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Ch 33: The Nature and Propagation of Light
Chapter 33, Problem 33

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. (b) If the pipe is immersed in water of refractive index 1.33, what is the largest that u can be?

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Identify the critical angle for total internal reflection at the interface between the plastic pipe and water. Use the formula for the critical angle, \( \theta_c = \sin^{-1}(\frac{n_2}{n_1}) \), where \( n_1 \) is the refractive index of the plastic (1.60) and \( n_2 \) is the refractive index of water (1.33).
Calculate the critical angle using the values provided. This angle represents the maximum angle at which light can strike the interface and still be totally internally reflected.
Understand that the angle \( u \) in the problem is the angle of incidence inside the plastic pipe, relative to the normal at the interface. For total internal reflection to occur, \( u \) must be greater than or equal to the critical angle.
Set up the inequality \( u \geq \theta_c \) to find the range of possible values for \( u \).
Interpret the result to determine the largest possible value for \( u \) that ensures all light reflects back into the pipe when it strikes face AB.

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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 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. A higher index indicates that light travels slower in that medium, affecting how light bends when entering or exiting different materials.
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Total Internal Reflection

Total internal reflection occurs when a light wave traveling in a medium hits a boundary with a less dense medium at an angle greater than the critical angle. Instead of refracting, the light is completely reflected back into the denser medium. This phenomenon is crucial for applications like fiber optics and is dependent on the indices of refraction of the two media involved.
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Critical Angle

The critical angle is the minimum angle of incidence at which total internal reflection occurs. It can be calculated using the formula sin(θc) = n2/n1, where θc is the critical angle, n1 is the index of refraction of the denser medium, and n2 is that of the less dense medium. Understanding the critical angle is essential for determining the conditions under which light will reflect rather than refract at a boundary.
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Related Practice
Textbook Question
A horizontal, parallelsided 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.(a) What angle does the ray refracted into the water make with the normal to the surface?
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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 (a) the red part of the spectrum.
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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. (a) What is the largest that u can be if the pipe is in air?
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Textbook Question
The critical angle for total internal reflection at a liquid– air interface is 42.5°. (a) 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?
336
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Textbook Question
The glass rod of Exercise 34.22 is immersed in oil (n = 1.45). An object placed to the left of the rod on the rod's axis is to be d 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the ?
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