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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
Chapter 32, Problem 24

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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Identify the given values: the angle of incidence θ₁ = 53.5°, the index of refraction for violet light n₁ = 2.46, and the index of refraction for red light n₂ = 2.41. The index of refraction for air is approximately n₀ = 1.00.
Use Snell's Law to find the angle of refraction for each color of light. Snell's Law is given by: n0sin(θ1) = n1sin(θv) for violet light and n0sin(θ1) = n2sin(θr) for red light.
Rearrange Snell's Law to solve for the angle of refraction for each color: θv = sin1(n0sin(θ1)n1) for violet light and θr = sin1(n0sin(θ1)n2) for red light.
Calculate the angle of refraction for both violet and red light using the rearranged Snell's Law equations. This involves substituting the known values into the equations and using a calculator to find the angles.
Determine the angular separation between the two colors of light by subtracting the angle of refraction for red light from the angle of refraction for violet light: Δθ = θv - θr.

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Key Concepts

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

Index of Refraction

The index of refraction, denoted as 'n', is a measure of how much light slows down as it passes 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. Different wavelengths of light have different indices of refraction, which causes dispersion.
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Index of Refraction

Snell's Law

Snell's Law describes how light bends when it passes from one medium to another. It states that n1 * sin(θ1) = n2 * sin(θ2), 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. This law is crucial for calculating the angle of refraction for different wavelengths.
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Dispersion

Dispersion occurs when different wavelengths of light refract at different angles due to varying indices of refraction. This phenomenon causes the separation of light into its constituent colors, as seen in prisms. In this context, dispersion explains the angular separation between violet and red light as they pass through diamond.
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Related Practice
Textbook Question

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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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

Light enters a solid pipe made of plastic having an index of refraction of 1.60. The light travels parallel to the upper part of the pipe (Fig. E33.15). You want to cut the face AB so that all the light will reflect back into the pipe after it first strikes that face. If the pipe is immersed in water of refractive index 1.33, what is the largest that u can be?

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Textbook Question

The critical angle for total internal reflection at a liquid–air interface is 42.5°. If a ray of light traveling in the liquid has an angle of incidence at the interface of 35.0°, what angle does the refracted ray in the air make with the normal?

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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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