# Diffraction Grating fringes

1. Sep 9, 2010

### pat666

1. The problem statement, all variables and given/known data
Monochromatic light with wavelength 580nm passes through a single slit 2.20um wide and 1.80m from a screen.
Find the distance between the first- and second-order dark fringes on the screen.

2. Relevant equations

3. The attempt at a solution
y=n$$\lambda$$L/2d
y1=0.237237m
y3=0.711818m
delta y = 0.4745m
This doesn't work-I think possible because there arn't two slits but I'm not sure of what to do if this is the case??
Thanks for any help.

2. Sep 9, 2010

### ehild

It is not a diffraction grating but a single slit. Look after the Physics of "single-slit diffraction" in your textbook.

ehild

3. Sep 9, 2010

### pat666

the topics after single slit diffraction are diffraction limit,x-ray diffraction and polarization???? I can't find anything else on single slits?

Thanks

4. Sep 9, 2010

5. Sep 9, 2010

### pat666

I read that and am still not totally sure.
I found theta min to be 15.3 degrees
then the path difference to be 2.9*10^-7 m. (assuming d is the width of the slit?)
I don't think this is right since I didn't use the length at all?

Thanks

6. Sep 9, 2010

### rl.bhat

Try

y1 = λL/d and

y2 = 2λL/d.

7. Sep 9, 2010

### pat666

I effectively did that originally - except I used the formula for dark fringes but apparently it makes no numeric difference which formula you use. anyway from what you said
y1=0.4745
y2=0.949
delta y=0.47

ehild did point me in the direction of some different formula that produce different results but as I said earlier I'm not sure if what I did from that is right???

Thanks

8. Sep 9, 2010

### ehild

Patt, you need to know the meaning or the symbols in the equation you use.
d is the width of the slit. λ is the wavelength of light. L is the distance between the slit and the screen. y is the position of the dark fringe with respect to the centre of the pattern. You get minimum intensity for a diffracted beam if it travels at an angle θ with respect to the normal of the slit:

d sinθ =n λ, n is an integer.

if θ is a small angle its tangent is the same as its sine and you can approximate sinθ by y/L.
But this is not valid if the angle is not small enough. So try to get y from the tangent of θ.

ehild

Last edited: Sep 9, 2010
9. Sep 9, 2010

### pat666

Thanks ehild, I thought I was using the symbols correctly. So now I know the minimum angle what do I do with it? I need to know the distance between the 1st and 2nd order dark fringes which I thought was all to do with y???? none of the equations in the wiki article have y though.

10. Sep 9, 2010

### ehild

You know the first angle already. Calculate the second one, too, and determine the y values from

y= L tan(θ).

ehild

11. Sep 9, 2010

### pat666

Thanks heaps ehild problem solved!
I have another question that rl.bhat has already guided me through, there is just one thing i'm not totally sure on and he hasn't posted back yet. Don't worry about it if you don't have time etc, as I said I've already had rl.bhat look.

Light consisting of two wavelengths which differ by 160 nm passes through a diffraction grating with 2.50 x 105 lines per metre. In the diffracted light, the third order of one wavelength coincides with the fourth of the other. What are the two wavelengths and at what angle of diffraction does this coincidence occur?

d*sin(θ) = mλ

at the point of coincidence θ is same for both the wavelengths. So

m1λ1 = m2λ2
3*λ1 = 4(λ1- 160)
$$\lambda$$1=6.4*10-7m
$$\lambda$$2=4.8*10-7m
I think those two should be correct???
d=1/2.5*10^5 = 4*10-6m This calculation rl.bhat didn't look at so I'm not sure on it?
assuming d is right 3*6.4*10-7=4*10-6 sin ($$\theta$$)
$$\theta$$=28.69o

That question is for an assignment which is why I wan't to be sure about my answer....

12. Sep 9, 2010

### rl.bhat

13. Sep 9, 2010

### ehild

It is correct.

ehild

14. Sep 10, 2010

### pat666

Thanks heaps!!!