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tony873004

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## 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?

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