PHL 201 Midterm: Creation and Destruction

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lavendermole636

16 May 2016

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We are accustomed to thinking that if a person pulls one hair from her head, then that person existed before she pulled out the hair, and she still exists after she pulls out the hair. Despite a change in the number of hairs on her head, something remains the same: she the same person exists both before and after the change. Thus far, this appears to be only a common sense view of change. We can put this point in more general terms: if an object undergoes a change, then it has one characteristic before the change and it has a different characteristic after the change. But if an object has different characteristics at two different times, then it must be present both times. On this view, an object does not undergo a change unless the same object exists before and after the change-- unless it persists through the change.

Related Questions

The ciliary muscles in a horse's eye can make only small changes tothe shape of the lens, so a horse can't change the shape of thelens to focus on objects at different distances as humans do.Instead, a horse relies on the fact that its eyes aren't spherical.As the figure shows, different points at the back of the eye are atsomewhat different distances from the front of the eye. We say thatthe eye has a "ramped retina"; images that form on the top of theretina are farther from the cornea and lens than those that form atlower positions. The horse uses this ramped retina to focus onobjects at different distances, tipping its head so that light froman object forms an image at a vertical location on the retina thatis at the correct distance for sharp focus.

Part A
In a horse's eye, the image of a close object will be infocus
A. at the top of the retina.
B. at the bottom of the retina.

Part B
In a horse's eye, the image of a distant object will be infocus
A. at the top of the retina.
B. at the bottom of the retina.

Part C
A horse is looking straight ahead at a person who is standing quiteclose. The image of the person spans much of the vertical extent ofthe retina. What can we say about the image on the retina?
A. The person's head and feet are both in focus.
B. The person's feet are in focus; the head is out of focus.
C. The person's head is in focus; the feet are out of focus.
D. The person's head and feet are both out of focus.

Part D
Certain medical conditions can change the shape of a horse'seyeball; these changes can affect vision. If the lens and corneaare not changed but all of the distances in the figure areincreased slightly, then the horse will be
A. unable to focus clearly at any distance.
B. nearsighted.
C. farsighted.

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 t_B 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=t_B 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 (d_C).

Give your answer in terms of the time t_B, 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 t_B when the wavefront at point B reached the interface.

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

greenduck607

MAC, Inc. (the company you work for) has decided to move forwardwith a feasibility
study for a product it wishes to manufacture and sell. This productis a particle launcher that should have the capability, for everyactivation, of launching two particles into the air at twodifferent times and at two different angles of elevation, in anautomated fashion so that they both land at the same distance fromthe launch point at the same time. For example, let P1 and P2 betwo particles, let T1 and T2 be two times, and let theta1 andtheta2 be two angles of elevation. Once activated, the launchershould be able to operate so that particle P1 is launched at timeT1 with an angle of elevation theta 1, and particle P2 islaunched
at time T2 with an angle of elevation theta2.
The launcher must be configured by a technician, prior to eachactivation, so that the
initial speed is set for the launcher, and the distance from thelauncher to where both
particles will land is set. For each activation of the launcher,the initial speed will be the
same for both particles. Once the technician has set the initialspeed and desired distance,
the launcher must calculate the angles of elevation to use and howmuch time to wait in
between particle launches. Then, when the technician activates thelauncher, both
particles will be launched at the appropriate angles of elevationand at the appropriate
times so that they both land at the configured distance from thelauncher at the same time.
The launcher, based on its calculations, will change the angle ofelevation to be ready for
the second particle launch. The launcher, however, must flash awarning for the
technician if the information the technician used to configure thelauncher does not result
in a two-particle launching solution.
Certain assumptions must be made. It is assumed that that launchercan always change
the angle of elevation to get ready for the second particle launchfaster than the time
delay calculated for the second particle launch. It is assumed thatthe units for speed are
meters per second, that the units for distance are meters, and thatgravity is 9.8 meters per
second squared. It is assumed that gravity is the only force actingon the motion of the
particles after launch, and that there is no wind to be considered,so that the particles land
at a point exactly in line with the direction the launcher ispointed.
The questions that must be answered and written up in anengineering report are as
follows:
1. Given the initial speed and desired landing distance for theparticles, find the
mathematical expressions that must be used to find 1, 2, and thetime delay
between particle launches.
2. Determine under which conditions the launcher will not be ableto perform the
operation requested, and will consequently flash a warning to thetechnician.
3. Each particle launched will follow a path of motion that isparabolic. Show that
the vertex of the parabola for each particleâs path occurs at thesame distance from
the launcher.
4. Along the path of each launched particle, there will be amaximum curvature.
Find that maximum curvature for each launched particle.

PHL 201 Midterm: Creation and Destruction

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