- #1
Chris18
- 12
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350nm falls on a single slit of width of 0.20mm. What is the angular width of the central diffraction peak?
I think that the width should be equal to 2Θ, where Θ=arcsin(m*λ/d)... m=1 and we have λ=350*10^-9 and d= 0.20*10^-6...but when i do the calculations I get 1.75 and the arcsin is a maths error...Could anyone help me please?
I think that the width should be equal to 2Θ, where Θ=arcsin(m*λ/d)... m=1 and we have λ=350*10^-9 and d= 0.20*10^-6...but when i do the calculations I get 1.75 and the arcsin is a maths error...Could anyone help me please?