Double-slit interference

  • Thread starter tony873004
  • Start date
  • #1
tony873004
Science Advisor
Gold Member
1,751
143

Homework Statement


I did part A & B and got the right answer. Question C is "Is the separation the same for all maxima?"


Homework Equations


[tex]\sin \theta = m\lambda /d[/tex]


The Attempt at a Solution


In the homework problem, sin theta can be computed with y/D, where is the separation of the maxima, and D is the distance from the slits to the screen. So [tex]\frac{y}{D} = \frac{{m\lambda }}{d}\,\,\,\, \Rightarrow \,\,\,\,y = \frac{{Dm\lambda }}{d}[/tex].

But in the book's example, they don't use this formula. They use the small-angle approximation, and state in the example: "This result is valid near the center of the screen, where the small-angle approximation is valid." This is consistent with the answer in the back of the book.

But without using the small-angle approximation, I get a separation of 1.2 mm between the m=1 and m=2 fringes. But I also get a separation of 1.2 mm between the m=1,000,000 and m=1,000,001 fringes, where theta would be huge.

So why is the book concluding that this is only valid for small angles?

** Edit, Tex is seriously malfunctioning. In the preview window, it's as if it mixed up my tex with somewone else's work, I was seeing stuff about kinetic energy and electron Volts. And in the final window, it calls it invalid.

To sum up my question without Tex:

y = Dm lambda / d


(2*2*600e-9/1e-3) - (2*1*600e-9/1e-3) = 0.0012

and

(2*1000001*600e-9/1e-3) - (2*1000000*600e-9/1e-3) = 0.0012

Without using the small angle approximation,
When I choose 1 & 2 for m, I get 0.0012
When I choose 1000000 & 1000001 for m, I get 0.0012

So why does the book say the maxima separation are only constant near the center of the screen?
 
Last edited:

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi tony873004,

Homework Statement


I did part A & B and got the right answer. Question C is "Is the separation the same for all maxima?"


Homework Equations


[tex]\sin \theta = m\lambda /d[/tex]


The Attempt at a Solution


In the homework problem, sin theta can be computed with y/D, where is the separation of the maxima, and D is the distance from the slits to the screen. So [tex]\frac{y}{D} = \frac{{m\lambda }}{d}\,\,\,\, \Rightarrow \,\,\,\,y = \frac{{Dm\lambda }}{d}[/tex].

But in the book's example, they don't use this formula. They use the small-angle approximation, and state in the example: "This result is valid near the center of the screen, where the small-angle approximation is valid." This is consistent with the answer in the back of the book.

But without using the small-angle approximation, I get a separation of 1.2 mm between the m=1 and m=2 fringes. But I also get a separation of 1.2 mm between the m=1,000,000 and m=1,000,001 fringes, where theta would be huge.

So why is the book concluding that this is only valid for small angles?

** Edit, Tex is seriously malfunctioning. In the preview window, it's as if it mixed up my tex with somewone else's work, I was seeing stuff about kinetic energy and electron Volts. And in the final window, it calls it invalid.

To sum up my question without Tex:

y = Dm lambda / d


(2*2*600e-9/1e-3) - (2*1*600e-9/1e-3) = 0.0012

and

(2*1000001*600e-9/1e-3) - (2*1000000*600e-9/1e-3) = 0.0012

Without using the small angle approximation,
When I choose 1 & 2 for m, I get 0.0012
When I choose 1000000 & 1000001 for m, I get 0.0012

So why does the book say the maxima separation are only constant near the center of the screen?

I don't believe these calculations are valid; you are using the formula that assumes the small angle approximation, but the angles associated with large m values are not small. (In fact, there is no line for m=1000000.)

To test these, go back to the original formulas:

[tex]
d \sin(\theta) = m \lambda
[/tex]

[tex]
\tan(\theta) = y / L
[/tex]


Then test these for various values of m and m+1 by plugging in an m value in the first equation to find theta, and using that in the second equation to find y.

For m=1000 and 1001, I'm finding the change in the y is about 0.002m; for 1600 and 1601 the change is about 0.06m, and for 1665 and 1666 the change is about 26m.

(m=1666 is the highest value of m that is allowed; any higher would require the sine function to be greater than 1.)

(I'm assuming that lambda is 600nm, slit distance d is 1mm, and distance to screen L is 2m.)
 
  • #3
tony873004
Science Advisor
Gold Member
1,751
143
Thanks!

I think I see my error. Sin does not equal y/D. Tan = y/D, but for small angles, sin=tan, so you're right, by considering y/D = sin, I am using the small angle apprx.
 

Related Threads on Double-slit interference

  • Last Post
Replies
5
Views
854
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
833
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
7K
Top