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course Phy 202
3/14 122.23.2011
`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).
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The focal point always lies halfway between the mirror and the center of the circle. This is due to the angle at which the ray refracts.
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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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The meeting place will be closer to the central axis the further away from the mirror the flame is.
@& It is possible to reason this out using similar triangles.
It turns out that the formula 1 / f = 1 / i + 1 / o gives the correct result, but without all the reasoning.*@
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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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It will be approx 2 cm from the axis and 21 cm from the lens. Given we compare the 2 similar triangles.
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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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It will hit the axis first. As it passes the other side, will it be further deflected as it leave the cylindrical lens?
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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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No, the two waves will cancel out the effects of each other when it reaches 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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Yes, a wave will be created in the third chain because the paths differ by a half a wavelength, 180 degrees out of phase
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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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There will be a wave created in the third wave as the wavelengths of the left hand are more tense and the wavelengths of the right hand are more slack as the chain is more slack.
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03-16-2011
@& The propagation velocity will be affected by the increased tension. If the propagation velocity wasn't affected you would have destructive interference, just as in the first case. Since propagation velocity is affected the interference won't be destructive.*@
&#This looks good. See my notes. Let me know if you have any questions. &#