Number of bright fringes given by diffraction grating on a screen

  • #1
lorenz0
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Homework Statement:
A diffraction grating with ##N## lines per unit length is used initially with monochromatic light of wavelength ##\lambda##.
How many bright fringes are seen in total:
(1) On a screen of width ##W## distant ##L## from the grating;
(2) On a screen of infinite width?
(3) The monochromatic light is replaced with white light:
say what appears now on the screen of infinite width, indicating the difference(s) with respect to the previous setup with monochromatic light.
Relevant Equations:
##\Delta y=\frac{\lambda}{d}##
(1) In the book I am using the separation of bright fringes is indicated as being ##\Delta y=\frac{\lambda}{d}##, where ##d## is the separation of the slits so on a screen of width ##W## I would see ##\frac{W}{\frac{\lambda}{\frac{1}{N}}}## bright fringes. I don't see why the text of the exercise mentions the distance from the grating since it doesn't appear in the formula for the distance between bright fringes.

(2) Although the intensity of the maxima gradually diminishes with the distance from the center, it doesn't become ##0## so, in theory, there should be an infinite number of maxima on an infinite screen.

(3) If white light is used the diffraction maxima are separated into different wavelength components.

I am still trying to wrap my head around the concept of diffraction so I would appreciate if someone would give me some feedback on my solution (and on how to better understand diffraction) thanks.
 

Answers and Replies

  • #2
kuruman
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There is something not quite right for the book formula that you quoted. It predicts that if the distance to the screen is doubled, the separation between the lines will stay the same. Doesn't that bother you? The equation that I know is that the angular position ##\theta_m## of the ##m##th order maximum is such that $$\sin\theta_m=\frac{m\lambda}{d}.$$ Does this look familiar? Clearly, ##m## cannot become infinite because the sine on the left side cannot be greater than 1.

What is the context of the formula that you quoted in your book?
 
  • #3
lorenz0
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There is something not quite right for the book formula that you quoted. It predicts that if the distance to the screen is doubled, the separation between the lines will stay the same. Doesn't that bother you? The equation that I know is that the angular position ##\theta_m## of the ##m##th order maximum is such that $$\sin\theta_m=\frac{m\lambda}{d}.$$ Does this look familiar? Clearly, ##m## cannot become infinite because the sine on the left side cannot be greater than 1.

What is the context of the formula that you quoted in your book?
Ah, I see, so if I understand correctly since ##\sin(\theta_m)=m\frac{\lambda}{d}## for the infinite screen I have to impose that ##\sin(\theta_m)\leq 1\Leftrightarrow m\frac{\lambda}{d}\leq 1\Leftrightarrow m\leq \frac{d}{\lambda}## so I can get at most ##\frac{d}{\lambda}## maxima hence ##2\frac{d}{\lambda}## bright sposts.
For the screen of width ##W## I have that, relative to the center of the apparatus, it spans an angle ##\theta=2\arctan(\frac{W}{2L})## so I have to impose that ##\sin(\theta_m)\leq \sin(2\arctan(\frac{W}{2L}))\leftrightarrow m\frac{\lambda}{d}\leq\sin(2\arctan(\frac{W}{2L}))## so ##m\leq \frac{\sin(2\arctan(\frac{W}{2L}))d}{\lambda}## and we can get at most ##\frac{\sin(2\arctan(\frac{W}{2L}))d}{\lambda}## maxima hence at most ##2\frac{\sin(2\arctan(\frac{W}{2L}))d}{\lambda}## bright spots. Is this correct? Thanks.
 
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  • #4
kuruman
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Is this correct? Thanks.
It's a good first try but only the infinite screen part is correct. An infinite screen subtends a maximum angle of 90° on each side for a total of 180°. Because it's infinite, it can be any distance from the diffraction grating and the maximum angle will stay the same. Is this true for a finite screen? Make a drawing, place a finite screen of width ##W## at distance ##L## and then at distance ##2L## and see what the angle is in each case. Besides the argument of the arctangent must be dimensionless. The one in your expression is not.
 
  • #5
lorenz0
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It's a good first try but only the infinite screen part is correct. An infinite screen subtends a maximum angle of 90° on each side for a total of 180°. Because it's infinite, it can be any distance from the diffraction grating and the maximum angle will stay the same. Is this true for a finite screen? Make a drawing, place a finite screen of width ##W## at distance ##L## and then at distance ##2L## and see what the angle is in each case. Besides the argument of the arctangent must be dimensionless. The one in your expression is not.
Thanks, I forgot to divide by ##L##, it should be correct now.
 
  • #6
kuruman
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I don't see why the text of the exercise mentions the distance from the grating since it doesn't appear in the formula for the distance between bright fringes.
Actually, it would be more appropriate to provide a new post with the corrected equation instead of editing the old one. Now you see why the exercise mentions the distance. However, you still have not addressed the issue that the distance from the grating "doesn't appear in the formula for the distance between bright fringes." That is why I asked you about the context of that formula as presented in your book. I think this issue needs to be resolved.
 

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