Calculating Wavelength and Central Maximum Width Using Single Slit Equations

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Homework Statement



A beam of light of unknown wavelength is incident on a screen with a single-slit. Observing the interference pattern on the projection screen that is 1.7 m away, the 4th minimum is 27 cm away from the central line. The width of the slit is 15 μm.
a) What is the wavelength (in nm) of the light used?

b) How wide(in cm)is the central maximum?

Homework Equations


Single slit equations


The Attempt at a Solution


I plugged everything in, stuck on the first step.

Xn= n*Lλ/w
= 4*1.7λ/1.5x10-5


Help?
 

Discussion

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Hey TSny, actually that is what I'm having trouble with. My brain is absolutely fried. It's probably so simple, but I can't seem to figure it out right now.
 
divide a by b to isolate x?
 
TSny said:
No. Consider a simple equation where you can easily see what the answer should be. Such as 4 = x/2

Well for that I would just multiply the 4 and 2 to find x...Sorry, I misread. I'm a little tired. I just don't know how to rearrange this equation to find λ.
 
TSny said:
Good. So to solve 4 = x/2 you multiply both sides by the denominator 2. Going back to your equation for λ, what would you multiply both sides by?

by 1.5x10-5?

So it should look like this then?

(4)(1.5x10-5) = (4)(1.7)(1.5x10-5)λ
 
Not quite. Let's work in symbols before plugging in numbers.

You have the equation X = n*L*λ/w. What does this become after you multiply both sides by w? (Note that I wrote your symbol Xn as just X because I think Xn looks like X multiplied by n which could get confusing.)
 
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TSny said:
Not quite. Let's work in symbols before plugging in numbers.

You have the equation X = n*L*λ/w. What does this become after you multiply both sides by w? (Note that I wrote your symbol Xn as just X because I think Xn looks like X multiplied by n which could get confusing.)

You would get X*w= n*L*λ right?
 
alittlelost said:
You would get X*w= n*L*λ right?

Yes. Good. Now you need to isolate λ. How are you going to (legally!) "get rid" of n and L on the right?
 
TSny said:
Yes. Good. Now you need to isolate λ. How are you going to (legally!) "get rid" of n and L on the right?

Could I move n and L to the other side of the equation?
 
Divide both sides by n and L which would cause them to disappear?
 
Since I'm bringing over n and L they would become negative on the left side correct?
 
alittlelost said:
Since I'm bringing over n and L they would become negative on the left side correct?

No, dividing both sides by n and L does not cause n and L to change sign.

If you wanted to solve a = x*b for x, you would divide both sides by b to get a/b = x. Note that b does not change sign.
 
after everything gets canceled out, should I only have X*w =λ? If so, I got 60000nm
 
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alittlelost said:
after everything gets canceled out, should I only have X*w =λ? If so, I got 60000nm

No that's not the correct formula for λ. Remember, if you solve a = x*b for x by dividing both sides by b, you get a/b = x. So, the b disappears on the right, but it now shows up on the left.

So, when you divide both sides of X*w = λ*L*n by L and n, L and n will then appear somewhere on the left side. On the left side you will have an expression that represents dividing X*w by L and n.
 
TSny said:
No that's not the correct formula for λ. Remember, if you solve a = x*b for x by dividing both sides by b, you get a/b = x. So, the b disappears on the right, but it now shows up on the left.

So, when you divide both sides of X*w = λ*L*n by L and n, L and n will then appear somewhere on the left side. On the left side you will have an expression that represents dividing X*w by L and n.

Ugh, so close to giving up on this one :p too tired.. I divided 4*1.5*10-5 by 4*1.7 and I got 8820nm.. is that correct?
 
alittlelost said:
Ugh, so close to giving up on this one :p too tired.. I divided 4*1.5*10-5 by 4*1.7 and I got 8820nm.. is that correct?

Ok, I think you are very close! So, after dividing both sides of X*w = λ*L*n by L and n you get X*w/(n*L) = λ, or

λ = X*w/(n*L)

Looks like you substituted 4 for X, as shown in red above. That's not correct. Do you know what the symbol X stands for? It's value is stated in the problem. Otherwise, your calculation looks correct.
 
According to my notes it's the 4th minimum that's 27cm from the central line.. I was supposed to plug in 0.27m wasn't I...
 
Okay, I got 596nm... did I finally get it?
 
TSny said:
Yes, that's what I got, too. Good work.

Awesome! Thanks a lot TSny, I appreciate the help.

for b) I didn't know the formula but after looking around online, I found the equation

a(central maximum)= 2Lλ/w

a=2(1.7)(5.96*10-7) / 1.5*10-5

I got 0.13m = 13cm... Did I do it right?BTW, this isn't related to what you were helping me with, but I have another question..

Consider the following polarization problem.The original light wave is completely un-polarized.The first screen is oriented vertically, while the last (third) screen is oriented horizontally. The middle screen is angled at 30° from the vertical. Assume the original intensity of the unpolarized light is 100%.

I know that after going through the first screen, 50% of the original intensity remains and I also figured out the second screen too. I missed the notes for this so I'm a little stumped with what I would do for the last screen. Would it's % of intensity be half of what came out of the second screen?
 
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alittlelost said:
for b) I didn't know the formula but after looking around online, I found the equation

a(central maximum)= 2Lλ/w

a=2(1.7)(5.96*10-7) / 1.5*10-5

I got 0.13m = 13cm... Did I do it right?

That looks good. The width of the central maximum is twice the distance on the screen between the center of the central maximum and the first minimum. So you can derive the equation you used from your equation X = nLλ/w with n = 1 and remembering to double the distance.

BTW, this isn't related to what you were helping me with, but I have another question..

Consider the following polarization problem.The original light wave is completely un-polarized.The first screen is oriented vertically, while the last (third) screen is oriented horizontally. The middle screen is angled at 30° from the vertical. Assume the original intensity of the unpolarized light is 100%.

I know that after going through the first screen, 50% of the original intensity remains and I also figured out the second screen too. I missed the notes for this so I'm a little stumped with what I would do for the last screen. Would it's % of intensity be half of what came out of the second screen?

Here's an old thread that is similar:
https://www.physicsforums.com/showthread.php?t=591677

If you still have questions about this, you should post the question as a new thread with an appropriate title.

Thanks.
 

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