## Related questions with answers

Rayleigh’s criterion is used to determine when two objects are barely resolved by a lens of diameter d. The angular separation must be greater than $\theta_{R}$ where $\theta_{R} =1.22 \lambda /d$ . In order to resolve two objects 4000 nm apart at a distance of 20 cm with a lens of diameter 5 cm, what energy (a) photons or (b) electrons should be used? Is this consistent with the uncertainty principle?

Solution

Verified$\textbf{Given:}$

Diameter of the lens (d) = $5$ cm

The distance at which the two objects is resolved ($x$) = $20$ cm

The resulting distance in separation ($y$) = $4000$ nm

$\theta_R=1.22\dfrac{\lambda}{d}$

Since the minimum angle, at which the two objects are resolved, is given as follows:

$\begin{equation*} \tan \theta _R =\dfrac{y}{x} \;\;\;\;\;\text{and}\;\;\;\;\;\theta _R \simeq \tan \theta _R \;\;\;\;\;\;\textbf{for small\;\; $\theta$} \end{equation*}$

therefore,

$\begin{align*} \theta_R&=\dfrac{y}{x}\\\\ &=\dfrac{4000\times {10}^{-9}\;\cancel{\text{m}}}{20\times {10}^{-2}\;\cancel{m}}\\\\ &=2\times {10}^{-5} \end{align*}$

and since

$\begin{align*} \theta_R=1.22\dfrac{\lambda}{d} \end{align*}$

therefore, the minimum wavelength for a given particle is given as follows:

$\begin{align*} \lambda_{\text{min}} &=\dfrac{d\cdot \theta_R}{1.22}\\\\ &=\dfrac{5 \times {10}^{-2}\;\text{m}\times 2 \times {10}^{-5}}{1.22}\\\\ &\simeq 8.2 \times {10}^{-7}\;\text{m} \end{align*}$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651Randall D. Knight#### Modern Physics for Scientists and Engineers

4th Edition•ISBN: 9781133103721Andrew Rex, Stephen T. Thornton#### Physics for Scientists and Engineers

9th Edition•ISBN: 9781133947271 (9 more)John W. Jewett, Raymond A. Serway## More related questions

1/4

1/7