Calculating Distance to Screen in Double Slit Diffraction Experiment

In summary, when light of wavelength 460nm falls on two slits spaced .3 mm apart, the required distance from the slit to a screen is 2.609 meters if the spacing between the first and second dark fringes is 4mm.
  • #1
solomar
6
0

Homework Statement


Light of wavelength 460nm falls on two slits spaced .3 mm apart. What is the required distance from the slit to a screen if the spacing between the first and second dark fringes is to be 4mm?

Homework Equations



dsin(theta)=(m+1/2)lambda
where d = .3mm
y=4mm
lambda = 460nm
sin(theta)=y/L

The Attempt at a Solution


My problem is I don't understand what to use for m...delta m between the two dark fringes? (3/2-1/2), what am I supposed to use for m?
Using delta m I get:
L=dy/deltam*lambda
L=2.609 meters
Basically all I would really like to know is what in the world do I use for m?
also I am unsure if I am approaching this correctly, do I need to somehow incorporate approximation with small angles? If so, how?
 
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  • #2
dsin(theta) = (m+1/2)(lambda)

So at m = 0, you get the first dark fringe. At m = 1, you get the second dark fringe.

I don't think "delta m" has any meaning in this context.
 
  • #3
quanticism said:
dsin(theta) = (m+1/2)(lambda)

So at m = 0, you get the first dark fringe. At m = 1, you get the second dark fringe.

I don't think "delta m" has any meaning in this context.

So then using those two values for m I would receive two different values for L. I don't see how having those two lengths solved would help me find the distance to the screen at all. I'm sorry I just don't understand how this leads to a solution...
 
  • #4
dsin(theta) = (m+1/2)(lambda)
d(y/L) = (m+1/2)(lambda)

Here, m gives you the "order" of the fringe, L is the distance from the slit and y is the vertical distance from the center of the screen to the dark fringe.

When you sub m=0 and all your known data in, you obtain an expression for y and L. Here, y is the distance to the first dark fringe.

Similarly, sub m=1 in and you obtain another expression involving y and L. This time, y is the distance to the second dark fringe.

The question gives you the separation between the first and second dark fringe so I'm sure you can work it out from there.
 
  • #5
There we go! For some reason it wasn't quite clicking with me...but I definitely got it now, thank you very much :)
 

1. What is double slit diffraction?

Double slit diffraction is a phenomenon that occurs when a beam of light or a wave passes through two parallel slits and produces an interference pattern on a screen or surface behind the slits.

2. How does double slit diffraction work?

Double slit diffraction works by causing the waves passing through the two slits to interfere with each other, creating an interference pattern on the surface behind the slits. This pattern is a result of constructive and destructive interference of the waves.

3. What factors affect the double slit diffraction pattern?

The factors that affect the double slit diffraction pattern include the distance between the two slits, the wavelength of the wave passing through the slits, and the size of the slits.

4. What are some real-life applications of double slit diffraction?

Double slit diffraction is commonly observed in the natural world, such as in the patterns created by ocean waves passing through a gap in a barrier. It is also used in various technologies, such as in diffraction gratings for optical instruments, and in the study of quantum mechanics.

5. How can double slit diffraction be used to determine the wavelength of a wave?

Double slit diffraction can be used to determine the wavelength of a wave by measuring the distance between the interference fringes on the screen or surface and using the known distance between the slits. This can be done by using the equation λ = ds/L, where λ is the wavelength, d is the distance between the slits, and L is the distance between the slits and the screen or surface.

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