Question

Describe the displacement method to determine focal length of Convex Lens under the following headings:
(i) Ray diagram
(ii) Formula derivation
(iii) Why the distance between two pins should be greater than $4f$?

Answer 1

(ii) Formula derivation:

Let $AB$ and $CD$ be the two pins so placed such that the distance between then is more than $uf$.

Let $A^{{}^{\prime }}{B}^{{}^{\prime }}$ be the image formed on pin $CD$ when lens is placed at $L$,

$\therefore$ $OA=u$, $O{A}^{{}^{\prime }}=v$ and $d=v+u$ .........(i)

In the second position of lens $L_2$ image $A^{{}^{\prime }}{B}^{{}^{\prime \prime }}$ is once again formed on $CD$ because $A$ and $A^{{}^{\prime }}$ are conjugate foci.

$O^{{}^{\prime }}{A}^{{}^{\prime }}=u$ and $O^{{}^{\prime }}A=v$

If the displacement of lens is $x$, then

$x=v-u$ ........(ii)

from equation (i) and (ii)

$u=\frac{d-x}{2}$ and $v=\frac{d+x}{2}$

from lens equation

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

Putting the values of $v$ and $u$

$\frac{1}{f}=\frac{1}{\frac{d+x}{2}}-\frac{1}{\frac{d-x}{2}}$

By cartesian sign convention, $u$ will be negative and $v$ will be positive.

Then $\frac{1}{f}=\frac{1}{\left(\frac{d+x}{2}\right)}-\frac{1}{\{-\frac{d-x}{2}\}}$

$=\frac{1}{\frac{d+x}{2}}+\frac{1}{\frac{d-x}{2}}$

$=\frac{2}{d+x}+\frac{2}{d-x}$

$=\frac{2d-2x+2d+2x}{(d+x)(d-x)}$

$f=\frac{4d}{d^2-x^2}$

This is required formula.

(iii) Need for the separation between the two pins to be more than $4f$: To obtain real image in two positions of lens, the value of $d$ should be more than $4f$.

From relation, $f=\frac{d^2-x^2}{4d}$

we get, $d^2-4df-x^2=0$

$\therefore$ $d=\frac{4f\pm \sqrt{{(-4f)}^2-4\times 1\times (-x^2)}}{2}$

or $d=\frac{4f\pm \sqrt{16f^2+4x^2}}{2}$

But, $\sqrt{16f^2+4x^2}$ is always greater than $4f$. Hence the value of $d$ should be more than $4f$, otherwise the value of $d$ will be negative which is meaningless.