Maximizing Single Slit Diffraction: Solving for the Largest Width Without Minima

[bcjochim07]

Homework Statement

Light from a helium neon laser with wavelength 633nm is incident on a single slit. What is the largest slit width for which there are no minima in the diffraction pattern?

Homework Equations

The Attempt at a Solution

For single slit diffraction, asin(theta)=p*lambda, where a= slit width, and p= 1,2,3...


So I know that when the slit width is smaller than the wavelength, no minima occur, but ?
what about when it is equal?

I'll try out some numbers: say the slit width is 6.33e-7 m, the same as the wavelength. Then I will find the angular position of the first minimum:

(6.33e-7)sin(theta)=(1)(6.33e-7)t
theta has to be equal to 90 degrees, which I don't think can be possible. So the wavelength has to be less than 633 nm.

Is this correct?

 

[bcjochim07]

Actually, I think that when the slit width is equal to the wavelength, then the light spreads to fill the region behind the opening. Therefore, I think that the largest the wavelength can be is 633 nm.

 

[baba89]

Yes, your approach is correct. In order to maximize the diffraction pattern without any minima, the slit width should be equal to or less than the wavelength of the incident light. In this case, the largest slit width without any minima would be 633 nm. If the slit width is larger than the wavelength, there will be minima in the diffraction pattern.