#$&* course PHY 232 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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): 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: The power of the lens=1/(0.205m)=4.88diopeters And since the length is a positive number then it is converging. Also the lens has -6.25 diopters then the length 1/(-6.25m^-1)=-0.16m and because the number is negative it means the lens is diverging confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: Since we have been given the angle of incidence at 45deg then we use a similar idea as previous question as sin(angle of incidenc)/ sin(angle of refraction)=1/1.52=0.66 hence to find the angle of emerging light =0.66sin(45) =27.8deg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating:Ok ********************************************* Question: `q**** query univ phy problem (absent from 13th edition) / 34.92 11th edition 34.86 (35.62 10th edition) Consider a spherical surface, on the 'incident side' of which the index of refraction is n_a, on the other side of which the index of refraction is n_b. We assume that n_b > n_a. The 'first focal length' f at the spherical surface is the distance from the surface at which a point source (located in the material for which the index of refraction is n_a) must be located so that its light will be refracted to form a parallel beam, i.e., the distance of an object from the surface which will result in an image at infinity (the distance f for which the image forms at s ' = infinity). The 'second focal length', designated f ', is the distance at which parallel rays incident on the surface, having been refracted at the surface, will converge. Thus f ' is the focal distance for an object at distance s = infinity. How do you prove that the ratio of indices of refraction na / nb is equal to the ratio f / f ' of focal lengths? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the formula below; na/s+nb/sprime=(nb-na)/R once for s and another time for s' 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/f' and do a little algebra, you end up with f/f'=na/nb confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:Ok ********************************************* Question: `q **** univ phy How did you prove that f / s + f ' / s' = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we start from nb / f ' = (nb - na) / R, obtained in a previous part, we get f ' = nb * R / (nb - na). 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** The series of figures below represent the geometric solution. These figures are physically realistic, in that for real materials both f and f ' would be expected to be much larger relative to the radius of the sphere. In the next figure the 'green' triangle is similar to the 'blue' triangle, and the image h ' is depicted. The two triangles are not exactly right triangles and not quite similar: since A is further from the central axis that B the segment from A to B is not quite vertical, hence not quite perpendicular to the axis. However for rays which are close to the axis, the error in assuming these to be right triangles, and therefore similar, is not significant. In the next figure the two 'purple' triangles are also very nearly similar. The larger ('green') triangle is represented at the top of the figure below, along with the smaller ('blue') triangle. The distances h and h ' are indicated. The 'green' triangle has legs s and h + h ', while the 'blue' triangle has corresponding legs f and h '. Thus s / (h + h ') = f / h ' The two 'purple' triangles are represented in the lower half of the figure. The smaller triangle has legs f ' and h, while the larger has legs s ' and h + h ', where i is the distance at which the image forms. Thus f ' / h = s ' / (h + h '). The two larger triangles are depicted below, with their respective legs s and s ', and their common leg h + h ' so labeled. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:Ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!