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I'm having some trouble finding the answer to this homework question:

A wide beam of light from a laser with a wavelength of 450.0 nm is incident on a long narrow opening that is 0.300 mm wide. What will be the width, in mm, of the central maximum on a screen 2.00 m from the opening (i.e what will be the distance between the nodes adjacent to the central maximum?) (Use the small angle approximation.)

So far I've attempted two solution from the formulas that I know, but I'm not getting the right answer. Both of my solutions are shown below. Is there something that I'm missing, either conceptually or mathematically?


$ \lambda = 4.5^{-4} mm$

$ a= 0.300 mm$

$ L= 2000 mm$

Find, $w = ?$

Solution #1:

$a = \frac{\lambda L}{w}$

$w = \frac{\lambda L}{a}$

$w = \frac{4.5^{-4} \times 2000 mm }{0.300 mm}$

$w = 3.00 mm$

Solution #2:

$\theta = \frac{2.44 \lambda}{a}$

$\theta = 0.00366 rad$

$\theta = \frac{w}{L}$

$w = \theta \times L$

$w = 7.32$

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In your solution #2, the formula $\theta = 2.44\lambda/a$ would hold for a circular slit with radius $a$. But if the slit is long and narrow, it is a rectangle; and diffraction on a rectangle is not given by a Bessel function but by a sinc function, see, e.g., Wikipedia. The constant will therefore not be 2.44 but something else, connected to the first zero of the sinc function.

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I'm not understanding what a sinc function is... I think it may be beyond my level. The only other constant I can think of using is 1.22, but that didn't work either... is there any other suggestion you can give me? – Vanessa Dec 2 '12 at 1:24
@Vanessa The sinc function is explained shortly in the Wikipedia article; basically, $\mathrm{sinc}(x) = \sin(\pi x)/(\pi x)$. – Ondřej Černotík Dec 2 '12 at 12:48

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