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12 Nov 2019

A flat sheet of crown glass has a thickness of 3.7 cm. It is on top of a flat sheet of diamond that has a thickness of 2.6 cm. Light strikes the crown glass perpendicularly and travels through it and then through the diamond. In the time it takes the light to travel through the two sheets, how far would it have traveled in a vacuum?

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Nelly Stracke
Nelly StrackeLv2
4 Nov 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A ray of light strikes a flat, 2.00-cm-thick block of glass (n = 1.50) at an angle of 30.0˚with respect to the normal. (Fig P22.18.) (a) Find the angle of refraction at the top surface. (b) Find the angle of incidence at the bottom surface and the refracted angle. (c) Find the lateral distance d by which the light beam is shifted. (d) Calculate the speed of light in the glass and (e) the time required for the light to pass through the glass block. (f) Is the travel time through the block affected by the angle of incidence? Explain.

 

Huygens' principle (first described by the Dutch scientist Christiaan Huygens in 1678) uses geometry to determine the shape of a wavefront at a time t, given some initial wavefront at an earlier time. This can be constructed by imagining that the initial wavefront is a source of wavelets that propagate from each point at the speed of light. From this principle, one can determine the angles of reflection and refraction by considering the speed of light at each point on the wavefront. In this problem, we will explore some of these concepts.

Part A

First, let's look at a plane wave that is incident on a flat piece of material with an index of refraction n. Part of the light is transmitted through the material and part of it is reflected. For the moment, let's just look at the part that is transmitted. At time t=0, the wavefront is a distance d away from the surface of the material. At time t=t1, the wavefront is at the position of the material interface. Use the Huygens' principle to determine how far into the material (d') the wavefront has propagated by time t=2t1.

Express your answer in terms of the variables n, t1, and the speed of light in a vacuum c.

Note that there would be no change in the direction of the wavefront at any point because the wavefront encountered the material interface at the same time at all points. Thus, all of the individual wavefronts all propagated at the same speed, thereby maintaining the flat wavefront.

Now, instead of having a flat wavefront propagating normal to the material interface, we have a flat wavefront propagating toward the material at an angle of 55° to the axis perpendicular to the material interface. In this part, we will look at the relative positions of a few points—A, B, and C—on the wavefront to illustrate Huygens' principle. (Intro 1 figure) Point C touches the vacuum/material interface at time t=0 whereas point B is a distance d and point A is a distance 2d away from the interface.

Part B

What is the time tB it will take for point B of the wavefront to encounter the vacuum/material interface?

Express your answer numerically to two decimal places in units of t1 (the time it takes light to travel a distance d in a vacuum).

It should be noted that the accuracy of the method increases the smaller the distance is before making a new wavefront and redoing the wavelets. If you tried to just draw a large wavelet from point B, then it would hit the surface after only traveling a distance d. This large wavelet would make it difficult to visualize how to add up all of the other wavelets to make a coherent wavefront. As a result, one should just draw a line perpendicular to the wavefront at point B until it hits the surface, at which point the wavelets will change owing to the new index of refraction.

 

Part C

How far did point C go into the material interface in the time t=tB that it took for point B to get to the interface? For this part we are looking for the distance traversed by the point C in the material (dC).

Give your answer in terms of the time tB, c, and n.

 

Part D

What is the new angle θ at which point C of the wavefront is propagating (relative to a line perpendicular to the vacuum/material interface)? Try to use the fact that you have a spherical wavefront propagating from t=0 at the point where C met the vacuum/material interface until time tB when the wavefront at point B reached the interface.

Express your answer in terms of inverse trigonometric functions and n.

 

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