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
The correct answer is: A
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.