Maxima of Diffraction Grating with 300 Lines/mm and Wavelength of 450 nm

  • Thread starter Thread starter Viviana
  • Start date Start date
  • Tags Tags
    Grating
AI Thread Summary
The discussion revolves around calculating the number of diffraction maxima for light with a wavelength of 450 nm incident on a grating with 300 lines/mm. The equation used is d sin(theta) = n(lambda), where d is the grating spacing and n is the order of diffraction. The user initially calculated n as 7, but the correct answer is 15, indicating a misunderstanding of the maximum order of diffraction. It is clarified that maxima occur on both sides of the central maximum and that m can take integer values, including negative numbers and zero. The conversation emphasizes the importance of considering all possible orders of diffraction to arrive at the correct answer.
Viviana
Messages
5
Reaction score
0

Homework Statement



Light of wavelength 450 nm is incident normally on a grating with 300 lines per millimeter. How many orders of diffraction maxima can be obtained?



Homework Equations


d sin(theta) = n(lambda)

The Attempt at a Solution


300 lines per millimeter is 3.33 x 10^-6. And 450 nm should be 4.5 x 10^ -7 m.
This is my equation:
3.33 x 10^-6 sin 90 = n (4.5 x 10^ -7)
I get n= 7.4 = 7
The correct answer is 15. But i just can't get this answer. =( Can someone please guide me?
 
Physics news on Phys.org
What is in the middle point, just opposite to the centre of the grating, minimum or maximum? And are maxima at one side only? What values can m have?

ehild
 
Does that mean there are two maxima for each order?
 
ehild said:
What is in the middle point, just opposite to the centre of the grating, minimum or maximum? And are maxima at one side only? What values can m have?

ehild

Maxima are at both sides. But i still don't know how to get values of m.
 
Can you please give some more hints
 
m is integer, including negative numbers and zero. So the farthest maxima are either at 90°or -90°.

ehild
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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...

Similar threads

Replies
2
Views
919
Replies
3
Views
781
Replies
2
Views
3K
Replies
3
Views
2K
Back
Top