An object is placed in front of a concave mirror whose radius of curvature is 12cm. If the magnification of the Image produced is 1.5, how far is the object from the mirror?
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Correct Answer: Option B
Explanation:
Radius of curvature(r)=12cm
focal length(F) = r ÷ 2 → 12 ÷ 2 = 6cm
therefore F = 6cm
magnification(M) = v ÷ u
1.5 = v ÷ u
:v = 1.5u
Using mirror formula
1/f =1/v + 1/u
\(\frac{1}{6}\) =\(\frac{1}{1.5u}\) + \(\frac{1}{u}\)
\(\frac{1}{6}\) = \(\frac{1 + 1.5}{1.5u}\)
\(\frac{1}{6}\) = \(\frac{2.5}{1.5u}\)
1 * 1.5u = 2.5 * 6
u = \(\frac{15}{1.5}\) = 10
therefore object distance =10cm.
Radius of curvature(r)=12cm
focal length(F) = r ÷ 2 → 12 ÷ 2 = 6cm
therefore F = 6cm
magnification(M) = v ÷ u
1.5 = v ÷ u
:v = 1.5u
Using mirror formula
1/f =1/v + 1/u
\(\frac{1}{6}\) =\(\frac{1}{1.5u}\) + \(\frac{1}{u}\)
\(\frac{1}{6}\) = \(\frac{1 + 1.5}{1.5u}\)
\(\frac{1}{6}\) = \(\frac{2.5}{1.5u}\)
1 * 1.5u = 2.5 * 6
u = \(\frac{15}{1.5}\) = 10
therefore object distance =10cm.