Question

An object is located 12.5 cm in front of a convex mirror, the image being 6.90 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) = 12.5*(-6.9)/(12.5 + -6.9) = -15.4cm

Now m= -di/do for the first object m = 6.9/12.5 = 0.55

Since m = also = yi/yo we have yi/yo = 0.55 so yi = 0.55*yo

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

m = -di/do = 0.275 or di = -0.275*do

We now have 1/(-15.4) = 1/(do) + 1/di = 1/do + 1/(-0.275*do)

-0.0649 = (-0.275 + 1)/(-0.275*do) = 0.71/(-0.29*do)

so do = 37.765cm

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