Diffraction Grating fringes

In summary, the problem involved a monochromatic light with a wavelength of 580nm passing through a single slit 2.20um wide and 1.80m from a screen. The distance between the first- and second-order dark fringes on the screen was calculated using the formula y=nλL/2d. After some confusion about the correct formula to use, the correct values for y1, y2, and delta y were determined to be 0.4745m, 0.949m, and 0.47m respectively. The use of the tangent of the minimum angle θ allowed for a more accurate calculation of y. In addition, a separate question involving a diffraction grating with 2
  • #1
pat666
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0

Homework Statement


Monochromatic light with wavelength 580nm passes through a single slit 2.20um wide and 1.80m from a screen.
Find the distance between the first- and second-order dark fringes on the screen.

Homework Equations





The Attempt at a Solution


y=n[tex]\lambda[/tex]L/2d
y1=0.237237m
y3=0.711818m
delta y = 0.4745m
This doesn't work-I think possible because there arn't two slits but I'm not sure of what to do if this is the case??
Thanks for any help.
 
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  • #2
It is not a diffraction grating but a single slit. Look after the Physics of "single-slit diffraction" in your textbook.

ehild
 
  • #3
the topics after single slit diffraction are diffraction limit,x-ray diffraction and polarization? I can't find anything else on single slits?

Thanks
 
  • #5
I read that and am still not totally sure.
I found theta min to be 15.3 degrees
then the path difference to be 2.9*10^-7 m. (assuming d is the width of the slit?)
I don't think this is right since I didn't use the length at all?

Thanks
 
  • #6
Try

y1 = λL/d and

y2 = 2λL/d.
 
  • #7
I effectively did that originally - except I used the formula for dark fringes but apparently it makes no numeric difference which formula you use. anyway from what you said
y1=0.4745
y2=0.949
delta y=0.47

ehild did point me in the direction of some different formula that produce different results but as I said earlier I'm not sure if what I did from that is right?

Thanks
 
  • #8
Patt, you need to know the meaning or the symbols in the equation you use.
d is the width of the slit. λ is the wavelength of light. L is the distance between the slit and the screen. y is the position of the dark fringe with respect to the centre of the pattern. You get minimum intensity for a diffracted beam if it travels at an angle θ with respect to the normal of the slit:

d sinθ =n λ, n is an integer.

if θ is a small angle its tangent is the same as its sine and you can approximate sinθ by y/L.
But this is not valid if the angle is not small enough. So try to get y from the tangent of θ. ehild
 
Last edited:
  • #9
Thanks ehild, I thought I was using the symbols correctly. So now I know the minimum angle what do I do with it? I need to know the distance between the 1st and 2nd order dark fringes which I thought was all to do with y? none of the equations in the wiki article have y though.
 
  • #10
You know the first angle already. Calculate the second one, too, and determine the y values from

y= L tan(θ).

ehild
 
  • #11
Thanks heaps ehild problem solved!
I have another question that rl.bhat has already guided me through, there is just one thing I'm not totally sure on and he hasn't posted back yet. Don't worry about it if you don't have time etc, as I said I've already had rl.bhat look.

Light consisting of two wavelengths which differ by 160 nm passes through a diffraction grating with 2.50 x 105 lines per metre. In the diffracted light, the third order of one wavelength coincides with the fourth of the other. What are the two wavelengths and at what angle of diffraction does this coincidence occur?

answer

d*sin(θ) = mλ

at the point of coincidence θ is same for both the wavelengths. So

m1λ1 = m2λ2
3*λ1 = 4(λ1- 160)
[tex]\lambda[/tex]1=6.4*10-7m
[tex]\lambda[/tex]2=4.8*10-7m
I think those two should be correct?
d=1/2.5*10^5 = 4*10-6m This calculation rl.bhat didn't look at so I'm not sure on it?
assuming d is right 3*6.4*10-7=4*10-6 sin ([tex]\theta[/tex])
[tex]\theta[/tex]=28.69o

That question is for an assignment which is why I wan't to be sure about my answer...
Thanks for all the help you've already given me.
 
  • #12
Your calculations are correct.
 
  • #13
It is correct.

ehild
 
  • #14
Thanks heaps!
 

1. What is a diffraction grating?

A diffraction grating is a device made up of a large number of equidistant parallel slits or grooves, which cause light to diffract and produce a pattern of interference known as diffraction grating fringes. It is used to separate light into its component wavelengths.

2. How do diffraction grating fringes form?

Diffraction grating fringes form when light passes through a diffraction grating and the waves are diffracted by the slits or grooves. This results in constructive and destructive interference, producing a pattern of bright and dark fringes on a screen or detector.

3. What factors affect the spacing of diffraction grating fringes?

The spacing of diffraction grating fringes is affected by the distance between the slits or grooves on the grating, the wavelength of the incident light, and the angle of incidence at which the light strikes the grating. The spacing of the fringes can also be adjusted by changing the distance between the grating and the screen or detector.

4. What is the difference between a transmission and a reflection diffraction grating?

A transmission diffraction grating is made up of transparent material, while a reflection diffraction grating is made up of a reflective material. In a transmission grating, light passes through the slits, while in a reflection grating, light is reflected off the slits.

5. What are the applications of diffraction grating fringes?

Diffraction grating fringes have various applications in the fields of spectroscopy, astronomy, and optics. They are used in spectrophotometers to analyze the composition of substances, in telescopes to study the spectra of stars and galaxies, and in laser technology to produce specific wavelengths of light.

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