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JAMB Physics 2018 Paper

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Question : 81

Total: 87

A lens of focal length 15cm forms an erect image which is three times the size of the object. The distance between the object and the image is ___.

10cm
20cm
30cm
40cm

Solution:

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

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