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course PHY 232
3/16 11:28 am
`q001. A ray is directed parallel to the central axis of a circular mirror, 2 cm from the central axis. The mirror has radius 10 cm. A radial line (i.e., a line starting at the center of the mirror) through the point at which the ray strikes the mirror is normal (perpendicular) to the mirror's surface at that point. Show why the reflected ray strikes the axis at a point which is close to 5 cm from the center (and also 5 cm from the mirror). (This point is called the 'focal point' of the mirror).****
Drawing the setup gives us some triangles with theta_i being the angle that the ray makes with a radial line that strikes the surface at the same point as the ray and with theta_r being the angle the reflected ray makes with the same radial line. Draw 2 triangles with the upper one formed by the ray, the radial line, and a third line drawn perp. to the ray and the center of the circle. The second one being formed by the reflected ray, the radial line, and a third side perp. to reflected to ray connected to the center. These 2 triangles share a common hypotenuse of 10 cm and both their left sides are 2 cm. The distance from the center point along the central axis to the point where the reflected ray crosses that central axis is given to be 5 cm and if make that a hypotenuse of the lower triangle we have 2 similar triangles and 2/5 is 2*2/10 meaning that the lowest side is approximately twice the distance of the ray line that begins at the radial distance from the mirror and 2 cm above that central axis. Much easier to get this from a picture than describing it here and no one could really understand what I just said without a good diagram.
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The tip of a flame 'sends out' rays in all directions (see the above figure). Assume that the flame is 2 cm high and consider two of these: the ray parallel to the axis, which reflects through the focal point, and the ray which strikes the mirror along its axis and hence reflects as if the mirror was flat instead of curved. These two rays meet at a point, as shown below. If the flame is 20 cm from the mirror, how far will that meeting point be from the mirror, and from the central axis? If the flame is moved further from the mirror, will the meeting point move further from or closer to the mirror, and will it move further from or closer to the central axis?
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It seems to matter what the radius of curvature of the mirror is and it's not given so I'll assume it's the same mirror as before where r = 10 cm so that f = 5 cm. A formula readily gives the distance of the point to be 6.67 cm from the mirror but I can't seem to get this to work out geometrically. I've drawn it out like before and used a lot of geometry to find the distances but they aren't consistent with the formula, which is correct. Also, using y'/y = s'/s, y' is found to be 2/3. These seem to be consistent but I can't work them out graphically as I think they could be. Actually I just realized what I've doing wrong in my diagrams: I have been assuming the point is in front of the focal point but it is behind it as the value suggests but I'm not going to redo what I just wrote only to add that I did verify this. Also, as the candle moves out to inf the image point approaches 5 cm as a limit.
@& Good. No need to rewrite. And I agree that only someone familiar with the diagram would be able to follow a verbal description.*@
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`q002. A laser ray directed from the left parallel to the central axis of a lens, but 2 cm from the axis, is refracted by the lens and passes through the axis at a point 10 cm to the right of the lens. A second ray originates from a point 20 cm to the left of the lens, strikes the lens on the central axis and passes through with practically no deflection from its original path.
How far to the right of the lens will this ray intersect the first, and how far will it be from the axis when that occurs?
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Setting this up with similar triangles easily leads to the fact that the point of intersection is x= 20 cm and y = -2 cm. The process is hard to explain but follows right out of the class notes pictures.
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`q003. A laser ray enters a cylindrical lens of radius 10 cm, along a path parallel to but 2 cm from the central axis. It is deflected at that point toward a radial line, in such a way that its angle with the radial line is decreased by 50%. Will the ray get to the opposite surface of the mirror before it reaches the central axis? Where will it go then?
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More similar triangles. What you end up with is 2 triangles to share a 2 cm side with one having 10 cm hypotenuse. Looking at the geometry and the fact that the angle was reduced by 50% means that the reflected angle is 1/2 the original one means that the second hypotenuse is 20 cm which means it's 3rd side is 19.9 cm which is very close to the edge of the cylinder.
@& Good. Note that the sine of half of a small angle is a little greater than the sine of the angle, due to the way the sine differs from linear (also verifiable by considering the Taylor expansion x - x^3 / 3 ...). *@
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`q004. I send two waves with wavelengths 12 cm are sent down two chains, held in my hands like the reins of a horse. At some distance from my hands the two chains meet a third chain at a common point, forming a narrow Y. I create the two waves by moving my hands in opposite directions, alternately moving them further apart and closer together. The waves travel at the same speed toward their meeting point and travel the same distance before they get there.
Will this create a wave in the third chain?
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The alternating motion makes the waves 180º OOP, out of phase, and they are 1/2 wavelength path diff since one is opposite the other at any given time and this adds destructively meaning there will be no wave in the 3rd chain.
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If I now move my left hand 6 cm closer to the point where the chains meet, will a wave be created in the third chain?
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Here they are still OOP by 180º but the 1/2 wavelength difference is counteracted by the 1/2 wavelength change in distance for one chain, which means when they meet they will both be moving in the same direction and so will constructively interfere and a wave will be created in the 3rd chain.
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If without changing my grip I pull my left hand back to its original position, stretching the chain so the distances to the third chain are now equal, and repeat the same motion, how might this affect what happens in the third chain?
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First the 1st and 2nd chains will destructively interfere again and so the connection point of the 3 bands will be stationary but the third chain will be sending back waves of 2x the A of the other 2 so they will destructively interfere but won't completely cancel since A-A+2A ≠ 0. I'm completely certain about this though.
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@& The propagation velocity in the chain that's being pulled back will change, causing an effect on the phases that can't be determined without further information (i.e., effect of stretch on tension, and mass density).*@
&#Good responses. See my notes and let me know if you have questions. &#