Question

In the displacement method, a convex lens is placed in between an object and a screen.If the magnification in the two positions are, $m_1$ and $m_2$ ($m_1>m_2$) and the distance between the two positions of the lens is x, 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

Answer: (B)

$\frac{x}{m_1-m_2}$

We know in displacement method,

$m_1=\frac{v}{u}$

and

$m_2=\frac{u}{v}$

Thus, $m_1\times m_2=1$

$m_1-m_2=\frac{v}{u}-\frac{u}{v}$

$m_1-m_2=\frac{v^2-u^2}{uv}$

From the figure,

Given, $v-u=x$

$\Rightarrow m_1-m_2=\frac{(v+u)(v-u)}{uv}=\frac{(v+u)x}{uv}$

Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
For Position 1, we have

Object Distance $=-u$

Image Distance $=v$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{-u}$

$\frac{1}{f}=\frac{u+v}{uv}$

$\Rightarrow m_1-m_2=\frac{(v+u)x}{uv}=\frac{1}{f}\times x$

$\Rightarrow m_1-m_2=\frac{x}{f}$

$f=\frac{x}{m_1-m_2}$

Hence, the correct answer is OPTION B.