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A 0
A 0.50-mm-wide slit is illuminated by light of wavelength 500 nm. What is the width of the central maximum on a screen 2.0 m behind the slit? Please answer in mm.
Expert Solution
given
d = 0.5 mm = 0.5*10^-3 m
lamda = 500 nm = 500*10^-9 m
R = 2.0 m
we know, widht of central maximum, w = 2*lamda*R/d
= 2*500*10^-9*2/(0.5*10^-3)
= 0.004 m
= 4.0 mm
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