A Very, Very Interesting Problem

  • Thread starter Thread starter thedjoker
  • Start date Start date
  • Tags Tags
    Interesting
AI Thread Summary
A diffraction grating with 150 lines per mm is analyzed for its interference pattern when illuminated by light of wavelength 275 nm. The problem requires calculating the number of bright spots, or maxima, that appear on a screen positioned at a distance from the grating. The solution involves applying the diffraction formula, which relates the wavelength, grating spacing, and the angle of diffraction. Participants are encouraged to show their work and reasoning rather than simply providing the final answer. Engaging in the solution process is emphasized as a key aspect of the discussion.
thedjoker
Messages
1
Reaction score
0
A diffraction grating has 150 "lines" per mm etched upon it. When light of wavelength 275 nm is normally incident upon the grating, how many bright spots appear on a screen a short distance away?:smile:
 
Physics news on Phys.org
Please provide some attempt at the solution. We do not just provide answers here.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top