Question

An object is located 17.1 cm in front of a convex mirror, the image being 4.49 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

Answer #1

so f = do*di/(do + di) = 17.1*(-4.49)/(17.1 + -4.49) =
-6.088cm

Now m= -di/do for the first object m = 4.49/17.1 = 0.262

Since m = also = yi/yo we have yi/yo = 0.26 so yi = 0.26*yo

Since the second object is twice the first but the image height is
the same we get

m = -di/do = 0.13 or di = -0.13*do

We now have 1/(-6.088) = 1/(do) + 1/di = 1/do + 1/(-0.13*do)

-0.164 = (-0.13 + 1)/(-0.13*do) = 0.87/(-0.13*do)

so do = 40.804cm

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