Is My Thought Process Correct for Interference Patterns?

[echoi11]

1. The problem statement, all variables, and given/known data
The problem is attached.

Homework Equations

I do not know any relevant equations to this matter but I believe that a maximum has the length of λ and a minimum has the length of λ/2.

The Attempt at a Solution

I see one maximum and one minimum so I decided to add (λ) and (λ/2) together to get (3/2)*λ.
I was wondering if my thought process was correct or not.

 

[haruspex]

1. The problem statement, all variables, and given/known data
The problem is attached.

Homework Equations

I do not know any relevant equations to this matter but I believe that a maximum has the length of λ and a minimum has the length of λ/2.

The Attempt at a Solution

I see one maximum and one minimum so I decided to add (λ) and (λ/2) together to get (3/2)*λ.
I was wondering if my thought process was correct or not.

Looks like the right answer to me, but your reasoning seems a bit odd. Can you try to be a bit clearer? What do you mean by a maximum having a "length" of λ?

 

[echoi11]

Looks like the right answer to me, but your reasoning seems a bit odd. Can you try to be a bit clearer? What do you mean by a maximum having a "length" of λ?

I thought a bright maximum has a length of λ while a dark minimum has a length of λ/2. Is my justification of the answer incorrect?

 

[haruspex]

I thought a bright maximum has a length of λ

I still do not know what you mean by the length of a maximum or minimum. I would say each has a width and a distance from some reference point.

 

[echoi11]

I still do not know what you mean by the length of a maximum or minimum. I would say each has a width and a distance from some reference point.

Oh ok. How do you determine the answer from using width and distance from a reference point in the problem?

 

[haruspex]

Oh ok. How do you determine the answer from using width and distance from a reference point in the problem?

You don't. Those are not what matters here.
But.. just realized I misread the question. I now think your answer is wrong. Give me a minute or two...

 

[echoi11]

You don't. Those are not what matters here.
But.. just realized I misread the question. I now think your answer is wrong. Give me a minute or two...

Ok. no problem.

 

[haruspex]

Ok. no problem.

Ok, I'm back.
Draw a diagram, top view. On the left, you have the two slits, one above the other; on the right you have the screen with the maximally bright band, below that a dark band, below that another bright band, then the dark band of interest.
For reference, label the centres of these bands A, B, C, D respectively, and label the two slits U (upper) and L (lower).
What can you say about the distances UA, LA? What about the distances UC, LC?

 

[echoi11]

You don't. Those are not what matters here.
But.. just realized I misread the question. I now think your answer is wrong. Give me a minute or two...

Ok, I'm back.
Draw a diagram, top view. On the left, you have the two slits, one above the other; on the right you have the screen with the maximally bright band, below that a dark band, below that another bright band, then the dark band of interest.
For reference, label the centres of these bands A, B, C, D respectively, and label the two slits U (upper) and L (lower).
What can you say about the distances UA, LA? What about the distances UC, LC?

Ok I drew the two diagrams. From my diagram, it appears that the distances form triangles and therefore angles.

 

[echoi11]

Ok I drew the two diagrams. From my diagram, it appears that the distances form triangles and therefore angles.

Ok, I'm back.
Draw a diagram, top view. On the left, you have the two slits, one above the other; on the right you have the screen with the maximally bright band, below that a dark band, below that another bright band, then the dark band of interest.
For reference, label the centres of these bands A, B, C, D respectively, and label the two slits U (upper) and L (lower).
What can you say about the distances UA, LA? What about the distances UC, LC?

The distances are 2λ apart?

 

[haruspex]

The distances are 2λ apart?

Your use of "apart" makes me doubtful. Please post your diagram.

 

[echoi11]

Your use of "apart" makes me doubtful. Please post your diagram.

This is what I got.

 

[haruspex]

This is what I got.

Ok, but what in that are you saying are "2λ apart"? Can you write that statement as an equation using the lengths UA, LA etc. or mark it on the diagram?

 

[echoi11]

Ok, but what in that are you saying are "2λ apart"? Can you write that statement as an equation using the lengths UA, LA etc. or mark it on the diagram?

 

[haruspex]

I think you are saying your L+nλ are distances from the marked dot on the screen, right?
If so, no.
Let's back up a bit... where should the brightest spot on the screen be in relation to the slits?

 

[echoi11]

I think you are saying your L+nλ are distances from the marked dot on the screen, right?
If so, no.
Let's back up a bit... where should the brightest spot on the screen be in relation to the slits?

ohhh. The middle is where the brightest spot should be.

 

[haruspex]

ohhh. The middle is where the brightest spot should be.

Yes. And think more about the distances UA, LA etc. How are they related through λ?

 

[echoi11]

Yes. And think more about the distances UA, LA etc. How are they related through λ?

im not too sure, but I believe it is a λ to go up each time and a -λ to go down each time.

 

[haruspex]

im not too sure, but I believe it is a λ to go up each time and a -λ to go down each time.

You are still being too vague for me to know what you mean. Do please try to be clearer. What is going up and down by λ?

 

[echoi11]

You are still being too vague for me to know what you mean. Do please try to be clearer. What is going up and down by λ?

To get to more max or min, you either move 1λ or (λ/2).

 

[echoi11]

You are still being too vague for me to know what you mean. Do please try to be clearer. What is going up and down by λ?

Sorry I am confused as well.

 

[haruspex]

To get to more max or min, you either move 1λ or (λ/2).

Move what, where?
If you mean move λ across the screen, no.

 

[echoi11]

Move what, where?
If you mean move λ across the screen, no.

Im sorry. I just don't get it. Could you explain it to me?

 

[haruspex]

Im sorry. I just don't get it. Could you explain it to me?

What causes a dark band?

 

[echoi11]

What causes a dark band?

Loss of energy? I am not sure. I am sure the answer is (3/2)λ, but I am unsure how to get there.

 

[haruspex]

Loss of energy?.

Interference.
Where two waves of the same frequency are in phase, they add, producing a bright area. Where they are out of phase they cancel, making a dark area.
At the slits, the two are in phase. Why would they become out of phase at some places on the screen?

 

[echoi11]

Interference.
Where two waves of the same frequency are in phase, they add, producing a bright area. Where they are out of phase they cancel, making a dark area.
At the slits, the two are in phase. Why would they become out of phase at some places on the screen?[/QUOTe
Because of destructive interference.

 

[haruspex]

Because of destructve interference

No, that is the result. I am asking about the cause. If they start off in phase, why can they be out of phase at some point on the screen?
Draw a ray from each slit to some point on the screen. Show the rays as sine curves of the same wavelength. If each is at the top of a wave at its slit, why might they be at different parts of the wave at the point on the screen?
Diagram 87 at
http://farside.ph.utexas.edu/teaching/302l/lectures/node151.html might help.

 

[Dadface]

echoii
May I suggest that you forget the question for a while and go back and revise interference. The question seems to be at A level standard and if you do some searches you should find the information you need. Have you seen the CGP revision guides? I think they are very good and can help you a lot if you get hold of one. Try googling. Once you've got the basics of interference behind you come back to the question. Good luck.