Question

An object is placed in front of a convex mirror. Draw the convex mirror (radius of curvature  15 cm) to scale, and place the object 25 cm in front of it. Make the object height 4 cm. Using a ray diagram, locate the image and measure its height. Now move the object closer to the mirror, so the object distance is 5 cm. Again, locate its image using a ray diagram. As the object moves closer to the mirror, (a) does the magnitude of the image distance become larger or smaller, and (b) does the magnitude of the image height become larger or smaller? (c) What is the ratio of the image height when the object distance is 5 cm to its height when the object distance is 25 cm? Give your answer to one significant figure.

Answer 1

It's pretty much impossible for me to draw a decent picture to scale on this program, but I'll work it out with calculations. To draw the rays, draw one ray from the top of the object parallel to the surface into the mirror, then bend it through the focus. The other ray should go from the top of the object through the mirror's center, where the surface intersects. The image is then from the surface to the intersection of the two rays.

r = 2f
15 = 2f
f = -7.5 cm      because the mirror is convex,the focal length is negative

1/p + 1/q = 1/f
1/25 + 1/q = 1/-7.5
q = -5.8 cm      so the image is behind themirror (virtual)

M = -q/p = h'/h
M = -(-5.8)/25
M = 0.23x
0.23 = h'/4
h' = 0.92 cm      the image is 0.92 cm tall

Moving the image closer,
1/p + 1/q = 1/f
1/5 + 1/q = 1/-7.5
q = -3 cm      the image is 3 cm behind themirror (virtual)

M = -q/p = h'/h
M = -(-3)/5
M = 0.6x
0.6 = h'/4
h' = 2.4 cm      the image is 2.4 cm tall

As the object is moved closer to the mirror, the image distance becomes smaller (-5.8 to -3 cm).
The image height becomes larger (0.92 to 2.4 cm)
The ratio of 5 cm to 25 cm is: 2.4/0.92 = 2.6