5.4 Displacement method to find focal length.

For thin lens we can write,

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

Substituting the above situation we get,

$$\begin{equation} \begin{aligned} \frac{1}{{\left( {D - u} \right)}} - \frac{1}{{ - u}} = \frac{1}{{ + f}} \\ \frac{1}{{\left( {D - u} \right)}} + \frac{1}{u} = \frac{1}{f} \\ \frac{{u + D - u}}{{(D - u)u}} = \frac{1}{f} \\ fD = - {u^2} + Du \\ {u^2} - Du + fD = 0 \\ u = \frac{{D \pm \sqrt {{D^2} - 4fD} }}{2} \\\end{aligned} \end{equation} $$

or

$$u = \frac{{D \pm \sqrt {D(D - 4f)} }}{2}$$

1. If $D<4f$, then $u$ is imaginary . So, physically no position of lens is possible.

2. If $D=4f$, then $u = \frac{d}{2}$ or $u=2f$

(In this case, the minimum distance between an object and its real image for convex lens is $4f$).

3. If $D>4f$, then ${u_1} = \frac{{D + \sqrt {D(D - 4f)} }}{2}$ and ${u_2} = \frac{{D - \sqrt {D(D - 4f)} }}{2}$

(In this case, for two positions of convex lens real image is formed on the screen).

Let $d$ be the distance between two lens position.

So, $$d = {u_1} - {u_2}$$ $$\begin{equation} \begin{aligned} d = \left( {\frac{{D + \sqrt {D(D - 4f)} }}{2}} \right) - \left( {\frac{{D - \sqrt {D(D - 4f)} }}{2}} \right) \\ d = \sqrt {D(D - 4f)} \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} {d^2} = {D^2} - 4Df \\ 4Df = {D^2} - {d^2} \\\end{aligned} \end{equation} $$ $$f = \frac{{{D^2} - {d^2}}}{{4D}}$$

${u_1} = \frac{{D + \sqrt {D(D - 4f)} }}{2}$

$$\begin{equation} \begin{aligned} {v_1} = D - {u_1} \\ {v_1} = D - \left( {\frac{{D + \sqrt {D(D - 4f)} }}{2}} \right) \\ {v_1} = \left( {\frac{{D - \sqrt {D(D - 4f)} }}{2}} \right) \\\end{aligned} \end{equation} $$

Magnification for a lens is given by,

$$\begin{equation} \begin{aligned} m_1 = \frac{{{v_1}}}{{{u_1}}} \\ m_1 = \left( {\frac{{\frac{{D - \sqrt {D(D - 4f)} }}{2}}}{{\frac{{D + \sqrt {D(D - 4f)} }}{2}}}} \right)\quad ...(i) \\\end{aligned} \end{equation} $$

Similarly,

${u_2} = \frac{{D - \sqrt {D(D - 4f)} }}{2}$

$$\begin{equation} \begin{aligned} {v_2} = D - {u_2} \\ {v_2} = D - \left( {\frac{{D - \sqrt {D(D - 4f)} }}{2}} \right) \\ {v_2} = \left( {\frac{{D + \sqrt {D(D - 4f)} }}{2}} \right) \\\end{aligned} \end{equation} $$

Magnification for a lens is given by,

$$\begin{equation} \begin{aligned} m_2 = \frac{{{v_2}}}{{{u_2}}} \\ m_2 = \left( {\frac{{\frac{{D + \sqrt {D(D - 4f)} }}{2}}}{{\frac{{D - \sqrt {D(D - 4f)} }}{2}}}} \right)\quad ...(ii) \\\end{aligned} \end{equation} $$

Multiplying equation $(i)$ and $(ii)$ we get,

$${m_1}{m_2} = 1\quad ...(iii)$$

As we know, $$m = \frac{{{\text{height of image}}}}{{{\text{height of object}}}}$$ So,

\[\left. \begin{gathered} {m_1} = \frac{{{I_1}}}{{{I_O}}} \hspace{1em} \\ \hspace{1em} \\ {m_2} = \frac{{{I_2}}}{{{I_O}}} \hspace{1em} \\ \end{gathered} \right\}\quad ...(iv)\]

From equation $(ii)$ and $(iv)$ we get,

$$\frac{{{I_1}{I_2}}}{{I_O^2}} = 1$$ or $${I_O} = \sqrt {{I_1}{I_2}} $$

where,

$I_o$: Size of the object

$I_1$ & $I_2$: Size of image formed in two different cases