An object is placed against the center of a converging lens and then moved along the central axis until it is $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\text{m}}$from the lens. During the motion, the distance between the lens and the image it produces is measured. The procedure is then repeated with a diverging lens. Which of the curves in Fig. 34-28 best gives versus the object distance p for these lenses? (Curve 1 consists of two segments. Curve 3 is straight.)

• Fig.34-28 is given.

Using the lens formula, you can get the expression for image distance $\left(i\right)$. Then, putting different values of object distance $\left(p\right)$, you will get corresponding. By analyzing the given figure accordingly, you can find the curve in Fig.34-28 which best gives versus the object distance $\left(p\right)$for the given lenses.

Formula:

The lens formula for the object and image distance is,

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$…..(i)

The convex lens produces an image at infinity when the object is at the focal point. From the figure, the image distance becomes infinity for curve 1 at one point.

Therefore curve 1 corresponds to the convex lens. But for the concave lens the relation between$\left|i\right|$and $p$is a curved one and$\left|i\right|$never becomes infinity for a finite value of $p$. Therefore curve 2 corresponds to the concave lens.

Using equation (i), the image distance formula can be given as follows:

$\begin{array}{rcl}\frac{1}{i}& =& \frac{1}{f}-\frac{1}{p}\\ & =& \frac{p-f}{fp}\\ i& =& \frac{fp}{p-f}\end{array}$

When,$p=0,i=0$

When,$p=\infty ,i=f$

This is depicted in curve 2.

Therefore, the curve 2 in Fig.34-28 best gives $\left|i\right|$versus the object distance$p$ for the given lenses.