Suppose an object has thickness du so that it extends from object distance u to `u+du.` Prove that the thickness dv of its image is given by `(-v^2/u^2)du,` so the longitudinal magnification `(dv)/(du)=-m^2,` where m is the lateral magnification.

Explanation

To solve the problem, we need to prove that the thickness \( dv \) of the image is given by

\[ dv = -\frac{v^2}{u^2} du \]

and that the longitudinal magnification \( \frac{dv}{du} = -m^2 \), where \( m \) is the lateral magnification.

### Step-by-Step Solution:

1. **Understand the Lens Formula**:
The lens formula is given by

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.

2. **Differentiate the Lens Formula**:
Since \( f \) is a constant, differentiating both sides with respect to \( u \) gives:

\[ 0 = -\frac{d(v^{-1})}{du} + \frac{d(u^{-1})}{du} \]

This can be rewritten as:

\[ 0 = -\left(-\frac{1}{v^2} dv\right) + \left(-\frac{1}{u^2} du\right) \]

Simplifying this, we have:

\[ 0 = \frac{dv}{v^2} - \frac{du}{u^2} \]

3. **Rearranging the Equation**:
Rearranging gives:

\[ \frac{dv}{v^2} = \frac{du}{u^2} \]

Multiplying both sides by \( v^2 \) and \( u^2 \):

\[ dv = -\frac{v^2}{u^2} du \]

Here, we introduce a negative sign because the image distance \( v \) is considered negative in the context of real images formed by converging lenses.

4. **Longitudinal Magnification**:
The longitudinal magnification \( M_L \) is defined as:

\[ M_L = \frac{dv}{du} \]

Substituting our expression for \( dv \):

\[ M_L = \frac{-\frac{v^2}{u^2} du}{du} = -\frac{v^2}{u^2} \]

Since the lateral magnification \( m \) is defined as:

\[ m = \frac{h'}{h} = \frac{v}{u} \]

We can express \( M_L \) in terms of \( m \):

\[ M_L = -m^2 \]

5. **Conclusion**:
Thus, we have shown that:

\[ dv = -\frac{v^2}{u^2} du \]

and

\[ \frac{dv}{du} = -m^2 \]