# Homework Help: Fringe spacing for a diffraction grading (wave optics)

1. Oct 27, 2008

### spaghed87

1. The problem statement, all variables and given/known data
A 550 lines/mm diffraction grating is illuminated by light of wavelength 550nm.
How many bright fringes are seen on a 3.1-m-wide screen located 2.1m behind the grating?

2. Relevant equations
$$\Delta$$Y = $$\lambda$$*L/d

where Y is equal to the fringe spacing... lamda is equal to the wavelength. d is equal to the separation... which is 0.001m/550lines in this case.

3. The attempt at a solution
I did $$\Delta$$Y=(550*10^-9m*2.1m)/(~1.81*10-6) = 0.63525m

So then the width of 3.1m divided by the fringe spacing ~1.81*10^-6 gives me 4.87 which is unitless because it is n and equal to the number of gaps... now there are two fringes per gap so shouldn't the answer be 6? well... the last one is cut off so i can't round... so that would be 5 right? but mastering physics (my online homework website) still says i'm wrong. any ideas?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 28, 2008

### Redbelly98

Staff Emeritus
I looks like a small-angle approximation for a trig term was used here. But the screen subtends a rather large angle.

Perhaps using the exact formula, with the trig term, would work better.

3. Oct 28, 2008

### spaghed87

"d*sin theta = mt
t = wavelength

m = d*sin theta/t
I don't have the angle so I just omitted it assuming it would be very small and I did...
m = d/t = 3.4 so 3 diffraction orders

Is this correct? "

I also got around ~3 diffraction orders for my problem which turned out to be the number of bright fringes. Is that really true that the diffraction order is equal to the number of bright fringes in this case? I thought you had to add one to acount for the m=0 central maximum or something...

4. Oct 29, 2008

### Redbelly98

Staff Emeritus
Yes, the number of diffraction orders is the same as the number of bright fringes.

The angle is not small and must be accounted for. And what you actually did was take sin(theta)=1, for which the angle would be 90 degrees.

And yes, you need to include the m=0 central maximum, as well as negative values of m.

For theta, use the angle at the edge of the screen.

5. Mar 24, 2011

I know this is an old post, but I wanted to tack onto it. I just did this problem for my physics course, except my problem had a wavelength of 530nm.

The number of bright fringes on the screen is calculated from the formula:

d*sinθ = mλ (distance between successive slits = sine of angular distance to bright fringe m times wavelength)

m = (d*sinθ)/λ

This is a maximum when θ = π/2 (sinθ = 1)

So, m = d/λ = 1.67e-6m/530e-9m = 3.15 (round to nearest integer) = 3.

This gives you the number of bright fringes above the central maximum. Double that to get the number of bright fringes on both sides of the central maximum and you now have 6 bright fringes. Add one to that (for the m = 0 [central] maximum) and you get an answer of 7 bright fringes.

My question, though, is how would an angle of π/2 even show anything on the viewing screen? Wouldn't an angle of π/2 be parallel to the screen, and thus the bright fringe at that angle would never be able to be physically located on the screen?

6. Mar 24, 2011

### Redbelly98

Staff Emeritus
True. θ must be strictly less than π/2, not ≤ π/2.

But this does not change the result, which really says that m<3.15. So m=3 is still the largest that m can be.

7. Mar 24, 2011