3
$\begingroup$

The equation for angular positions of dark fringes due to single slit diffraction is $a \sin(x) = n \lambda$ correct? My question is on that $n$. I read that it takes integer values which give the order of the dark fringe. But in deriving that equation they said the path difference is given by $0.5 a \sin(x)$ and that it should equal a whole number of HALF wavelengths for there to be destructive interference resulting in a dark fringe. But if the $n$ in the final equation takes an even value, e.g 2 or 4, won't the path difference be a whole number of EVEN wavelengths instead? Causing a bright fringe?

$\endgroup$
3
  • $\begingroup$ You can write different expressions for this, with the constraints on $n$ depending on how you write it. I usually use one in which $n \in \mathbb{Z}$. $\endgroup$ May 9, 2018 at 21:39
  • $\begingroup$ So in that form where "asin(x) = n*lambda" what are the constraints on n? Can it take an even value? $\endgroup$ May 9, 2018 at 21:52
  • $\begingroup$ You should edit that into your question, but you should also be able to answer it yourself. $\endgroup$ May 9, 2018 at 21:55

1 Answer 1

2
$\begingroup$

Finally found the answer. The argument of splitting the slit width into 2 halves is used when the "n" in "n * lambda" is odd. In this case, the rays from the top half interfere destructively with their counterparts in the bottom half with path differences of half a wavelength. When n is even, however, instead of splitting the slit width into 2 halves, we argue by splitting it into four regions instead. Rays in the first quarter now interfere destructively with their counterparts in the 2nd quarter, while those in the third interfere with those in the 4th. A dark fringe ergo arises. We can use these arguments for all integers both odd and even and therefore dark fringes occur at all integral values of "n", odd and even.

Found the solution here: http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/krzak/

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.