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Mirrors problem help?
So the problem goes like this:
An object is located 13.5 cm in front of a convex mirror, the image being 7.35 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?
I know that the magnification equation is involved, but the thing is that it seems to contradict.
m = hi/ho = -di/do
Based on the problem, the convex mirror's magnification is 7.35/13.5 = 49/90. However, in the other situation, the convex mirros's magnification is 2, which contradicts the 1st magnification.
Can someone tell me what I did wrong and how to solve it please?
1 Answer
- AndresLv 41 decade agoFavorite Answer
Image 1
Magnification 1= m1 = y1' / y1 = - s1' / s1
s1 = 13.5cm & s1' = -7.35cm
m1 = 0.544
Image 2
y2 = 2 y1 & y2' = y1
m2 = y2' / y2 = y1' / 2 y1 = m1 / 2 = - s2' / s2 = 0.2722
Besides:
1/s2 + 1/s2' = - 1/focus = 1/s1 - 1/s1'
s1 = 13.5cm & s1' = 7.35cm
Then:
1/s2 + 1/s2' = -0,0619
- s2' / s2 = 0.2722 --> 1/s2' = - 1 / (.2722 s2)
1/s2 - 1/ (.2722 s2) = -.0619
s2 = 43.19cm fron of the mirror- answer -
