wavelength λ=725nm
Distance from slit to screen L=1.25m
Width of the slit d=18.5μm
Find the angle for the first minimum:
dsinθ=λ
It is small, therefore:
sinθ=dλ≈Lx⇒x=dλL The width of the central maximum :
b)
Δ=2x=d2λL=9.8cm
a)
2θ=d2λrad=d2λ⋅180/π=4.5∘
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