course Phy 201
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21:20:15 **** Query experiment 26 **** How nearly did your four rays come to converging? Did each ray reflect at the same angle from the normal as the angle of the incoming ray?
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RESPONSE --> From my results the rays were close to converging, a few degrees separted them. To me each ray reflected the same angle from the normal as the angle of the incoming ray.
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21:22:53 GOOD STUDENT RESPONSE: I had a trouble on this experiment, although the procedure seems quite simple. I used a diet coke can as my 'mirror'. I used a ruler to measure the radius of the half-circle. Then I used a compass to draw the circle of the can on my paper. The radius fo the can was 2.8 cm.According to the video clips and the class notes, if I aim the laser beam to the left or to the right of the straight line drawn from the center of the can keeping the beam perpendicular to this center line, all reflected angles would cross that line at approximately R/2 or 1.4 cm. Before taking data for this experiment, I tested this idea. I was able to see that all the reflected angles appeared to cross the central line at the focal point of 1.4cm. I used trig to calculate the angle values. I measured the approriate sides of the triangle (all distances in cm). By dividing I think that it is adj/opp and taking the inverse tangent of this value, I calculated the angles. When I aimed the beam 1/8 of the container's radius((.35) to the right, I determined that the angle was approximately 28.5 degrees. When I aimed the third beam parallel to the central beam and approximately 2.8 of the container's radius to the left of the central beam, the angle was 33.7 degrees. When I aimed the fourth beam parallel to the central beam and approximately 3/8 of the container's radius to the right of the central beam the angle was 38.9 degrees. I am not sure how all of these angles are related. I know that there is probably great experimental error because I had a hard time estimating the reflected angle.
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RESPONSE --> ok i understand, I also had a hard time estimating the reflected angle.
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21:25:01 **** how did the radius of the 'mirror' cut from the can and the distance of the focal point from thie mirror compare?
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RESPONSE --> The radius of the can was around 2.8 cm and the focal point of the mirror was 1.4 cm.
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21:27:06 STUDENT RESPONSE: Once again, I am not totally confident with this experiment. But, the radius of the can was 2.8 cm and the focal point of the mirror was R/2 or 1.4cm. INSTRUCTOR COMMENT: Sounds like you did this part right and that you got a result that agrees with the theory
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RESPONSE --> ok i understand
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21:31:12 **** For the circular lens were the rays entering the lens the diverted toward the normal?
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RESPONSE --> From the experiment as I moved the laser in one direction the laser beam would move in the opposite direction. So as my screen was moved closer the movement of the point on the screen stops. As the screen gets closer it starts moving in the same direction as the laser rather then opposite.
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21:35:19 STUDENT RESPONSE: In this experiment I used a 2-liter Mountain Dew softdrink bottle. I used my compass and placed the pointer in the center of the bottom of the container, which was already marked. This enabled me to draw my bottle on my paper. The radius was 3.7cm. I was able to observe that as I would move the laser in one direction the laser beam would move in the opposite direction. I used the cassette with the ruler taped to it as my screen. I believe that because the bottle was so large, I was unable to accurately measure the angles as I moved to the left and to the right of the center line. I had no other container at my apartment, so I was unable to finish this experiment. You should understand that as the 'screen' is moved closer the movement of the point on the screen stops, then as you get the screen closer it starts moving in the same direction as the laser, rather than in the opposite direction. This shows that at the 'stopped' point the beams passing through the bottle will converge. **
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RESPONSE --> ok i understand
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21:38:06 **** query univ phy problem 34.86 (35.52 10th edition) f when s'=infinity, f' when s = infinity; spherical surface. How did you prove that the ratio of indices of refraction na / nb is equal to the ratio f / f' of focal lengths?
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RESPONSE --> dna
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21:38:09 ** The symbols s and s' are used in the diagrams in the chapter, including the one to which problem 62 refers. s is the object distance (I used o in my notes) and s' the image distance (i in my notes). My notation is more common that used in the text, but both are common notations. Using i and o instead of s' and s we translate the problem to say that f is the object distance that makes i infinite and f ' is the image distance that makes o infinite. For a spherical reflector we know that na / s + nb / s' = (nb - na ) / R (eqn 35-11 in your text, obtained by geometrical methods similar to those used for the cylindrical lens in Class Notes). If s is infinite then na / s is zero and image distance is s ' = f ' so nb / i = nb / f ' = (nb - na) / R. Similarly if s' is infinite then the object distance is s = f so na / s = na / f = (nb - na) / R. It follows that nb / f ' = na / f, which is easily rearranged to get na / nb = f / f'. THIS STUDENT SOLUTION WORKS TOO: All I did was solve the formula: na/s+nb/sprime=(nb-na)/R once for s and another time for sprime I took the limits of these two expressions as s and s' approached infinity. I ended up with f=-na*r/(na-nb) and fprime=-nb*r/(na-nb) when you take the ratio f/fprime and do a little algebra, you end up with f/fprime=na/nb **
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RESPONSE --> dna
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21:38:12 **** univ phy How did you prove that f / s + f' / s' = 1?
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RESPONSE --> dna
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21:38:17 ** We can do an algebraic solution: From nb / f ' = (nb - na) / R, obtained in a previous note, we get f ' = nb * R / (nb - na). From na / f = (nb - na) / R we get f = na * R / (nb - na). Rearranging na/s+nb/s'=(nb-na)/R we can get R * na / ( s ( na - nb) ) + R * nb / (s ' ( na - nb) ) = 1. Combining this with the other two relationships we get f / s + f ' / s / = 1. An algebraic solution is nice but a geometric solution is more informative: To get the relationship between object distance s and image distance s' you construct a ray diagram. Place an object of height h at to the left of the spherical surface at distance s > f from the surface and sketch two principal rays. The first comes in parallel to the axis, strikes the surface at a point we'll call A and refracts through f ' on the right side of the surface. The other passes through position f on the object side of the surface, encounters the surface at a point we'll call B and is then refracted to form a ray parallel to the axis. The two rays meet at a point we'll call C, forming the tip of an image of height h'. From the tip of the object to point A to point B we construct a right triangle with legs s and h + h'. This triangle is similar to the triangle from the f point to point B and back to the central axis, having legs f and h'. Thus (h + h') / s = h / f. This can be rearranged to the form f / s = h / (h + h'). From point A to C we have the hypotenuse of a right triangle with legs s' and h + h'. This triangle is similar to the one from B down to the axis then to the f' position on the axis, with legs h and f'. Thus (h + h') / s' = h / f'. This can be rearranged to the form f' / s' = h' / (h + h'). If we now add our expressions for f/s and f'/s' we get f / s + f ' / s ' = h / (h + h') + h' / (h + h') = (h + h') / (h + h') = 1. This is the result we were looking for. **
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RESPONSE --> dna
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