Finding the largest angle from the central maximum (single slit diffraction)

[HenryHH]

Homework Statement

A single slit, 1400 nm wide, forms a diffraction pattern when illuminated by monochromatic light of 490 nm wavelength. The largest angle from the central maximum at which the intensity is zero is closest to:

A) 44° B) 38° C) 35°


d = 1400 nm, lambda = 490 nm.

Homework Equations

sin(theta) = wavelength/d

The Attempt at a Solution

Plugging in variables, I have sin(theta) = 490/1400 = .35. To get theta, I multiplied .35 by sin inverse (sin^-1). I keep getting approximately 20 degrees, but the correct answer is A.) 44 degrees. I don't understand what I'm doing wrong, unless there's some extra step I should be doing... ?

 

[Steely Dan]

You have found the first minimum, or the one closest to the central maximum. You are being asked to find the minimum farthest from the central maximum.

 

[HenryHH]

Thanks. Will I need to use a different formula since I don't have any other variables?

 

[Steely Dan]

No, it is the same formula:

a \text{sin}(\theta_n) = n \lambda

It is just more generalized, along you to find the nth minimum.

 

[HenryHH]

Thanks. This is probably a dumb question, but how would I figure out what to plug-in for n? I now know that I am finding the minimum farthest from the central maximum, but I don't know what number would correspond with that, if that statement makes any sense.

 

[Steely Dan]

You will have to do this by trial and error. Figure out what the maximum meaningful diffraction angle is in a single slit experiment, and then find the biggest value of n that results in an angle less than that maximum angle.

 

[HenryHH]

I'm kind of confused... when you say trial and error, will I just be plugging in random numbers for n? Also, just to be sure, is a in the formula you posted the same as d (the slit width)?

My confusion with doing it the trial and error way deals with the fact that I won't know when I've gotten the correct answer. The correct answer is 44 degrees, but I don't know how they determined that 44 degrees was the largest angle versus any other angle...

 

[Steely Dan]

I'm kind of confused... when you say trial and error, will I just be plugging in random numbers for n? Also, just to be sure, is a in the formula you posted the same as d (the slit width)?

Yes, they are the same.

My confusion with doing it the trial and error way deals with the fact that I won't know when I've gotten the correct answer. The correct answer is 44 degrees, but I don't know how they determined that 44 degrees was the largest angle versus any other angle...

If you use n = 3, you get \text{sin}(\theta) = 1.05. Is there any angle \theta such that that can be true? The same argument holds for all n > 2.

 

[HenryHH]

So basically, I'm just plugging in 1,2,3... for n until I get the biggest number for theta that also happens to be an answer choice? In other words, this particular problem wouldn't be solvable if it wasn't a multiple-choice question?

 

[truesearch]

The general way to solve this is to recognise that the largest angle is 90 degrees so this largest value of Sin∅ is 1
This gives the max value of n to be a/λ which must be rounded down tot find the number of minima