ps03_solns

# ps03_solns - Physics 1112 Fall 2015 PS#4 Solutions Page 1...

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Physics 1112 Spring 2008 PS #3 Solutions Page 1 of 10 Physics 1112 Problem Set #3 Solutions Problem 1: Determine whether each of the following statements is true or false. (a) A convex mirror never forms a real image of a real object. (b) The virtual image formed by a convex mirror is always smaller than the real object. (c) A reduced image formed by a concave mirror is always inverted relative to a real object. (d) A virtual image formed by a concave mirror is always smaller than the real object. (e) A concave mirror never forms an enlarged real image of a real object. (f) An enlarged image formed by a concave mirror is always upright. (a) True. A convex (diverging) mirror has negative f . By the mirror equation, since 1 /d i =1 /f - 1 /d o (and d o is positive for a real object), d i always has to be a negative number. (b) True. From part (a) we know that convex mirrors always form virtual images. Also, since f is negative and d o is positive, the expression 1 /d i /f - 1 /d o is more negative than either - 1 /d o or 1 /f , which means that | d i | is smaller than d o . This in turn implies, from M = - d i /d o , that the image is reduced compared to the object. (c) True. One way to prove this statement is to take the mirror equation and rearrange it to solve for the magni±cation, in terms of d o and f : 1 d o + 1 d i = 1 f = 1+ d o d i = d o f = 1 - 1 M = d o f 1 M - d o f = f - d o f = M = f f - d o . Now, to make a reduced image we need | M | < 1. But since d o has to be positive for a real object, the only way to get a reduced image is to make f - d o a large negative number by making d o large. (At least 2 f , in fact.) This implies that the magni±cation must be negative. (d) False. Actually, virtual images in concave (converging) mirrors are always enlarged; this is how makeup and shaving mirrors work. To see this, start with the mirror equation and solve for d i : 1 d o + 1 d i = 1 f = d i = fd o d o - f . This expression tells us that in order to have a virtual image (negative d i ), we need d o <f . In other words, the object is inside the focal length. Now use this expression for d i to get an expression for the magni±cation: M = - d i d o = f f - d o .

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Physics 1112 Spring 2008 PS #3 Solutions Page 2 of 10 Since d o <f , the denominator is a smaller positive number than the numerator. So M> 1 and the object is therefore always enlarged. (e) False. For example, let the object distance be d o =3 f/ 2. Then 1 /d i =1 /f - 1 /d o = 1 /f - 2 / 3 f / 3 f , so d i f (real image because d i > 0). The magni±cation in this case is M = - d i /d o = - 3 (3 2) = - 2, thus enlarged. For another way to see this, imagine you have a ray diagram for a converging mirror in which a real object produces a reduced real image. Then notice for your diagram that you are free to interchange the roles of image and object, and your ray tracing still happens to work in reverse. Poof! Now you have an example of an enlarged image.
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ps03_solns - Physics 1112 Fall 2015 PS#4 Solutions Page 1...

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