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.

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 × /2)×(10 / 3.50× )

Y = 2.61 m

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

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

0.19 m/s and 0.45 m/s