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Ch. 32 - Light: Reflection and Refraction
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 31, Problem 51

A light beam strikes a piece of glass at a 55.00° incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4831 and 1.4754, respectively. What is the angle between the two refracted beams?

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1
Identify the relevant physics principle: This problem involves Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. Snell's Law is expressed as: n1sinθ1=n2sinθ2, where n1 and n2 are the indices of refraction of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.
Apply Snell's Law for the first wavelength (450.0 nm): Substitute the given values into Snell's Law. Here, n1=1 (air), θ1=55.00°, and n2=1.4831. Solve for the refracted angle θ2 using: sinθ2=n1sinθ1n2.
Apply Snell's Law for the second wavelength (700.0 nm): Repeat the same process as in Step 2, but use the index of refraction for 700.0 nm, which is n2=1.4754. Solve for the refracted angle θ2 for this wavelength.
Determine the angle between the two refracted beams: The angle between the two refracted beams is the difference between the two refracted angles calculated in Steps 2 and 3. Subtract the smaller angle from the larger angle to find the result.
Summarize the process: The key steps involve applying Snell's Law for each wavelength, calculating the refracted angles, and then finding the difference between these angles to determine the angle between the two refracted beams.

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Key Concepts

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

Snell's Law

Snell's Law describes how light refracts when it passes from one medium to another. It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant and is equal to the ratio of the indices of refraction of the two media. This principle is essential for calculating the angles of refraction for different wavelengths of light as they enter the glass.
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Index of Refraction

The index of refraction is a dimensionless number that describes how much light slows down and bends when it enters a material. It varies with the wavelength of light, which is why different wavelengths can have different indices of refraction in the same medium. Understanding this concept is crucial for determining how much each wavelength of light will refract when passing through the glass.
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Wavelength Dependence of Refraction

Wavelength dependence of refraction refers to the phenomenon where different wavelengths of light are refracted by different amounts when passing through a medium. This occurs because the index of refraction varies with wavelength, leading to phenomena such as dispersion. In this problem, the differing indices of refraction for the two wavelengths will result in a measurable angle between the two refracted beams.
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Related Practice
Textbook Question

(III) A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32–24. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = tθ(n - 1)/n , where t is the thickness of the glass, θ is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32–24).

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Textbook Question

A parallel beam of light containing two wavelengths, λ₁ = 461 nm and λ₂ = 656 nm, enters the silicate flint glass of an equilateral prism as shown in Fig. 32–56. At what angle does each beam leave the prism (give angle with normal to the face)? See Fig. 32–28.

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Textbook Question

(II) A ray of light, after entering a light fiber, reflects at an angle of 14.5° with the long axis of the fiber, as in Fig. 32–57. Calculate the distance along the axis of the fiber that the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.55 and is 1.60 x 10-4 m in diameter.

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Textbook Question

The critical angle for a certain liquid–air surface is 52.6°. What is the index of refraction of the liquid?

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Textbook Question

(II) In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.8 m from his foot at the edge of the pool (Fig. 32–53). Where does the spot of light hit the bottom of the pool which is 2.1 m deep? Measure from the bottom of the wall beneath his foot.


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Textbook Question

A light beam strikes a 2.5-cm-thick piece of plastic with a refractive index of 1.62 at a 45° angle. The plastic is on top of a 3.8-cm-thick piece of glass for which n = 1.47. What is the distance D in Fig. 32–51?

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