Question

Describe the displacement method to determine the focal length of convex lens on the following points:
(i) Formula derivation
(ii) Ray diagram
(iii) Observation Table
(iv) Precautions (any two).

Answer 1

(i) Formula derivation: Let $AB$ and $CD$ be the two pins so placed such that the distance between them 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 $\frac{1}{f}=\frac{1}{\left(\frac{d-x}{2}\right)}-\frac{1}{-\left(\frac{d-x}{2}\right)}$
$=\frac{1}{\frac{d+x}{2}}+\frac{1}{\frac{d-x}{2}}$
$=\frac{2}{d+x}+\frac{2}{d-x}$
$=\frac{2d-2x+2d+2x}{(d+x)(d-x)}$
$f=\frac{4d}{d^2-x^2}$
This is required formula.
(iii) Observation Table

S.No.	Position of object pin $a$ $cm$	Position of image pin $b$ $cm$	First position of lens $m$ $cm$	Second position of lens $n$ $cm$	Distance between the pins $d=b-a$	Displacement of lens $x=n-m$	$F=\frac{d^2-x^2}{ud}cm$

(iv) Precautions:
(a) The distance between the pins should be greater than four times of focal length.
(b) The line joining the tip of the pins and optical centre must lie on horizontal line.