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The wave model of light was developed almost 200 years after Snell published his research An understanding that light interacts with atoms when traveling through a medium, such that it moves more slowly than in a vacuum, was achieved three hundred years after Snell s work The wave relationship that you learned in 16 for light traveling through a vacuum, 0 c/f, can be rewritten as v/f, where v is the speed of light in any medium and is the wavelength The frequency of light, f, does not change when it crosses a boundary That is, the number of oscillations per second that arrive at a boundary is the same as the number that leave the boundary and transmit through the refracting medium So, the wavelength of light, , must decrease when light slows down Wavelength in a medium is shorter than wavelength in a vacuum What happens when light travels from a region with a high speed into one with a low speed, as shown in Figure 18-2a The diagram in Figure 18-2b shows a beam of light as being made up of a series of parallel, straight wave fronts Each wave front represents the crest of a wave and is perpendicular to the direction of the beam The beam strikes the surface at an angle, 1 Consider the triangle PQR Because the wave fronts are perpendicular to the direction of the beam, PQR is a right angle and QRP is equal to 1 Therefore, sin 1 is equal to the distance between P and Q divided by the distance between P and R sin The angle of refraction, PSR In this case



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2, 1

What is the result A Compilation fails B The main thread will run mostly before t1 runs C The main thread will run after t1, but together with t2 D The main thread will run after t2, but together with t1 E F The main thread will run after both t1 and t2 are mostly done The main thread s execution will overlap with t1 and t2 s execution

By taking the ratio of the sines of the two angles, PR is canceled, leaving the following equation:





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Figure 18-2b is drawn such that the distance between P and Q is equal to the length of three wavelengths of light in medium 1, or PQ 3 1 In a similar way, RS 3 2 Substituting these two values into the previous equation and canceling the common factor of 3 provides an equation that relates the angles of incidence and refraction with the wavelength of the light in each medium

Because the frequency F is fixed by the generator, only the wavelength can change when the velocity changes In a microwave waveguide we can relate phase velocity to wavelength as the wave is propagated in the waveguide: Vp

Using v/f in the above equation and canceling the common factor of f, the equation is rewritten as follows:

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Answer (for Objective 42): F is correct The key to this question is join(500) This means that the main thread will try to join to the end of t1, but it will only wait 500 milliseconds for t1 to complete If t1 doesn t complete (and it will take at least 1000 milliseconds for t1 to complete), the main thread will start after waiting 500 milliseconds A, B, C, D, and E are incorrect based on the above

Index of refraction Using the transitive property of equality, the previous two equations lead to the following equation:

In a vacuum, n 1 and v c If either medium is a vacuum, then the equation is simplified to an equation that relates the index of refraction to the speed of light in a medium

c v The index of refraction of a medium is equal to the speed of light in a vacuum divided by the speed of light in the medium

Figure 18-3 Ray A is partially refracted and partially reflected (a) Ray B is refracted along the boundary of the medium and forms the critical angle (b) An angle of incidence greater than the critical angle results in the total internal reflection of Ray C, which follows the law of reflection (c)

47 Given:

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This definition of the index of refraction can be used to find the wavelength of light in a medium compared to the wavelength the light would have in a vacuum In a medium with an index of refraction n, the speed of light is given by v c/n The wavelength of the light in a vacuum is c/f Solve for frequency, and substitute f c/ 0 and v c/n into 0 v/f (c/n)/(c/ 0) 0/n, and thus the wavelength of light in a medium is smaller than the wavelength in a vacuum

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