course Phys 202 018. `Query 16
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Given Solution: `a**** The angle of incidence and the angle of refraction are both measured relative to the normal direction, i.e., to the direction which is perpendicular to the surface. For a horizontal water surface, these angles will therefore be measured relative to the vertical. 66 degrees is therefore the angle of refraction. Using 1.3 as the index of refraction of water and 1 as the index of refraction of air, we would get • sin(angle of incidence) / sin(angle of refraction) = 1 / 1.3 = .7, very approximately, so that • sin(angle of incidence) = 1.3 * sin(angle of refraction) = .7 * sin(66 deg) = .7 * .9 = .6, very approximately, so that • angle of incidence = arcSin(.6) = 37 degrees, again a very approximate result. You should of course use a more accurate value for the index of refraction and calculate your results to at least two significant figures. STUDENT QUESTION it should be 1/1.333 right? nb is where its going which is air sin(66)/sin (theta)=1/1.333=.75 INSTRUCTOR RESPONSE The incident beam is in the water (call this medium a, consistent with your usage), the refracted beam in the air (medium b). It's sin(theta_a) / sin(theta_b) = n_b / n_a, so sin(theta_a) = 1 / 1.333 * sin(theta_b) = .7 sin(66 dec) = .6 and theta_a = 37 degrees. Again all calculations are very approximate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qPrinciples of Physics and General College Physics 23.46 What is the power of a 20.5 cm lens? What is the focal length of a -6.25 diopter lens? Are these lenses converging or diverging? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = 1/f P = 1/ 0.205 m = 4.88 D -6.25 = 1/f f = -0.16 m = -16 cm This first has a positive focal length of 0.205 m so it’s a converging lens The second has a negative focal length of 16cm so it’s a diverging lens
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Given Solution: `aThe power of the 20.5 cm lens is 1 / (.205 meters) = 4.87 m^-1 = 4.87 diopters. A positive focal length implies a converging lens, so this lens is converging. A lens with power -6.25 diopters has focal length 1 / (-6.25 m^-1) = -.16 m = -16 cm. The negative focal length implies a diverging lens. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery gen phy problem 23.32 incident at 45 deg to equilateral prism, n = 1.52; and what angle does light emerge? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N1 sin(angle1) = n2 sin (angle2) 1 sin (angle45) = (1.58) sin (angle2) Angle 2 = 26.6 degrees 26.6 + 120 + x = 180 degrees X = 33.4 degrees N2 sin (angle2) = n3 sin(angle3) (1.58)sin (33.4) = (1) sin (angle 3) Angle 3 = 60.5 degrees
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Given Solution: `aSTUDENT 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q**** 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? ********************************************* Question: `q **** univ phy How did you prove that f / s + f' / s' = 1? "