A lens of focal length 15cm forms on erect image which is three times the size of the object. The distance between the object and the image is ___.
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Correct Answer: Option D
Explanation:
F= 15cm
 M = 3
 M = \(\frac{v}{u}\)
 V =3u
 \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
 \(\frac{1}{15}\) = \(\frac{1}{u}\) - \(\frac{1}{3u}\)
 \(\frac{1}{15}\) =  \(\frac{2}{3u}\)
U =10cm
 D = u + v
 = 10 +30 = 40cm
F= 15cm
 M = 3
 M = \(\frac{v}{u}\)
 V =3u
 \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
 \(\frac{1}{15}\) = \(\frac{1}{u}\) - \(\frac{1}{3u}\)
 \(\frac{1}{15}\) =  \(\frac{2}{3u}\)
U =10cm
 D = u + v
 = 10 +30 = 40cm