Question

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.

Short answer

The distance between an object and its real image is always greater than or equal to four times the focal length of the lens, which is X_min=4f .

Step-by-step solution

The given data:

The focal length is f .

The Object distance is p .

The image distance is i .

Understanding the concept of the lens:

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,

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

Calculation of the object and real image:

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 .