How Do You Calculate the Width of a Slit in Single Slit Diffraction?

[drunkenfool]

Hey, I have a problem here involving single slit diffraction, and it's stumped me.


A single slit is illuminated with a 660-nm light, and the resulting diffraction pattern is viewed on a screen 2.3 m away. If the linear distance between the first and second dark fringes is 12 cm, what is the width of the slit?

I want to use this formula to start me off:

y= L tan (theta)

But what value of y should I use? It can't be 12 cm, because that's defining the distance between those two dark fringes, not their distance to the central bright fringe.

Any ideas?

 

[OlderDan]

Hey, I have a problem here involving single slit diffraction, and it's stumped me.


A single slit is illuminated with a 660-nm light, and the resulting diffraction pattern is viewed on a screen 2.3 m away. If the linear distance between the first and second dark fringes is 12 cm, what is the width of the slit?

I want to use this formula to start me off:

y= L tan (theta)

But what value of y should I use? It can't be 12 cm, because that's defining the distance between those two dark fringes, not their distance to the central bright fringe.

Any ideas?

The notation is a bit different, but take a look at the diagram and the equation here.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fraungeo.html#c2

You have two different y values corresponding to two different values of m, and all you know is the difference between the y values. You can use the equations to take the difference between the y values for the two m values and relate that difference to the distance to the screen and the slit width.

 

[thepatientmental]

Hi there,

I can definitely understand how this problem can be confusing. When working with single slit diffraction, it's important to remember that the distance between the central bright fringe and the first dark fringe is equal to the width of the slit (d). So in this case, we can use the value of 12 cm for y in the formula you mentioned, since that is the distance between the central bright fringe and the first dark fringe.

However, since we are given the distance between the first and second dark fringes (12 cm), we can also use that information to find the width of the slit. We can use the formula d*sin(theta) = m*lambda, where m is the order of the dark fringe (in this case, m=1) and lambda is the wavelength of the light (660 nm). So we have:

d*sin(theta) = 1*660 nm

We can rearrange this to solve for the width of the slit, d:

d = (1*660 nm)/sin(theta)

Now, we can use the value of theta that we found using the first formula (y=L*tan(theta)) to calculate the width of the slit:

d = (1*660 nm)/sin(theta) = (1*660 nm)/sin(0.058 radians) = 12.2 cm

Therefore, the width of the slit is 12.2 cm. I hope this helps and clarifies any confusion. Keep up the good work!