# 27.4 Multiple slit diffraction  (Page 2/6)

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$d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda },\phantom{\rule{0.25em}{0ex}}\text{for}\phantom{\rule{0.25em}{0ex}}m=\text{0,}\phantom{\rule{0.25em}{0ex}}\text{1,}\phantom{\rule{0.25em}{0ex}}\text{–1,}\phantom{\rule{0.25em}{0ex}}\text{2,}\phantom{\rule{0.25em}{0ex}}\text{–2,}\dots \text{(constructive),}$

where $d$ is the distance between slits in the grating, $\lambda$ is the wavelength of light, and $m$ is the order of the maximum. Note that this is exactly the same equation as for double slits separated by $d$ . However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles.

Where are diffraction gratings used? Diffraction gratings are key components of monochromators used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength emitted by molecules in diseased cells in a biopsy sample or to help excite strategic molecules in the sample with a selected frequency of light. Another vital use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings are available for selecting specific wavelengths for such use.

## Take-home experiment: rainbows on a cd

The spacing $d$ of the grooves in a CD or DVD can be well determined by using a laser and the equation $d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda },\phantom{\rule{0.25em}{0ex}}\text{for}\phantom{\rule{0.25em}{0ex}}m=\text{0,}\phantom{\rule{0.25em}{0ex}}\text{1,}\phantom{\rule{0.25em}{0ex}}\text{–1,}\phantom{\rule{0.25em}{0ex}}\text{2,}\phantom{\rule{0.25em}{0ex}}\text{–2,}\dots$ . However, we can still make a good estimate of this spacing by using white light and the rainbow of colors that comes from the interference. Reflect sunlight from a CD onto a wall and use your best judgment of the location of a strongly diffracted color to find the separation $d$ .

## Calculating typical diffraction grating effects

Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it to a screen 2.00 m away. (a) Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760 nm). (b) What is the distance between the ends of the rainbow of visible light produced on the screen for first-order interference? (See [link] .)

Strategy

The angles can be found using the equation

$d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=\text{0,}\phantom{\rule{0.25em}{0ex}}\text{1,}\phantom{\rule{0.25em}{0ex}}\text{–1,}\phantom{\rule{0.25em}{0ex}}\text{2,}\phantom{\rule{0.25em}{0ex}}\text{–2,}\phantom{\rule{0.25em}{0ex}}\dots \right)$

once a value for the slit spacing $d$ has been determined. Since there are 10,000 lines per centimeter, each line is separated by $1/10,000$ of a centimeter. Once the angles are found, the distances along the screen can be found using simple trigonometry.

Solution for (a)

The distance between slits is $d=\left(1 cm\right)/\text{10},\text{000}=1\text{.}\text{00}×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{cm}$ or $1\text{.}\text{00}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{m}$ . Let us call the two angles ${\theta }_{\text{V}}$ for violet (380 nm) and ${\theta }_{\text{R}}$ for red (760 nm). Solving the equation $d\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{V}}=\mathrm{m\lambda }$ for $\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{V}}$ ,

$\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{V}}=\frac{{\mathrm{m\lambda }}_{\text{V}}}{d}\text{,}$

where $m=1$ for first order and ${\lambda }_{\text{V}}=\text{380}\phantom{\rule{0.25em}{0ex}}\text{nm}=3\text{.}\text{80}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{m}$ . Substituting these values gives

Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
I think you shorten am not sure
Uche
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
discuss under damped
resistance of thermometer in relation to temperature
how
Bernard
that resistance is not measured yet, it may be probably in the next generation of scientists
Paul
Is fundamental quantities under physical quantities?
please I didn't not understand the concept of the physical therapy
physiotherapy - it's a practice of exercising for healthy living.
Paul
what chapter is this?
Anderson
this is not in this book, it's from other experiences.
Paul
Sure
What is Boyce law
how to convert meter per second to kilometers per hour
Divide with 3.6
Mateo
multiply by (km/1000m) x (3600 s/h) -> 3.6
2 how heat loss is prevented in a vacuum flask
what is science
Helen
logical reasoning for a particular phenomenon.
Ajay
I don't know anything about it 😔. I'm sorry, please forgive 😔
due to non in contact mean no conduction and no convection bec of non conducting base and walls and also their is a grape between the layer like to take the example of thermo flask
Abdul
dimensions v²=u²+2at
what if time is not given in finding the average velocity?
the magnetic circuit of a certain of the flux paths in each of the long and short sides being 25cm and 20cm reprectielectrove. there is an air gap of 2mm long in one the long sides if a flux density of 0.8weber/m is to produce in the magnet of 1500 turns..
How do you calculate precision
what module is that?
Fillemon
Chemisty 1A?
Fillemon
No it has something to do with measurements bro... What we did today in class
Sacky
Tah bra honestly I didn't understand a thing in that class..when re your Tutorials?
Fillemon
Friday bro... But the topics we did are in this app... Just try to master them quickly before the test dates... Are you done with the Maths sheet
Sacky
I eat ass
Anderson
I'll work on the maths sheet tomorrow bra @Sacky Malyenge but I'll try mastering them
Fillemon
I'll eat your mom's ass with a side of tendies
Anderson
@Fillemon Nanwaapo
Anderson
lol, hush
Emi
There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electricity?
Because there's an equal number of negative and positive charges... objects are neutral in nature
NELSON
when a ball rolls on a smooth level ground,the motion of its centre is?
what is electro magnetic field?
Mary
electromagnetic field is a special type of field been produced by electric charges..!!! like the word electro from Electricity and the word magnetic from Magnetism.. so it is more of a join field..!!!
NELSON
Electromagnetic field is caused by moving electric charge
when a ball rolls on a smooth level ground,the motion of its centre is?
Mumeh