# Diffraction of light and orders equation?

1. ### Sabrina_18

8
Diffraction of light and orders equation??

1. The problem statement, all variables and given/known data

A particular grating has slits of width 600nm and a slit separation of 1800nm. Will there be any missing orders if this is used to observe a line spectrum consisting of 450nm, 600nm and 650nm?

2. Relevant equations

I'm not sure what formula to use but think it may be either

n=d sin theta/lamda or
a sin theta/n lamda

3. The attempt at a solution

As I'm not sure even what equation to use I'm not going to attemp it. Can someone just tell me what equation I need in order to answer the question. I'm sure I could figure it out if I just know what the equation is.

Thanks

2. ### Volcano

129
Suggestion, learn the conditions from your book about slit widths and wave lengths. Because 450 lower, 600 exact, 650 bigger than slit width(600). Sorry if misunderstood.

3. ### Sabrina_18

8
I've tried doing the question and heres an answer for the first wavelength.

n = d sin theta/lambda

1800 X 10^-9 X sin 90/450 X 10^-9
= 4

Someone please tell me if its right or wrong
Thanks

4. ### Volcano

129
Hi, i am not sure what are 450nm, 600nm and 650nm. Are those wavelangth, d or slit width? I suppose wavelength. If so, all of them very close to slit separation. To observing the diffraction, slit separation must be very bigger than wavelength. Else line spectrum can not be observed. Also L must be very bigger than slit separation too.

Last edited: May 16, 2008
5. ### alphysicist

2,247
Hi Sabrina_18,

That's the first step for the 450 nm wavelength: you have found that there are four possible orders (although one of these is at 90 degrees). You can find these four angles; it might be a good idea to go ahead and list them (use your equation with m=1,2,3,4).

Now, still for the 450 nm wavelength, you must take into account the effects of the width of the slits. You do this by considering single slit diffraction. With single slit diffraction you have a formula that gives you the minima if you know the wavelength and slit width. What are the angles of these minima?

Now the point of the question is that if one of the angles of these minima line up with one of the four angles you found earlier, then that order will be missing for the 450 nm wavelength spectrum. What do you get?

6. ### Volcano

129
sorry, what does "order" mean here? lines or data,... my language not good and also not familiar with some physics terms

7. ### Volcano

129
alphysicist, the slits are not exactly opposite side of light source. They are 900nm faraway from central line. After all, this is double slit diffraction. In this respect, single slit lines shifted both sides of central line on the screen doen't it?

8. ### alphysicist

2,247
Hi Volcano,

By order, I am using it this way: The first order line is that spectral line whose angle results from using m=1 in the condition $d \sin\theta=m\lambda$.

No, part of the assumption is that the single slit diffraction pattern is centered on the axis. If you were setting up this experiment in the real world, this would not be strange at all. One thing I think they don't emphasize very well is that the usual practice is to use a lens to focus the (plane wave) light on a screen (instead of having the screen be physically very far away). (That is, the lens is between the grating and the screen.)

But the lens will focus even that light that is off axis onto it's optical axis; the net result is that the center of both single slit diffraction patterns will be centered at the same place.

In many cases, the thinking behind the problems is the same whether the lens is there or not; here it can be a little confusing. However, I think that you would get the same result even if there were no lens. What matters it the angular separation of the lines; even if you had no lens, the centers of the lines would be a constant distance, and the angular separation would decrease as you made the screen get farther away.

9. ### rosstheboss23

37
Bragg's Law = n(wavelength)= 2dsin(theta)

10. ### alphysicist

2,247
Hi rosstheboss23,

I don't believe Bragg's law applies to this situation; I think this is a combination of single slit diffraction minima ($a\sin\theta=m\lambda$) and two-slit interference maxima ($d\sin\theta=n\lambda$).

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