A0.50 mm-wide slit is illuminated by light of wavelength 500 nm. what is the width of the central maximum on a screen 2.0m behind the slit?

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

where

is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

where we have

is the wavelength

D = 2.0 m is the distance of the screen

is the width of the slit

Substituting, we find

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.

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