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!
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!
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.
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 ?
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*m_{max}.
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?
Right. Just think in terms of counting them up. 3 fringes on each side of the center plus one in the middle.
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?