Lens Displacement Method

The lens displacement method is one of the most accurate and commonly used techniques in optics to determine the focal length of a convex lens. In this method, an object and a screen are kept fixed at a certain distance, and the lens is moved between them to obtain sharp images at two different positions. The distance between these two positions helps in calculating the focal length precisely.

This method is widely used in physics experiments, practical exams, and optical instrument design because it reduces errors and does not require direct measurement of object and image distances. Due to its simplicity and high accuracy, the lens displacement method is an important topic for students preparing for board exams, JEE, and other competitive exams.

Displacement Method

Consider an object and a screen fixed at a distance $D$ apart. Let a lens of focal length $f$ be placed between the object and the screen.

From figure we observe that

$
u+v=D \quad \Rightarrow \quad v=D-u
$

Also from Lens formula

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

$\begin{aligned} & \frac{1}{D-u}-\frac{1}{-u}=\frac{1}{f} \\ & \frac{1}{D-u}+\frac{1}{u}=\frac{1}{f} \\ & D f=(D-u) u \\ & u^2-D u+D f=0 \\ & u=\frac{D \pm \sqrt{D^2-4 f D}}{2}\end{aligned}$

For $u$ to be mathematically real,

$
\begin{aligned}
& D^2-4 f D \geq 0 \\
\Rightarrow \quad & D \geq 4 f
\end{aligned}
$

CASE-1: For $D=4 f$

$
u=v=\frac{D}{2}
$

i.e. the lens is placed exactly between the object and the screen.

CASE-2: For $D>4 f$
We get two different position of lens ( $L_1$ and $L_2$ ) for which the image of object on the screen is distinct and clear.

The object distances for these two positions are given by

$
\begin{aligned}
& u_1=\frac{D-\sqrt{D^2-4 f D}}{2} \\
& u_2=\frac{D+\sqrt{D^2-4 f D}}{2}
\end{aligned}
$

Since $u+v=D$, so

$
\begin{aligned}
& v_1=\frac{D+\sqrt{D^2-4 f D}}{2} \\
& v_2=\frac{D-\sqrt{D^2-4 f D}}{2}
\end{aligned}
$

We observe that

$
\begin{aligned}
& u_1=v_2=u(\text { say }) \\
& v_1=u_2=v(\text { say })
\end{aligned}
$

Let the lens be displaced through $x$, then we observe from figure that

$
\begin{aligned}
& x=v_1-u_1=\sqrt{D^2-4 f D} \\
\Rightarrow & x^2=D^2-4 f D \\
\Rightarrow & f=\frac{D^2-x^2}{4 D}
\end{aligned}
$