#$&* course phy 242 june 2 Question: `q**** query univ phy problem (absent from 13th edition) / 34.92 11th edition 34.86 (35.62 10th edition)
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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): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** univ phy How did you prove that f / s + f ' / s' = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From previous problem we have nb/s’ = (nb-na)/ R and na/s = (nb-na)/ R and s = f and s’= f’. So , nb/f’ = (nb-na)/ R And na/f = (nb-na)/ R Plugging this into f / s + f ' / s' = 1, we get : na/ (s*(nb-na)/ R) + nb/(s’*(nb-na)/ R) = 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): ok ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!