Light of wavelength 610 nm falls on a slit that is 3.50×10^−3 mm wide. How far the first bright diffraction fringe is from the strong central maximum if the screen is 10.0 m away.

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-09-03T06:37:22+02:00

Answer:

first bright diffraction fringe is from the strong central maximum is 2.6 m

Explanation:

wavelength λ = 610 nm

wide d = 3.50 ×10^−3 mm

screen distance D = 10.0 m

to find out

How far the first bright diffraction fringe

solution

we know here screen 10.0 m away

so for path difference is express as for nth bright fringe is

d × sinθ = ( 2n + 1 ) ( λ /2 ) ...............1

we consider here sinθ = θ

so for distance between central maximum and 1st bright fringe is express as given below

Y = ( 2n + 1 ) ( λ /2 ) ( D/d) .................2

so for n = 1 put all these all we get

Y = ( 2(1) + 1 ) ( 610 ×[tex]10^{-9}[/tex] /2)×(10 / 3.50×[tex]10^{-6}[/tex] )

Y = 2.61 m

so here first bright diffraction fringe is from the strong central maximum is 2.6 m