Question

Consider an object of thickness ds (parallel to the axis) in front of a lens or mirror. The image of the object has thickness ds′. Define the longitudinal magnification as M = ds′/ds. Prove that M = $-m^2$, where m is the lateral magnification

Short answer

$$\begin{gathered} \frac{1}{f}=\frac{1}{s}+\frac{1}{s\text{'}} \\ s\text{'}=\frac{fs}{s-f} \\ m=-\frac{s\text{'}}{s} \\ m=-\frac{f}{s-f} \\ M=\frac{ds\text{'}}{ds} \\ M=\frac{d}{ds}\left(\frac{fs}{s-f}\right) \\ =\frac{(s-f)f-fs}{{(s-f)}^2} \\ =\frac{sf-f^2-fs}{{(s-f)}^2} \\ =-\frac{f^2}{{(s-f)}^2} \\ =-m^2 \end{gathered}$$

Step-by-step solution

Step 1. Concepts and Principles 

The distance s of an object from a lens, the distance s′ of the image from the lens, and the focal length f of the lens are related by the thin lens equation:

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}\left(1\right)$

Several sign conventions are important when using the thin lens equation:

1. The focal length $f$is positive for convex lenses and negative for concave lenses.

2. The image distance$s\text{'}$ is positive for real images and negative for virtual images.

The lateral magnification M of the image due to a mirror or lens is defined as the negative of the ratio of the image distance s′ to the object distance s:

$M=\frac{s}{s\text{'}}\left(2\right)$

Step 2.Given Information

• The thickness of the object is: ds.

• The thickness of the image is: ds′.

• The longitudinal magnification is defined as: M=ds'/ds

Step 3.Simplify

We are asked to prove that $M=-m^2$, where mm is the lateral magnification.

The focal length of the lens is related to the object and image distances by the thin lens equation ''Equation (1)":

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s\text{'}}$

Rearrange and solve for s'

role="math" localid="1650036599612" $$\begin{gathered} \frac{1}{s\text{'}}=\frac{1}{f}-\frac{1}{s} \\ \frac{1}{s\text{'}}=\frac{s-f}{fs} \\ s\text{'}=\frac{fs}{s-f}\left(3\right) \end{gathered}$$

Step 4.

The lateral magnification is found from Equation (2):

$m=-\frac{s\text{'}}{s}$

Substitute for s' from Equation (3):

$$\begin{gathered} m=-\frac{{\displaystyle \frac{fs}{s-f}}}{s} \\ =-\frac{f}{s-f} \end{gathered}$$

Step 5.

Using the given definition of the longitudinal magnification, we have

$M=\frac{ds\text{'}}{ds}$

Substitute for s's′ from Equation (3):

$$\begin{gathered} M=\frac{d}{ds}\left(\frac{fs}{s-f}\right) \\ =\frac{(s-f)f-fs}{{(s-f)}^2} \\ =\frac{sf-f^2-fs}{{(s-f)}^2} \\ =-\frac{f^2}{{(s-f)}^2} \\ =-m^2 \end{gathered}$$

Hence, we have proved that $M=-m^2$.