# Homework Help: Bright Bands in Single Slit Diffraction

1. Nov 17, 2013

### Joshb60796

1. The problem statement, all variables and given/known data
A single slit forms a diffraction pattern with monochromatic light. The 4th minimum of the pattern occurs at an angle of 35 degrees from the central maximum. How many bright bands are on each side of the central maximum?

2. Relevant equations
Number of minimum brightness bands = ((Width of slit)(sinθ))/λ
Sinθ maxes out at sin90 = 1

3. The attempt at a solution
I drew a pictures of the central maximum and then four bright spots on each side. I numbered the minimums one through four and then drew a line from the slit to the central maximum and another line from the slit to the center of the fourth minimum and labeled that angle 35 degrees. Now the problem doesn't give the width of the slit and it doesn't give the wavelength but does state that it's monochromatic. From the above formula it seems that the number of minimums is inversely proportional to the the wavelength so as the wavelength gets closer and closer to red, the number of minimums drops and as it gets closer to violet, it increases. The distance from the slit to the screen where these minimums and maximums are viewed, isn't give. I thought I could have used that length and the angle to figure out the width of the slit. I feel like there are more unknowns that I can deal with. Someone please help.

Last edited: Nov 17, 2013
2. Nov 17, 2013

### Joshb60796

I want to add that I'm aware that theta in the above formula doesn't coincide with the angle as I have it labeled in my drawing but only a quarter of it. (or maybe 1/3.5 of it?) I was merely trying to figure out the distance in which I had 4 minimums and that would have been sin(theta). Then I could've used that distance and the number of minimums to figure out fringe lengths and then maybe used that to figure out wavelength?

3. Nov 17, 2013

### ehild

A diffracted ray deviates from the original direction of the light at angle θ. At those angles where the intensity is minimum, there is destructive interference between the diffracted rays emerging from the different places of the slit. You need only use the equation for the minima:

sinθ = ±mλ/d (d is the slit width and λ is the wavelength and m is the order of the minimum.)

At the fourth minimum (m=4), θ=35° . The angle of deviation can not be greater than 90°. What is m for a minimum at 90° angle?

There is a maximum between two subsequent minima. How many maxima can be present?

ehild

4. Nov 17, 2013

### Joshb60796

The m for theta equals 90, is d/lambda. which would represent the greatest width the slit could be at a particular wavelength? or the smallest wavelength given a particular slit width? I've been working on this problem for hours (watching youtube and reading the textbook and trying to find problems which helped connect the dots) I feel very burned out and mildly retarded. Please help me connect the dots you have presented me with.

5. Nov 17, 2013

### Joshb60796

"How many maxima can be present?"
If I have four minima and there is a maxima between two minima then there is a maxima on each side of the fourth minima which means four maxima on each side of the central bright spot for a total of 8 maxima not including the central maximum.

6. Nov 17, 2013

### Joshb60796

So is that the answer? That there are four bright bands on each side of the central maximum? Did I not need the angle 35? Is this much simpler than I originally considered?

7. Nov 17, 2013

### ehild

The slit is illuminated with light of certain wavelength. You get a a diffraction pattern, like in the picture

http://tsgphysics.mit.edu/pics/Q Diffraction/Q2-Single-Slit-Diffraction.jpg

The equation sin(θ)=±m λ/d refers to the dark spots.

sin(35°) corresponds to the diffraction order m=4. That means λ/d=sin(35°)/4.

What is the order of diffraction m if θ=90°? (off course, with the same light and slit) It is more than 4 four minima, at both sides of the central maximum. How many?

ehild

#### Attached Files:

• ###### singleslit.JPG
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8. Nov 17, 2013

### Joshb60796

If theta is 90 then sinθ=1 and you just have 1/m or 1/4? Maybe I should be a janitor instead of an engineer? haha....(sigh) ok so the ratio of: the length of the opposite side, from the central maximum to the fourth dark spot, and the path to the screen, becomes one, and 1 equals m times the ratio of wavelength to width of slit. So if the width of the slit and lamba don't change and just m changes then lamba/d could just be a constant, k. then I'd have 1=k(m) then 1 divided by k would give me m....so now I just gotta figure out what k is then I'll know m? haha I'm sooooo lost. It seems that as the angle went to 90 I'd have infinite minima and maxima but at some point they would smear and become infinite in length there by contradicting any ability to be infinite in number. Thank you for you help and the picture, would you mind still trying to help me? :)

Last edited: Nov 17, 2013
9. Nov 17, 2013

### Joshb60796

ok so I thought about it some more. when the angle is 90, the difference in the length of the path between the two waves would be the width of the slit. But then I'm troubled by the fact that there are an infinite number of wavelets and not just two. If the difference in the length of the paths was the slit width, then the amount that they are out of phase of each other would be the slit width and if it wasn't a half multiple of the wavelength then you wouldn't get destructive interference. Maybe I'm on to something? or maybe that was all just bogus thoughts?

10. Nov 18, 2013

### ehild

You may watch the video

Also see my figure. Every point of the wavefront of light falling on the screen is source of new waves according to Huyghens' Principle (Fig.a). The rays reaching a certain point of the screen make the average angle θ with the original direction of light. To every point on the slit emitting rays in the blue region, there is a point in the red region D/2 distance apart, so that the path difference is the same Δs between them (Fig. b). The screen is far away, so the rays arriving to a point from different places of the slit may be considered parallel (same θ)
Fig c, shows two rays travelling at angle θ, originating from the red and blue points, D/2 distance apart. The path difference between the rays is shown. It is D/2 sinθ . If it is equal to half the wavelength, λ/2 or odd number of λ/2, the waves interfere destructively, and you see a dark spot on the screen.

You have two equations in the problem. sin35°= 4 (λ/D), and 1=m (λ/D) . (λ/D) is just a number. Can you calculate it from the first equation and plug in to the second one to get the unknown, m?

ehild

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• ###### singleslittd.JPG
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11. Nov 18, 2013

### Joshb60796

Ok so the unknown m when the angle is 90 comes to 6.97, or roughly 7? and then that formula is for minima with a maxima on each side and if there are 7 minima then there are 8 maxima, right? and a total of 16?

12. Nov 18, 2013

### ehild

The question was: "How many bright bands are on each side of the central maximum?"
The central maximum does not count. There is a maximum between the first and second minima, and then between the second and third minima up to the sevenths minimum: There is nothing after the seventh minimum, as it corresponds to 90° angle - a light ray going totally sideways, not reaching the screen.

ehild

13. Nov 18, 2013

### Joshb60796

ahhh that makes sense, the minima occurs last because after that, there can be no maxima because the surface the slits are in would be in the way. There are 7 minima starting on each side of the central maxima which means there is one less maxima than there are minima for a total of 6 maxima on each side of the central maxima, right?

14. Nov 18, 2013

### ehild

Now, it is right.

ehild

15. Nov 18, 2013

### Joshb60796

Thank you so very much sir for helping me!

16. Nov 18, 2013

### Joshb60796

My book solution is 5. Where is my mistake? If I input 3 into the formula instead of 4 and run the whole process, I end up with 5.23 which is closer to the book answer. Is that formula for maxima and not minima and that's where the "1 off" error is?

17. Nov 18, 2013

### ehild

The problem is that you got 6.97 for the maximum m, which is less than 7. The sixth maximum is between the minima, but the 7th minimum is not present. As the maxima are about halfway between the minima, and the 7th minimum is "almost" there, the 6th maximum certainly would be present but it is not between two minima, so the book rejected it. It said that the last minimum that appeared is the 6th one.
If you see the diffraction pattern I sent, the maxima are broad, and you see only bright patches separated by dark spots. After the sixth dark spot there is an illuminated part with no end. The book does not count it "maximum".
So, when calculating the maximum m, and you get something not integer, round it always downward.

The formula refers to the minima. The exact position of the maxima is not easy to calculate.

ehild

Last edited: Nov 18, 2013
18. Nov 18, 2013

### Joshb60796

Ahhh... thank you for taking the time to help me understand :) I hope to one day understand this as well as you do. It's slow going to say the least.

19. Nov 18, 2013

### ehild

Sometimes the most difficult problem is to read the mind of the problem-writer:tongue2:

ehild

20. Nov 18, 2013

### ehild

Reading the problem again, it asks the number of bright bands. After the 6th maximum, there is an illuminated part with no end, so it is not a "band". Well, it is important to read the text carefully.

ehild