Suggested languages for you:

Americas

Europe

Q101P

Expert-verifiedFound in: Page 1043

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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**

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

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

**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)

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 $xx\text{'}={f}^{2}$.

94% of StudySmarter users get better grades.

Sign up for free