?????€???????Physics II 07-03-2006
......!!!!!!!!...................................
16:16:15 **** 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?
......!!!!!!!!...................................
RESPONSE --> did do this experiemtn and also don't have Notes #19 on any of my CDs
.................................................
......!!!!!!!!...................................
16:16:18 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:16:21 **** how did the radius of the 'mirror' cut from the can and the distance of the focal point from thie mirror compare?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:16:23 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
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:16:25 **** For the circular lens were the rays entering the lens the diverted toward the normal?
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
16:16:27 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:38:32 query gen phy problem 23.32 incident at 45 deg to equilateral prism, n = 1.52; and what angle does light emerge?
......!!!!!!!!...................................
RESPONSE --> n1*sintheta1 = n2*sintheta2 theta 2 = sin-1(n1*sintheta1/n2) theta2 = 1.0*sin45/1.52 theta 2 = 27.7 (90-theta2)+(90-theta3)+theta 4 = 180 (90-27.7)+(90-theta3)+theta 4 = 180 62.3 + 60 + theta4 =180 180-62.3-60 = 57.7degrees
.................................................
......!!!!!!!!...................................
16:38:36 STUDENT SOLUTION: To solve this problem, I used figure 23-51 in the text book to help me visualize the problem. The problem states that light is incident on an equilateral crown glass prism at a 45 degree angle at which light emerges from the oppposite face. Assume that n=1.52. First, I calculated the angle of refraction at the first surface. To do this , I used Snell's Law: n1sin'thea1=n2sin'thea2 I assumed that the incident ray is in air, so n1=1.00 and the problem stated that n2=1.52. Thus, 1.00sin45 degrees=1.52sin'theta2 'thea 2=27.7 degrees. Now I have determined the angle of incidence of the second surface('thea3). This was the toughest portion of the problem. To do this I had to use some simple rules from geometry. I noticed that the normal dashed lines onthe figure are perpendicular to the surface(right angle). Also, the sum of all three angles in an equilateral triangle is 180degrees and that all three angles in the equilateral triangle are the same. Using this information, I was able to calculate the angle of incidence at the second surface. I use the equation (90-'thea2)+(90-'thea3)+angle at top of triangle=180degrees. (90-27.7degrees)=62.3 degrees. Since this angle is around 60 degrees then the top angle would be approx. 60 degrees. ###this is the part of the problem I am a little hesitant about. Thus, 62.3 degrees+(90-'thea3)+60 degrees=180 degrees-'thea)=57.7degrees 'thea=32.3 degrees This is reasonable because all three angles add up to be 180 degrees.62.3+60.0+57.7=180degrees Now, I have determined that the angle of incidence at the second surface is 32.3 degrees, I can calculate the refraction at the second surface by using Snell's Law. Because the angles are parallel, nsin'thea3=n(air)sin''thea4 1.52sin32.3=1.00sin (thea4) 'thea 4=54.3 degrees INSTRUCTOR COMMENT: Looks great. Here's my explanation (I did everything in my head so your results should be more accurate than mine): Light incident at 45 deg from n=1 to n=1.52 will have refracted angle `thetaRef such that sin(`thetaRef) / sin(45 deg) = 1 / 1.52 so sin(`thetaRef) = .707 * 1 / 1.52 = .47 (approx), so that `thetaRef = sin^-1(.47) = 28 deg (approx). We then have to consider the refraction at the second face. This might be hard to understand from the accompanying explanation, but patient construction of the triangles should either verify or refute the following results: This light will then be incident on the opposing face of the prism at approx 32 deg (approx 28 deg from normal at the first face, the normals make an angle of 120 deg, so the triangle defined by the normal lines and the refracted ray has angles of 28 deg and 120 deg, so the remaining angle is 32 deg). Using Snell's Law again shows that this ray will refract at about 53 deg from the second face. Constructing appropriate triangles we see that the angle between the direction of the first ray and the normal to the second face is 15 deg, and that the angle between the final ray and the first ray is therefore part of a triangle with angles 15 deg, 127 deg (the complement of the 53 deg angle) so the remaining angle of 28 deg is the angle between the incident and refracted ray. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:38:41 **** 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?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:38:45 ** 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 **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:38:49 **** univ phy How did you prove that f / s + f' / s' = 1?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:38:55 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:39:00
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"