# Max # of bright fringes

1. Apr 30, 2011

### kevnm67

1. The problem statement, all variables and given/known data
A diffraction grating with 620 lines per mm is illuminated with light of wavelength 520 nm. A very wide viewing screen is 2.0 m behind the grating. (Note: 1 mm = 10-3 m; 1 nm =10-9 m)
a) What is the distance between the two m=1 bright fringes? (Express your answer in meters)
b) How many bright fringes can be seen on the screen

2. Relevant equations
m = d(sinθ)/λ
y=Ltanθ

3. The attempt at a solution

A)
d(sinθ)=mλ
=sin-1 (1x(520x10-9)/(1/620 x10-3))
=18.81

y=Ltanθ
tan(18.81)x2
=.34
.34 x 2 x 2 = 1.36 meters

B) ?
m = d(sinθ)/λ
but when I use this equation I do not get the correct answer. I am at a loss if someone can walk me through this part.

Thanks!

2. Apr 30, 2011

### Staff: Mentor

What did you use for θ?

3. Apr 30, 2011

### kevnm67

I tried the theta I solved for, 18.81, and 90. I wasn't sure about that part of the equation

4. Apr 30, 2011

### Staff: Mentor

Trying θ = 18.81 will just give you back m = 1, which you already know. θ = 90 is what you want.

5. Apr 30, 2011

### kevnm67

Alright. When I do that I get 3.1, which is 3.1 x 2 = 6.2 but the answer is 7. What I am doing wrong?

6. Apr 30, 2011

### Staff: Mentor

Getting m = 3.1 tells you that the highest order of fringe that you could see would be m = 3. (You won't see m = 4.)

Don't forget the center fringe, where m = 0.

7. Apr 30, 2011

### kevnm67

Ok, that's what I was thinking...so it's 3 above the central max and 3 below plus 1 b/c you include m=0 ?

8. Apr 30, 2011

### Staff: Mentor

You got it.

9. Apr 30, 2011

### kevnm67

10. Apr 30, 2011

### kevnm67

I have one more question:

To get the value of m=0 do you always add 1?

11. Apr 30, 2011

### Staff: Mentor

I don't understand the question. m=0 is the central maximum. (Add 1 to what?)

If you mean: To find the total number of fringes given a maximum value of m, then yes add 1 to 2*mmax.

12. Apr 30, 2011

### kevnm67

after solving for the no. of fringes per side of the central max you get 6.2 but must account for m=0, which is why 1 was added to 6.2 and you get 7?

13. Apr 30, 2011

### Staff: Mentor

Right. Just think in terms of counting them up. 3 fringes on each side of the center plus one in the middle.

14. Apr 30, 2011

### kevnm67

great, thanks Doc Al!

15. May 2, 2011

### kevnm67

Doc Al,

I want to run this by you if you get this post...
A 3600 line/cm diffraction grating produces a third-order bright fringe at a 31.0 degree angle.
(Recall: 1 cm = 10-2 m; 1 mm = 10-3 m; 1 nm =10-9 m)
A) What wavelength (in nm) of light is being used?
a. 477 nm b. 1430 nm c. 233 nm
B) What is the distance (in meters) between this third-order bright fringe and the central maximum on the screen 2.0 meters away?
a. 2.4 m b. 1.2 m c. 2.2 m
C) How many total bright fringes can be seen on the screen?
a. 3 b. 5 c. 11

For C) would you do the following:
(1/3600) x 10-2 sin(90)/(477 x 10-9)
= 5.8
so 5 fringes up and 5 fringes down for a total of 10 fringes but to account for the central max you add 1 to get 11, choice C?
And is 90 degrees what you use in every scenario?

16. May 2, 2011

### Staff: Mentor

Yes to all of that. Good!

17. May 2, 2011