Question

A convergent beam of light passes through a diverging lens of focal length $0.2\, m$ and comes to focus $0.3\, m$ behind the lens. The position of the point at which the beam would converge in the absence of the lens is

• A. $0.12\,m$
• B. $0.6\,m$
• C. $0.3\,m$
• D. $0.15\,m$

Answer 1

Answer: (A)

$0.12\,m$

Answer 2

Here, $f= - 0.2\, m, v = +0.3\, m$ The lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$ $= \frac{1}{0.3}+\frac{1}{0.2} = \frac{0.5}{0.06}$ $u = \frac{0.06}{0.5} = 0.12 m$