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

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1
Identify the critical angle for total internal reflection. The critical angle, \( \theta_c \), can be calculated using the formula \( \sin(\theta_c) = \frac{n_2}{n_1} \), where \( n_1 \) is the index of refraction of the plastic (1.60) and \( n_2 \) is the index of refraction of air (1.00).
Calculate the critical angle using the formula from step 1. This angle will determine the maximum angle at which light can strike the face AB and still be totally internally reflected.
Determine the angle of incidence, \( \theta_i \), of the light rays striking face AB. Since the light travels parallel to the upper part of the pipe, the angle of incidence is related to the angle at which face AB is cut.
Set the angle of incidence, \( \theta_i \), equal to the critical angle, \( \theta_c \), calculated in step 2. This ensures that all light hitting face AB at this angle will be totally internally reflected.
Solve for the angle \( u \), which is the angle at which face AB should be cut. This angle will be complementary to the critical angle, so \( u = 90^\circ - \theta_c \).

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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 (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 plastic with an index of refraction of 1.60, light travels slower than in a vacuum, which affects how it bends when entering or exiting the material.
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Total Internal Reflection

Total internal reflection occurs when light attempts to move from a medium with a higher index of refraction to one with a lower index (like from plastic to air) at an angle greater than the critical angle. This phenomenon ensures that all light is reflected back into the medium, which is essential for the design of optical fibers and other applications where light confinement is necessary.
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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 Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. For a plastic pipe in air, knowing the critical angle helps determine the maximum angle at which light can strike the face AB to ensure it reflects back into the pipe.
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Related Practice
Textbook Question
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in Fig. E33.1. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cm long) after reflecting from the first mirror?
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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. (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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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?
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