Number of Visible Diffraction Orders on a Grating

  • Thread starter Thread starter Sabrina_18
  • Start date Start date
  • Tags Tags
    Diffraction Light
AI Thread Summary
The discussion focuses on calculating the number of visible diffraction orders for light of wavelength 500 nm incident on a grating with 600,000 lines per meter. The initial calculation suggested that only one order of diffraction would be visible, but further analysis revealed that the correct approach involves determining the grating spacing and applying the diffraction equation. After recalculating with the correct values, it was concluded that three orders of diffraction would be visible. The final consensus confirmed that the calculations were accurate and the problem was resolved. Understanding the relationship between wavelength, grating spacing, and diffraction angles is crucial for solving such problems.
Sabrina_18
Messages
8
Reaction score
0

Homework Statement


How many orders of diffraction will be visible if light of wavelength 500nm fals on a grating with 600 000 lines per metre?


Homework Equations



n = dsinѲ / λ


The Attempt at a Solution



n = 6X10⁵m ÷ 5 X 10ˉ⁷m = 1.2
So 1 order of diffraction will be visible is this right??

I don't have the angle so I could put it into the equation.
 
Last edited:
Physics news on Phys.org
Note that d is really (1/600,000) m, so that 600,000 lines would fit in 1 m.
 
Another way to pose this question is: how many maxima do you see between theta=0 and theta=90 degrees?

Consider that for a maximum to appear on the wall or the screen in front of the grating the angle theta can never be more than 90 degrees. (Recall that sin(theta) ~ (distance of nth maxima from the 0th maxima)/(distance of the grating from the screen).)
 
Thanks for your help. Here is my improved answer:
λ = 500nm = 5 X 10ˉ⁷m
d = 6 X 10⁵ per metre = 1 ÷ 6 X 10⁵ = 16.7 X 10ˉ⁷m
n = 16.7 X 10ˉ⁷ X sin 90 ÷ 5 X 10ˉ⁷ = 3.34
There will be 3 orders of diffraction visible

Is this correct? If not please tell me where I am going wrong

Thanks
 
Looks good. Problem solved :smile:
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top