Solution

Statement

$13.3.6.$ An object in the form of a segment of length $l$ is located along the optical axis of a converging lens with focal length $f$, which produces a real image of all its points. The middle of the segment is located at a distance $a$ from the lens. Determine the longitudinal magnification of the object.

Solution

According to the figure, the distance from the left end of the object to the lens will be:

$$a_{1}=a+\frac{l}{2} \tag{1}$$

The distance from the right end of the object to the lens will be:

$$a_{2}=a-\frac{l}{2} \tag{2}$$

Using the thin lens formula, we find the distances between the lens and the images of these ends of the object:

$$\frac{1}{a_{1}}+\frac{1}{b_{1}}=\frac{1}{f}$$
$$b_{1}=\frac{a_{1}f}{a_{1}-f} \tag{3}$$

Similarly for $b_{2}$:

$$b_{2}=\frac{a_{2}f}{a_{2}-f} \tag{4}$$

Since $a_{1}> a_{2}$, then $b_{2}> b_{1}$.

The longitudinal magnification of the object, by definition, is:

$$k=\frac{b_{2}-b_{1}}{l} \tag{5}$$

Substituting (1) into (3), we find $b_{1}$:

$$b_{1}=\frac{(a+\frac{l}{2})f}{(a-f)+\frac{l}{2}}$$

Substituting (2) into (4), we find $b_{2}$:

$$b_{2}=\frac{(a-\frac{l}{2})f}{(a-f)-\frac{l}{2}}$$

Substituting the expressions for $b_{1}$ and $b_{2}$ into formula (5):

$$k=\frac{1}{l}\cdot \left( \frac{(a-\frac{l}{2})f}{(a-f)-\frac{l}{2}}-\frac{(a+\frac{l}{2})f}{(a-f)+\frac{l}{2}} \right)$$

Finding a common denominator and simplifying:

$$k=\frac{f}{l}\cdot \frac{(a-\frac{l}{2})(a+\frac{l}{2})-f(a-\frac{l}{2})-(a-\frac{l}{2})(a+\frac{l}{2})+f(a+\frac{l}{2})}{(a-f)^{2}-\frac{l^{2}}{4}}$$
$$k=\frac{f}{l}\cdot \frac{-f(a-\frac{l}{2})+f(a+\frac{l}{2})}{(a-f)^{2}-\frac{l^{2}}{4}}$$
$$k=\frac{f}{l}\cdot \frac{fl}{(a-f)^{2}-\frac{l^{2}}{4}}$$

Finally:

$$k=\frac{f^{2}}{(a-f)^{2}-\frac{l^{2}}{4}}$$