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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

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

See the step by step solution

## Step 1: 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" $=x\text{'}$.

## Step 2: 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)

## Step 3: 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:

$\frac{1}{i}=\frac{1}{f}-\frac{1}{p}\phantom{\rule{0ex}{0ex}}\frac{1}{i}=\frac{p-f}{pf}\phantom{\rule{0ex}{0ex}}$

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:

$i=\frac{f\left(f+x\right)}{f+x-f}\phantom{\rule{0ex}{0ex}}=\frac{f\left(f+x\right)}{x}\phantom{\rule{0ex}{0ex}}$

As,

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

$f+x\text{'}=\frac{f\left(f+x\right)}{x}\phantom{\rule{0ex}{0ex}}x\text{'}=\frac{f\left(f+x\right)}{x}-f\phantom{\rule{0ex}{0ex}}x\text{'}=\frac{{f}^{2}+fx-fx}{x}\phantom{\rule{0ex}{0ex}}x\text{'}=\frac{{f}^{2}}{x}\phantom{\rule{0ex}{0ex}}xx\text{'}={f}^{2}\phantom{\rule{0ex}{0ex}}$

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