. A second source of unknown wavelength produces its second-order fringe 1.23 mm closer to the central maximum than the 690-nm light. What is the wavelength of the unknown light?

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Textbook Question

(II) When a Newton’s ring apparatus (Fig. 34–18) is immersed in a liquid, the diameter of the tenth dark ring decreases from 2.92 cm to 2.54 cm. What is the refractive index of the liquid? [Hint: See Problem 37.]

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