# 0.4 2d and 3d wavefronts  (Page 5/7)

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A slit with a width of 2511 nm has green light of wavelength 532 nm impinge on it. The diffracted light interferers on a surface, at what angle will the first minimum be?

1. We know that we are dealing with interference patterns from the diffraction of light passing through a slit. The slit has a width of 2511 nm which is $2511×{10}^{-9}\phantom{\rule{4pt}{0ex}}\mathrm{m}$ and we know that the wavelength of the light is 532 nm which is $532×{10}^{-9}\phantom{\rule{4pt}{0ex}}\mathrm{m}$ . We are looking to determine the angle to first minimum so we know that $m=1$ .

2. We know that there is a relationship between the slit width, wavelength and interference minimum angles:

$sin\theta =\frac{m\lambda }{a}$

We can use this relationship to find the angle to the minimum by substituting what we know and solving for the angle.

3. $\begin{array}{ccc}\hfill sin\theta & =& \frac{532×{10}^{-9}\mathrm{m}}{2511×{10}^{-9}\mathrm{m}}\hfill \\ \hfill sin\theta & =& \frac{532}{2511}\hfill \\ \hfill sin\theta & =& 0.211867782\hfill \\ \hfill \theta & =& {sin}^{-1}0.211867782\hfill \\ \hfill \theta & =& 12.{2}^{o}\hfill \end{array}$

The first minimum is at 12.2 degrees from the centre peak.

From the formula $sin\theta =\frac{m\lambda }{a}$ you can see that a smaller wavelength for the same slit results in a smaller angle to the interference minimum. This is something you just saw in the two worked examples. Do a sanity check, go back and see if the answer makes sense. Ask yourself which light had the longer wavelength, which light had the larger angle and what do you expect for longer wavelengths from the formula.

A slit has a width which is unknown and has green light of wavelength 532 nm impinge on it. The diffracted light interferers on a surface, and the first minimum is measure at an angle of 20.77 degrees?

1. We know that we are dealing with interference patterns from the diffraction of light passing through a slit. We know that the wavelength of the light is 532 nm which is $532×{10}^{-9}\phantom{\rule{4pt}{0ex}}\mathrm{m}$ . We know the angle to first minimum so we know that $m=1$ and $\theta =20.{77}^{\mathrm{o}}$ .

2. We know that there is a relationship between the slit width, wavelength and interference minimum angles:

$sin\theta =\frac{m\lambda }{a}$

We can use this relationship to find the width by substituting what we know and solving for the width.

3. $\begin{array}{ccc}\hfill sin\theta & =& \frac{532×{10}^{-9}\phantom{\rule{4pt}{0ex}}\mathrm{m}}{a}\hfill \\ \hfill sin20.{77}^{o}& =& \frac{532×{10}^{-9}}{a}\hfill \\ \hfill a& =& \frac{532×{10}^{-9}}{0.354666667}\hfill \\ \hfill a& =& 1500×{10}^{-9}\hfill \\ \hfill a& =& 1500\phantom{\rule{4pt}{0ex}}\mathrm{nm}\hfill \end{array}$

The slit width is 1500 nm.

run demo

## Shock waves and sonic booms

Now we know that the waves move away from the source at the speed of sound. What happens if the source moves at the same time as emitting sounds? Once a sound wave has been emitted it is no longer connected to the source so if the source moves it doesn't change the way the sound wave is propagating through the medium. This means a source can actually catch up to the sound waves it has emitted.

The speed of sound is very fast in air, about $340\phantom{\rule{4pt}{0ex}}{\mathrm{m}·\mathrm{s}}^{-1}$ , so if we want to talk about a source catching up to sound waves then the source has to be able to move very fast. A good source of sound waves to discuss is a jet aircraft. Fighter jets can move very fast and they are very noisy so they are a good source of sound for our discussion. Here are the speeds for a selection of aircraft that can fly faster than the speed of sound.

 Aircraft speed at altitude ( ${\mathrm{km}·\mathrm{h}}^{-1}$ ) speed at altitude ( ${\mathrm{m}·\mathrm{s}}^{-1}$ ) Concorde 2 330 647 Gripen 2 410 669 Mirage F1 2 573 715 Mig 27 1 885 524 F 15 2 660 739 F 16 2 414 671

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