Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is a question directly from a homework sheet, however I can't find any decent answers online so hopefully someone can help!

In single slit diffraction, why is the central maximum double the width of the other bands and the brightest?

I understand vaguely that it is something to do with Fresnel zones, but I'm quite confused. Instead of making another question, I think this question also fits in with the above:

Why does the intensity of light bands decrease away from the central maximum in single slit diffraction?

Thanks!

share|improve this question
add comment

3 Answers 3

up vote 2 down vote accepted

A more physical way of restating Ron's answer: in diffraction theory you are adding up Huygens wavelets over the aperture with appropriate phase factors. The central fringe is the only one where the entire aperture is adding in phase, i.e. constructively. The next fringe will have 2/3 adding constructively and 1/3 destructively, then after that 3/5 - 2/5, etc. The farther out you go in fringes, the more of the aperture is just cancelling itself.

share|improve this answer
add comment

I think what you want to know is why the maximum bright spot is the central belt and from it begins to decrease in intensity? the answer is: because the path difference between the center point and the crack and the other parts of the screen have a greater distance.

One can formulate a relationship between the separation of the slits, s, the wavelength λ, the distance from the slits to the screen D, and the width of the interference bands (the distance between successive bright fringes), x $$λ / s = x / D$$

This expression is only an approximation and formulation depends on specific conditions. It is possible however to calculate the wavelength of incident light from the ultimate relationship. If s and D are known and x is observed then λ can be calculated, which is of particular interest in measuring the wavelength corresponding to beams of electrons or other particles.

I hope you serve.

share|improve this answer
add comment

In single slit diffraction, assuming small diffraction angles, the intensity profile is the magnitude-squared of the Fourier transform of the function which is constant between -1 and 1 (up to units of length), and this Fourier transform is $\frac{\sin(x)}{x}$. This has a peak at zero, because of the falloff of $1/x$ but most importantly, and this can be seen qualitatively, the first zero of $\sin(x)$ is absent, the central maximum is twice as wide as all the others.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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