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expeople1 [14]
3 years ago
5

Coherent light with wavelength 610 nm passes through two very narrow slits, and theinterference pattern is observed on a screen

a distance of 3.00{\rm m} from the slits. The first-order bright fringe is adistance of 4.84 {\rm mm} from the center of the central bright fringe.For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?Express your answer in micrometers(not in nanometers).
Physics
1 answer:
andriy [413]3 years ago
7 0

Answer:

= 1220 nm

= 1.22 μm

Explanation:

given data:

wavelength \lambda = 610 nm = 610\times  10 ^{-9} m

distance of screen from slits D = 3 m

1st order bright fringe is 4.84 mm

condition for 1 st bright is

d sin \theta =\lambda     ---( 1)

andtan \theta = \frac{y}{ D}

\theta = tan^{-1}\frac{(y }{D})

= 0.0924 degrees

plug theta value in equation 1 we get

d sin ( 0.0924) = 610 \times 10 ^{-9}

d = 3.78\times 10^{-4} m

condition for 1 st dark fringe

d sin \theta =\frac{λ'}{2}

\lambda '= 2 d sin\theta

= 2λ    since from eq (1)

= 1220 nm

= 1.22 μm

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