WAEC - Physics (2012 - No. 28)

An object is placed 10cm from a converging lens of foal length 15cm. Calculate the magnification of the image formed
3.0
1.5
0.6
0.3

Explanation

Concave/converging mirror:

* produces a virtual image when object distance(u) exceeds its focal length(f).

\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)

\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30}\)

\(\frac{1}{v} = \frac{-1}{30}\)

v = -30cm( virtual image) 

but m = \(\frac{v}{u} = \frac{30}{10} = 3\)

Note that the negative sign only shows the placement of the image. It does not affect the magnification.

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