JAMB - Physics (2016 - No. 42)
If the magnification of a virtual image formed by an object 10cm from a convex less is 3, then the focal length of the lens is
10cm
20cm
25cm
15cm
Explanation
M = \(\frac{v}{u}\) = 3 = \(\frac{v}{30}\). V = 30m
\(\frac{1}{u}\) + \(\frac{1}{v}\) = \(\frac{1}{f}\) = \(\frac{1}{10}\) - \(\frac{1}{30}\) = \(\frac{1}{f}\)
\(\frac{3 - 1}{30}\) = \(\frac{1}{f}
f = \(\frac{30}{2}\) = 15cm
\(\frac{1}{u}\) + \(\frac{1}{v}\) = \(\frac{1}{f}\) = \(\frac{1}{10}\) - \(\frac{1}{30}\) = \(\frac{1}{f}\)
\(\frac{3 - 1}{30}\) = \(\frac{1}{f}
f = \(\frac{30}{2}\) = 15cm
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