Single Slit Question: Linear Distance & Wavelength Calculation

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    Single slit Slit
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The discussion revolves around calculating the width of a single slit based on the diffraction pattern produced by a 660 nm wavelength light source. The initial calculation for the slit width was incorrect due to misunderstanding the relationship between the linear distance of dark fringes and their respective positions. It was clarified that the distance between the first and second dark fringes should be halved to find the correct width of the slit. Additionally, the impact of slit width on fringe spacing and the size of the central bright fringe was explored, with a suggestion that increasing the slit width affects these parameters. Understanding the equations and their variables is crucial for accurate calculations in diffraction problems.
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Homework Statement


A single slit is illuminated with 660 nm and the resulting diffraction pattern is viewed on a screen 2.3 m away.
a. If the linear distance between the 1st and 2nd dark fringes of the pattern is 12 cm. What is the width of the slit?


Homework Equations


W(X/L) = m(lambda)


The Attempt at a Solution


W(.012m/.0023m = 1(660x 10^-9)
W = 1.27 x 10^-7

Someone else tells me that's wrong and that I have to divide the linear distance by two to get 6 cm instead of 12. Is this true? And if so why? I really don't understand why we would have to do that.
 
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Explain me the meaning of the letters in the equation you used, please.

ehild
 
What happens to the distance between fringes as the width of the single slit become larger? Does the distance between fringes increases? What happens to the centre bright fringe? Does it become larger with an increasing single slit width? My hunch is: Increasing the width increases the size of the bright central fringe, but applets on the net show otherwise. Can someone help?
 
It is a single slit, so a pair of rays which cancel each other come out from the same slit. For a dark fringe, each rays emerging from the slit has to get an other one to cancel with. So the angles at which a dark fringe occurs are those for which

W/2 sin(α)=(2m+1)λ /2 ---->W sin(α)=(2m+1)λ

If the screen is at distance L from the slit and the fringe is at distance X from the centre, tan(a)=X/L, but tan(α)=sin(α) for small angles, so

W *X/L=(2m+1)λ , (m=0,1,2...)

for the dark fringes.

Find X for the first and second dark fringes and see what happens if the width of the slit W increases.

ehild
 

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