Ch 33: The Nature and Propagation of Light

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 33: The Nature and Propagation of Light

Problem 15a

Chapter 33, Problem 15a

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. What is the largest that u can be if the pipe is in air?

Verified step by step guidance

Understand the concept of total internal reflection: Total internal reflection occurs when light travels from a medium with a higher index of refraction to a medium with a lower index of refraction, and the angle of incidence is greater than the critical angle.

Identify the indices of refraction: The index of refraction for the plastic pipe is given as 1.60, and the index of refraction for air is approximately 1.00.

Calculate the critical angle using Snell's Law: Snell's Law states that \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). For total internal reflection, \( \theta_2 \) is 90 degrees, so \( \sin(\theta_2) = 1 \). Rearrange the formula to find the critical angle \( \theta_c \): \( \theta_c = \arcsin\left(\frac{n_2}{n_1}\right) \).

Substitute the values into the critical angle formula: \( \theta_c = \arcsin\left(\frac{1.00}{1.60}\right) \). This will give you the critical angle in degrees.

Determine the largest angle \( u \): Since the light must reflect back into the pipe, \( u \) must be less than or equal to the critical angle calculated in the previous step. This ensures that all light will undergo total internal reflection.

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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 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. In this problem, the plastic pipe has an index of refraction of 1.60, indicating that light travels slower in the pipe than in air.

Total Internal Reflection

Total internal reflection occurs when light traveling in a medium hits the boundary with a less dense medium at an angle greater than the critical angle, causing it to reflect entirely back into the original medium. This concept is crucial for ensuring that all light reflects back into the pipe, as the angle of incidence must be managed to exceed the critical angle.

Critical Angle

The critical angle is the minimum angle of incidence at which total internal reflection occurs. It is determined by the indices of refraction of the two media involved. For light traveling from plastic (n=1.60) to air (n=1.00), the critical angle can be calculated using Snell's Law, ensuring that light reflects back into the pipe rather than refracting out.

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Critical Angle

Related Practice

Textbook Question

As shown in Fig. E33.11, a layer of water covers a slab of material X in a beaker. A ray of light traveling upward follows the path indicated. Using the information on the figure, find the angle the light makes with the normal in the air.

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Textbook Question

A horizontal, parallel-sided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of 35.0° with the normal to the top surface of the glass. What angle does the ray refracted into the water make with the normal to the surface?

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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

(a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3° angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2° from the normal, what is the refractive index of the unknown liquid?

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