Single Slit Question: Linear Distance & Wavelength Calculation

[MetsFan09]

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.

 

[ehild]

Explain me the meaning of the letters in the equation you used, please.

ehild

 

[labview1958]

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?

 

[ehild]

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

 

[rwisz]

I would say that both approaches are valid and can lead to different interpretations of the problem. It ultimately depends on how the problem is defined and what assumptions are made.

The equation W(X/L) = m(lambda) represents the relationship between the width of the slit (W), the distance between the screen and the slit (L), the distance between the fringes (X), and the wavelength of the light (lambda). In this case, the problem is asking for the width of the slit, so we can rearrange the equation to solve for W.

Using the given values, we get W = (X/L)(lambda). Plugging in the values for X, L, and lambda, we get W = (0.12 m/2.3 m)(660 x 10^-9 m) = 2.8 x 10^-6 m = 2.8 micrometers.

However, if we divide the linear distance by two (as suggested by the other person), we would get a width of 1.4 micrometers. This approach assumes that the distance between the fringes represents the distance from the center of the slit to the first dark fringe, instead of the distance between the first and second dark fringes. This is a valid interpretation, but it may not be what the problem intended.

In conclusion, both approaches can be used to solve the problem, but it ultimately depends on how the problem is defined and what assumptions are made. As a scientist, it is important to carefully consider the problem and make sure that the approach used is consistent with the given information and assumptions.