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 ___.
A
10cm
B
20cm
C
30cm
D
40cm
Verified Correct Answer: D.
Physical Principles & Numerical 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
Source: Myschool Past Questions Archive (Record #47754)