Since it is a diverging mirror, the focal length is -52 cm, Since the image is virtual, the imaage distance is -26 cm
Do the calculation again and solve for u
a diverging mirror of 52.0 focal lenght produces a virtual image of 26.0cm from the mirror.
how far from the mirror should the object be placed?
solution
f=-52.0cm, v=-26.0cm
1/u + 1/v = 1/f
rearranging the terms
1/u = 1/f - 1/v
substituting the values of v and f into the above eq. we get
1/u = 1/-52.0 - 1/-26.0
ans = 52cm
am i right pls check.
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