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, ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ $({\mathrm{m}}_1>{\mathrm{m}}_2)$ and the distance between the two positions of the lens is $\mathrm{x}$, the focal length of the lens is,

• A. $\frac{\mathrm{x}}{{\mathrm{m}}_1+{\mathrm{m}}_2}$
• B. $\frac{\mathrm{x}}{{\mathrm{m}}_1-{\mathrm{m}}_2}$
• C. $\frac{\mathrm{x}}{{\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)}^2}$
• D. $\frac{\mathrm{x}}{{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}^2}$

Answer 1

The correct option is B

$\frac{\mathrm{x}}{{\mathrm{m}}_1-{\mathrm{m}}_2}$

Step 1. Given data

It is given that the magnification in the two positions are, ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ $({\mathrm{m}}_1>{\mathrm{m}}_2)$ and the distance between the two positions of the lens is $\mathrm{x}$.

We have to find the focal length of lens.

Step 2. Formula to be used

The lens formula is,

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

Here, $\mathrm{f}$ is the focal length measured in meter, $\mathrm{u}$ is object distance and $\mathrm{v}$ is image distance.

Step 3. Draw the figure which show image formation in the displacement method

We know in displacement method,

Let distance of the image from the lens $\mathrm{v}$, distance of the object from the lens be $\mathrm{u}$ and magnification of the lens $\mathrm{m}$.

So the displacement formula is,

${\mathrm{m}}_1=\frac{\mathrm{v}}{\mathrm{u}}$ --------- $\left(1\right)$

and
${\mathrm{m}}_2=\frac{u}{v}$ -------- $\left(2\right)$

So,

${\mathrm{m}}_1\times {\mathrm{m}}_2=1$

${\mathrm{m}}_1-{\mathrm{m}}_2=\frac{\mathrm{v}}{\mathrm{u}}-\frac{\mathrm{u}}{\mathrm{v}}$

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

Step 4. Find the focal length.

From the figure,

Given is, $\mathrm{v}-\mathrm{u}=\mathrm{x}$

$\Rightarrow$ ${\mathrm{m}}_1-{\mathrm{m}}_2=\frac{\left(\mathrm{v}+\mathrm{u}\right)\left(\mathrm{v}-\mathrm{u}\right)}{\mathrm{uv}}$

$=\frac{\left(\mathrm{v}+\mathrm{u}\right)\mathrm{x}}{\mathrm{uv}}$

Solve it by using lens formula is,

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

Here, $\mathrm{f}$ is the focal length measured in meter, $\mathrm{u}$ is object distance and $\mathrm{v}$ is image distance.

So,

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

$=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$=\frac{\mathrm{u}+\mathrm{v}}{\mathrm{uv}}$

We know that, ${\mathrm{m}}_1-{\mathrm{m}}_2=\frac{\left(\mathrm{v}+\mathrm{u}\right)\mathrm{x}}{\mathrm{uv}}$.

So,

${\mathrm{m}}_1-{\mathrm{m}}_2=\frac{1}{\mathrm{f}}\mathrm{x}$

${\mathrm{m}}_1-{\mathrm{m}}_2=\frac{\mathrm{x}}{\mathrm{f}}$

$\mathrm{f}=\frac{\mathrm{x}}{{\mathrm{m}}_1-{\mathrm{m}}_2}$

Hence, option $\mathrm{B}$ is correct answer.