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10 Nov 2019

Assume all angles to be exact.
A 13cm deep opaque container is empty except for a single coinresting on its bottom surface. When looking into the container at aviewing angle of 50 degrees relative to the vertical side of thecontainer, you see nothing on the bottom. When the container isfilled with water, you see the coin (from the same viewing angle)on the bottom of, and just beyond, the side of the container.
1.How far is the coin from the side of the container?

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Jean Keeling
Jean KeelingLv2
19 Oct 2019

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Related textbook solutions

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

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An opaque container is 15 cm deep. Itcontainsonly a single coin. When looking into the container at aviewingangle of 50o relative to the vertical side ofthecontainer, you see nothing on the bottom. When the containerisfilled with water, you see the coin (from the same viewingangle)on the bottom of, and just beyond, the side of the container.Howfar is the coin from the side of the container?

I see the formula and understand but I do not get why wearecalculating from a 90 degree angle and why we suddenly switchedtotangent in calculating the answer...

5) [8 pts] A friend of yours works in a bar as a bartender, and you decide to improve his life by making his job easier. In particular, he has to take empty beer mugs from the bar and put them in the sink (see the diagram below) - and of course, since students come in after exams for a drink, there are a lot of dirty beer mugs to move. You realize that he could send them sliding down the bar and they would land in the sink if they had the right initial speed. You wonder if you can find a single speed (so that he can easily learn to slide them correctly) for different conditions: in particular, if the bar is clean (larger coefficient of kinetic friction between the mug and the bar), or if there is some beer spilt on it (lower coefficient of kinetic friction). The vertical distance from the bar to the surface of the water in the sink is h=0.65m; the horizontal distances from the edge of the bar to the front and back of the sink are dmin=1.2m and dmax=1.8m. The coefficients of kinetic friction are pk=0.25 for a clean bar, and Uk=0.15 for a bar with beer spilt on it. Your friend stands at a point along the bar which is 5.6m from the edge of the bar, and an empty beer mug has a mass of 0.325kg. a) Find the minimum horizontal speed the mug must have when it leaves the bar so that it lands in the sink (at the near edge). b) Using the speed you found in a), calculate the bartender's "launch speed" (ie, the initial sliding speed) for the high-friction case (Uk=0.25). c) Using the speed you found in b) as the initial sliding speed, analyse the low- friction case: find the speed the mug will have when it leaves the bar, and determine whether it lands in the sink (or hits the far wall and shatters, thus making your friend lose his job). d) [for one extra point) Your bartender friend loves your new system, but the next time you visit him you realize he's still walking over to the sink every time he has a shot glass (a glass that's smaller than a beer mug - it's mass is only 0.125 kg) to put there. Does the mass of the glass matter, or can he launch them at the same speed as the mugs without worry (assume the same coefficients of friction as for the beer mug)? dmax beer mug bar sink

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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