Question

In lens displacement method lateral magnification obtained are 2 and 0.5. if separation between position of lenses is 30 cm, then focal length of lens is

• A. 30 cm
• B. 20 cm
• C. 15 cm
• D. 10 cm

Answer 1

The correct option is B 20 cm

$\begin{array}{c}\text{Since, from the question we get that the magnification are given}\\ \text{as}2\&0.5\text{, if we take the object distance is}u\text{and the}\\ \text{image distance is}V\text{.}\\ \text{So:}2=\frac{v}{u};0.5=\frac{4}{v}\end{array}$

$\begin{array}{c}\text{Hence}2-0.5=\frac{v}{u}-\frac{u}{v}\text{or}1.5=\frac{v^2-u^2}{vu}\\ \Rightarrow \frac{(v-u)(v+u)}{vu}=1\cdot 5\text{we know}v-u=30cm\\ \Rightarrow \frac{\left(30\right)(v+u)}{vu}=1\cdot 5\Rightarrow \frac{{10}^{-1}}{2}=\frac{(v+u)}{vu}\end{array}$

$\begin{array}{c}{\text{from eq}}^n\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\text{; vel Rnow that}f=\frac{uv}{v+v}\\ \therefore \frac{{10}^{-1}}{2}=\frac{1}{f}\Rightarrow f=\frac{2}{{10}^{-1}}\Rightarrow 20cm\\ \text{Hence obtion}B\text{is correct.}\end{array}$