Max # of bright fringes
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 = 103 m; 1 nm =109 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! 
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Re: Max # of bright fringes
I tried the theta I solved for, 18.81, and 90. I wasn't sure about that part of the equation

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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?
Thanks for your help help! 
Re: Max # of bright fringes
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. 
Re: Max # of bright fringes
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 ?

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Great! thanks for your help.

Re: Max # of bright fringes
I have one more question:
To get the value of m=0 do you always add 1? 
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If you mean: To find the total number of fringes given a maximum value of m, then yes add 1 to 2*m_{max}. 
Re: Max # of bright fringes
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?

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Re: Max # of bright fringes
great, thanks Doc Al!

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I want to run this by you if you get this post... A 3600 line/cm diffraction grating produces a thirdorder bright fringe at a 31.0 degree angle. (Recall: 1 cm = 102 m; 1 mm = 103 m; 1 nm =109 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 thirdorder 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? 
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Re: Max # of bright fringes
Thanks for your help!

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