Optics Homework: Blue Light Diffracted by Grating

[poohead]

Homework Statement

Blue light of wavelength 470nm is diffracted by a grating ruled 5000 lines/cm.

a) compute the angular deviation of the second-order image
b) what is the highest order-image theoretically possible with this wavelength and grating

Homework Equations

stumped

Ephoton = hc/lambda
Ee= hf
Lambda = h/mv

The Attempt at a Solution

i figured i could have used a waves, sound & light formula like

lambda = dsin@/n

but there arent any nodes,

stumped, sorry
help is very much appreciated considering this question is really pissing me off

 

[poohead]

i beg for help for i really cannot interpret the question, i am a poohead

 

[Lambduh]

Let me see if i can steer you the right way.

Light coming from a diffraction grating can be described by:

a * sin(θm) = mλ

Where a = space between gratings

θm is the angle from the normal (coming strait out of the grating)

m is the mth diffraction order

λ is the wavelength of light.

Read these pages for a nice diagram that will help make more sense.

I'll check back later and see if you ended up in the right place:)

http://en.wikipedia.org/wiki/Diffraction_grating
also read up on diffraction
http://en.wikipedia.org/wiki/Diffraction

http://www.csupomona.edu/~hcmireles/PreLabWebpages/Hydrogen%20Emission%20Spectrum/Hydrogen.htm

 

[poohead]

sorry I am still not interpreting your answer, what would i do with 5000 lines/cm in inserting it into helping me solve for an answer

 

[poohead]

like how would i get a

 

[Lambduh]

So basically you want to know how many cm / line right? That is how far is it inbetween each line? Well you have cm / line right now and you want line / cm.

So all you have to do is invert :) it's the same thing as going from a wavelength to a frequency, you have a certain frequency of lines per distance. You want the actual distance between lines.

http://en.wikipedia.org/wiki/Frequency

F = 1 / t

Note* doesn't apply to an electromagnetic wave such as light.
If you were given the frequency of the blue light in the problem it would be determined by Frequency of Light = c / Wavelength. Where c is the speed of light in the medium. But don't worry about that at this time. That was just a side note.

Back to the problem. Inverting gives you the distance between your lines, a.

1 / (5000 line/cm) = .0002 cm/line

 

[poohead]

so i did what you said and the angle i get is 27, so where could i have gone wrong, what answer do you get?

 

[Lambduh]

So put in different values for m to solve for the different diffraction orders.

m = 1 will give one answer

m = 2 will give another (27)

m = 3 and so on

The question you're supposed to answer is

a) compute the angular deviation of the second-order image
b) what is the highest order-image theoretically possible with this wavelength and grating

A.) I've never used angular deviation before so I'm trying to figure out the definition. Then i'll see if i can help with that part:) Or maybe someone else can.

b.) So at what point will the diffraction orders disappear? Another way to think about it is how much angular swing do you have before there is physically something in the way of the next diffraction order?

 

[poohead]

well i think A) your trying to find θ for the 2nd order image right which would be m, and you know the wavelength 470nm and you know m=0.0002 cm. would this get me 28 degrees tho?

 

[poohead]

okay so i understand A) you computed 1/5000 line/cm wrong, you should've used 1.0 x10^-2/ 5000 line/cm then plug 2.0x10^-6 in for the equation thank you for your help though

 

[poohead]

but still i don't understand the concepts of B)

 

[Lambduh]

For a all you did was convert to meters. Mine is also correct, just still in units of cm.

For B: Look at this picture:

http://www.zemax.com/kb/content_images/Binary2/diffraction_1.gif

how many diffraction orders can you go out before it's behind the grating? That's what it wants to know. M = ?

 

[astrorob]

Consider this:

http://upload.wikimedia.org/wikiped...71115002925!Diffraction_grating_principle.png

Q: At what angle, \beta (equal to \vartheta in your equation), would the diffracted light be no longer incident on the screen? Consider the sin of this angle and you may be able to simplify your diffraction equation!