Question

In the displacement method, a convex lens is placed in between an object and a screen. If the magnifications in the two positions are $m_1$ and $m_2$ and the displacement of the lens between the two positions is x, then the focal length of the lens is

• A. $\frac{x}{m_1+m_2}$
• B. $\frac{x}{m_1-m_2}$
• C. $\frac{x}{(m_1+m_2{)}^2}$
• D. $\frac{x}{(m_1-m_2{)}^2}$

Answer 1

The correct option is B $\frac{x}{m_1-m_2}$

$m_1=\frac{v}{u},m_2=\frac{u}{v}$
$m_1-m_2=\frac{v}{u}-\frac{u}{v}$
$m_1-m_2=\frac{v^2-u^2}{uv}=\frac{(v-u)(v+u)}{uv}$
Now, $v-u=x,\frac{1}{f}=\frac{1}{v}+\frac{1}{u}or\frac{1}{f}=\frac{u+v}{uv}$
$\therefore m_1-m_2=\frac{x}{f}$ or $f=\frac{x}{m_1-m_2}$