An object is placed 10cm from a converging lens of foal length 15cm. Calculate the magnification of the image formed
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30}\)
\(\frac{1}{v} = \frac{-1}{30}\)
v = - 30 cm
but m = \(\frac{v}{u} = \frac{30}{10} = 3\)
Note that the negative sign only shows the placement of the image. It has no effect on the magnification basically.
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30}\)
\(\frac{1}{v} = \frac{-1}{30}\)
v = - 30 cm
but m = \(\frac{v}{u} = \frac{30}{10} = 3\)
Note that the negative sign only shows the placement of the image. It has no effect on the magnification basically.