An object placed on the principle axis of a convex lens of focal length 10m produces a real image of double magnification. The image distance from the lens is
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Correct Answer: Option A
Explanation:
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) = +ve real image distance
where; U = +ve real object distance
f = real focus = 10
m = magnification
M = 2 = \(\frac{v}{u}\)
2u = V
\(\frac{1}{2u}\) + \(\frac{1}{u}\) = \(\frac{1}{10}\)
2u = 30
u = 15cm
v = 2u = 2 x 15
= 30cm
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) = +ve real image distance
where; U = +ve real object distance
f = real focus = 10
m = magnification
M = 2 = \(\frac{v}{u}\)
2u = V
\(\frac{1}{2u}\) + \(\frac{1}{u}\) = \(\frac{1}{10}\)
2u = 30
u = 15cm
v = 2u = 2 x 15
= 30cm