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Light of wavelength 550 nm falls on a slit that is 3.50 x 10^-3 mm wide. How far from the central . maximum will the first diffraction maximum fringe be if the screen is 10.0 m away?

1 Answer

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λ
= 550 * 10^(-9)\] m. D = 3.50*10^-6 m.

d = 10 m. m = 1 2x = ?

Then you need to find x when n=1
or

sinθ=λ/w=524x10^-9/0.0033=1.59x10^-4
≈θ for θ<<1 , so θ=
1.59x10^-4 radians

Then you can find rhe distance of bright fringe from the center:
d= θD=1.59x10^-4x10m ≈ 1.6 mm
User Surjit Joshi
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