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a_sh-v [17]
4 years ago
12

Coherent light with wavelength 490 nm passes through two very narrow slits that are separated by 0.200 mm, and the interference

pattern is observed on a screen 4.00 m from the slits. (a)what is the width (in mm) of the central interference maximum? (b) what is the width of the first-order bright fringe?
Physics
1 answer:
Sati [7]4 years ago
6 0
Double slit diffraction has a convenient formula for this of
s=wD/d
Where s is the distance between the fringes, w is the wavelength (should be lambda but this website doesnt seem to like it in the formula construction box), D is the distance slit to screen, and d is the distance between the centres of the two slits.

Substituting this all in:
s= \frac{(490E-9)*4}{2E-4} = 0.0098m
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