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Chapter 34: Q. 3 (page 989)

3. One problem with using optical fibers for communication is that a light ray passing directly down the center of the fiber takes less time to travel from one end to the other than a ray taking a longer, zig-zag path. Thus light rays starting at the same time but traveling in slightly different directions reach the end of the fiber at different times. This problem can be solved by making the refractive index of the glass change gradually from a higher value in the center to a lower value near the edges of the fiber. Explain how this reduces the difference in travel times.

Short Answer

Expert verified

This reduces the difference in travel times is Larger index of refraction will slow the light down and it will get to the end of the fiber closer to the time the other light gets there.

Step by step solution

01

 Step1:Definition of light ray

When a light ray travels at an angle into a medium with a different refractive index, it refracts. This change in speed causes a shift in direction. Consider the movement of air into water as an example. As light continues to travel at a different angle, its speed decreases.

02

find reduces the different in travel time

Larger index of refraction v=c/n.will slow the light down and it will arrive at the end of the fibre closer to the time the other light arrives.

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Most popular questions from this chapter

A lens placed10cm in front of an object creates an upright image twice the height of the object. The lens is then moved along the optical axis until it creates an inverted image twice the height of the object. How far did the lens move?

The mirror in FIGURE P34.65 is covered with a piece of glass. A point source of light is outside the glass. How far from the mirror is the image of this source?

Shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles θ1and θ2in the two media. Suppose that you did not know Snell’s law.

A. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n1 and n2

B. The time depends on x. There’s one value of x for which the light travels from A to B in the shortest possible time. We’ll call it xmin. Write an expression (but don’t try to solve it!) from which xmincould be found.

C. Now, by using the geometry of the figure, derive Snell’s law from your answer to part b.

You’ve proven that Snell’s law is equivalent to the statement that “light traveling between two points follows the path that requires the shortest time.” This interesting way of thinking about refraction is called Fermat’s principle.

An air bubble inside an8.0-cm--diameter plastic ball is 2.0cmfrom the surface. As you look at the ball with the bubble turned toward you, how far beneath the surface does the bubble appear to be?

A 2.0-cm-tall candle flame is 2.0 m from a wall. You happen to have a lens with a focal length of 32 cm. How many places can you put the lens to form a well-focused image of the candle flame on the wall? For each location, what are the height and orientation of the image?
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