Show that the distance between an object and its real image formed by a thin converging lens is always greater than or equal to four times the focal length of the lens.

The focal length is f .

The Object distance is p .

The image distance is i .

Using the lens equation and substituting the image distance in terms of object distance and x , where x is the total distance from an object to an image, you calculate the required value.

Formula:

The lens formula,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$

Consider the distance x to be the distance between the object and image.

$x=p+i$

Then, the image distance can be given as:

i = x - p

Using the above value of i into the thin lens equation (i) and solving for x, you can get that

$\frac{1}{p}+\frac{1}{(x-p)}=\frac{1}{f}$

$$\begin{gathered} \left(\right(x-p)+p)/p(x-p)=1/f \\ \frac{x}{(xp-p^2)}=1/f \\ fx=xp-p^2 \\ x(p-f)=p^2 \end{gathered}$$

$x=\frac{p^2}{p-f}$ ….. (ii)

To find the minimum value of x , differentiating x with respect to p and equating it to zero, we get the minimum x value as follows:

$$\begin{gathered} \frac{dx}{dp}=0 \\ \frac{d}{dp}\left(\frac{p^2}{p-f}\right)=0 \\ \frac{p^2{\displaystyle \frac{d}{dp}}(p-f)-(p-f){\displaystyle \frac{d}{dp}}{p}^2)}{(p-f{)}^2}=0 \end{gathered}$$

$\frac{(p^2-2(p-f\left)\right)}{(p-f{)}^2}=0$

$\frac{p(p-2f)}{(p-f{)}^2}=0$

$$\begin{gathered} (p-2f)=0 \\ p=2f \end{gathered}$$

Using the above value in equation (ii), you get the minimum separation x value as follows:

$\begin{array}{l}x=\frac{2f^2}{2f-f}\\ =\frac{4f^2}{f}\\ =4f\end{array}$

Hence, x should be equal to or greater than 4f .