Homework Solutions – Giancoli, Chapter 33 Problems: 10. Consider Figure 1 below, in which I’ve shown explicitly that, by the law of reflection, the angle that the reflected ray makes with the normal to the mirror is the same as the angle the incident ray makes with the normal, namely ? . 2. 30 m ? 1. 14 m 1.
54 m ? ? ? x Figure 1 0. 40 m From the figure, x= 0. 40 m tan? 1. 14 m , 2. 30 m But tan ? = so x= 0. 40 m = 0. 81 m ? 1.
14 m ? ? ? ? 2. 30 m ? 1 Physics 2426 Homework Solutions – Giancoli, Chapter 33 11. The situation is shown in Figure 2 below.The angle of incidence for the first reflection is 90 ? 40 = 50 , as shown. Therefore, by the law of reflection, the angle of reflection is also 50 , as shown. ? ? ? ? 50 40 40 Figure 2 50 135 The angle ? is therefore: ? = 180 ? 135 ? 40 = 5 Therefore, the angle ? is ? = 90 ? ? = 85 , and this is the angle of incidence for the second reflection.
By the law of reflection, then, the angle of reflection is also ? = 85 , and therefore the angle ? is, in fact, the same as ? : ? =? = 5 . 16. If the reflected rays are to make an image at infinity, then the reflected rays must be parallel to one another.
And in order for this to happen, the object must be placed at the focal point. Thus the object distance do is equal to f. But r f = , 2 so the object distance must be: r do = = 11.
0 cm . 2 18. The Christmas tree ball is a convex spherical mirror. If the diameter of the ball is 9. 0 cm, then r = 4. 5 cm, and the focal length is: r f = ? = ? 2. 25 cm .
2 (Note that the focal length is negative since the focal point is behind the mirror. ) So if you are 25. 0 cm from the surface of the ball, the ray diagram would look something like Figure 3 below. 2Physics 2426 Homework Solutions – Giancoli, Chapter 33 Ray 1 Ray 3 Ray 2 Figure 3 As you can see from the ray diagram, the image is virtual..
. It’s behind the mirror, so no light really emanates from the image. Also, as you can see from the ray diagram, the image is upright. To find the image distance di , we use the mirror equation: 1 1 1 + = do di f 1 1 1 1 1 = ? = ? di f do ( ? 2. 25 cm ) 25. 0 cm di = ? 2. 1 cm Thus, the image is 2.
1 cm behind the mirror… This is the meaning of the minus sign in the image distance. 21. For the mirror described, the image must clearly be upright. You wouldn’t want to see an inverted image of the car behind you.
) And the only time a concave mirror gives an upright image is when the object is between the focal point and the mirror, and then the concave mirror gives an image that is larger than the object. Since this is not what the mirror described does, this mirror cannot be a concave mirror… It must be convex. (By the way, note that a convex mirror always gives an image that is upright and smaller than the object, as is desired in this case. ) If the cars appear to be 0.
33 times their normal size, then h m ? i = 0. 3 . ho But hi d =? i , ho do so di = ? 0. 33 ? di = ? 0. 33d o do Therefore, the mirror equation gives: 1 1 1 + = di do f 1 1 1 + = ? 0. 33d o ) d o f ( Rearranging for f, I get: ? 1 ? f = do ? ? 1? 1 ? ? ? 0. 33 ? 3 Physics 2426 Homework Solutions – Giancoli, Chapter 33 And putting in do = 20. 0 m , I get: f = ? 9.
85 m So the radius of curvature is r = 2 f = ? 19. 7 m = ? 20 m , keeping two sig figs. 29. (a.
) The mirror must be a convex mirror. (The only time a concave mirror gives an erect image is when do < f , and then the concave mirror gives a magnified image.But this is not what we want, so we can’t use a concave mirror. ) Alternatively, to answer this question, you could think about the formula for the magnification: h d m= i =? i ho do In order to get an erect image, the magnification will have to be positive. But this will occur only if the image distance di is negative. (The object distance d o will always be positive in the cases we will consider. ) If di is negative, then di = ? di , and we can write the mirror equation as: 1 1 1 ? = do di f (1)Now, the fact that we want the image to be smaller than the object implies that we want hi < 1, ho and from the formula for the magnification, this implies that di d ? i n1 .
Applying Snell’s law at the left boundary: And at the right boundary: n1 sin ? 1 = n2 sin ? 2 . n2 sin ? 2 = n1 sin ? 3 . (1) (2) Notice that ? 2 , the angle of refraction at the left boundary, becomes the angle of incidence for the refraction at the right boundary. Equations (1) and (2) imply that: or It follows that n1 sin ? 3 = n1 sin ? 1 , sin ? 3 = sin ? 1 .
?3 = ? 1 , thus the beam emerging from the slab is parallel to the incident beam, as we wanted to show. 6