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mooshasta
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I've encountered a standard problem about single slit diffraction, but there's a slight change to the problem and I don't know how it should be dealt with.
The question is:
A plane wave of 400-nm light is incident on a 25-µm slit in a screen, as shown in the figure below. At what incident angle will the first null of the diffraction pattern be on a line perpendicular to the screen?
http://www.tubaroo.com/ssd.PNG
I know the equation for the minima in a single slit diffraction problem is [itex]\sin \theta = m \frac{\lambda}{d}[/itex] where [itex]m[/itex] is 1,2,3,... and [itex]d[/itex] is the width of the slit. I want to say that the question is not as tricky as I'm making it out to be, and that I should just solve for [itex]\theta[/itex] using that equation. But if this were the case, it seems to me that since the light is coming from an angle, the effective width of the slit should be smaller, or something along those lines. I've scoured the internet for this type of situation, but every time I see the single slit diffraction problem, the angle of incidence of the light is 0 degrees.
Can anyone help me out?
The question is:
A plane wave of 400-nm light is incident on a 25-µm slit in a screen, as shown in the figure below. At what incident angle will the first null of the diffraction pattern be on a line perpendicular to the screen?
http://www.tubaroo.com/ssd.PNG
I know the equation for the minima in a single slit diffraction problem is [itex]\sin \theta = m \frac{\lambda}{d}[/itex] where [itex]m[/itex] is 1,2,3,... and [itex]d[/itex] is the width of the slit. I want to say that the question is not as tricky as I'm making it out to be, and that I should just solve for [itex]\theta[/itex] using that equation. But if this were the case, it seems to me that since the light is coming from an angle, the effective width of the slit should be smaller, or something along those lines. I've scoured the internet for this type of situation, but every time I see the single slit diffraction problem, the angle of incidence of the light is 0 degrees.
Can anyone help me out?
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