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For what wavelength of light in micrometers will the first-order dark fringe be observed at this same point on the screen?

To solve this, I must use the formula for destructive interference:

d*sin(theta) = (m + 1/2)*lambda, where m= 0 or -1 since it is the first dark fringe,

but I am stuck on using small-angle approximations where sin(theta) = tan(theta) to find d.

Could someone please explain this method to me clearly?

After finding d, what formula must I use for the 2nd beam's interference pattern to find the 2nd wavelength?

Thanks.