Question

The formula $\frac{\mathbf{1}}{\mathit{p}}\mathbf{+}\frac{\mathbf{1}}{\mathit{i}}\mathbf{=}\frac{\mathbf{1}}{\mathit{f}}$ is called the Gaussian form of the thin-lens formula. Another form of this formula, the Newtonian form, is obtained by considering the distance $\mathit{x}$from the object to the first focal point and the distance$\mathit{x}\mathbf{\text{'}}$ from the second focal point to the image. Show that$\mathit{x}\mathit{x}\mathbf{\text{'}}\mathbf{=}{\mathit{f}}^{\mathbf{2}}$ is the Newtonian form of the thin-lens formula

Short answer

The Newtonian form of the thin-lens formula is$xx\text{'}=f^2$.

Step-by-step solution

Given data

• Distance from the object to the first focal point$=x$.
• Distance from the second focal point to the image role="math" localid="1663015192526" $=x\text{'}$.

Understanding the concept of thin-lens formula

In the given problem, we have to convert the Gaussian form of the thin-lens formula to the Newtonian form. So first we find the object's distance. The value of x is dependent on the position of the object. After that, we find the image distance where the value of x’ is dependent on the position of the image formed. We consider x and x’ as positive, i.e., the object is outside the focal point and the image is outside the focal point. Now by using the Gaussian formula, we solve for i and substituting the object distance and image distance, we prove the Newtonian form.

Formula:

The lens formula,

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

Calculation of the Newtonian thin-lens formula

Let, the object distance be $p=f+x$and the image distance be $i=f+x\text{'}$, where, $f$is the focal length, p is the object distance, and i is the image distance.

And x is the distance from the object to the first focal point, $x\text{'}$ is distance from the second focal point to the image.

Now, from equation (i), we get that the image distance as follows:

$$\begin{gathered} \frac{1}{i}=\frac{1}{f}-\frac{1}{p} \\ \frac{1}{i}=\frac{p-f}{pf} \end{gathered}$$

so,

$i=\frac{fp}{p-f}$ ...(1)

By substituting the value of $p=f+x$in the equation (1), we get the above equation as follows:

$$\begin{gathered} i=\frac{f\left(f+x\right)}{f+x-f} \\ =\frac{f\left(f+x\right)}{x} \end{gathered}$$

As,

$i=f+x\text{'}$

$$\begin{gathered} f+x\text{'}=\frac{f\left(f+x\right)}{x} \\ x\text{'}=\frac{f\left(f+x\right)}{x}-f \\ x\text{'}=\frac{f^2+fx-fx}{x} \\ x\text{'}=\frac{f^2}{x} \\ xx\text{'}=f^2 \end{gathered}$$

Hence, the thin-lens formula is $xx\text{'}=f^2$.