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Coherent light with wavelength of 580 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is at 4.81 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

Sagot :

nuhulawal20

Answer:

the required wavelength is 1.15815 μm

Explanation:

Given the data in the question;

The position of bright fringes [tex]y_m[/tex] on screen in double slit experiment is expressed as follows;

[tex]y_m[/tex] = mλD / d

solving for d, we substitute 1 for m

y₁ = (1)λD / d

d = λD / y₁

given that λ = 580 nm = 5.8 × 10⁻⁷ m,  D = 3.00 m and y₁= y₀ = 4.81 mm = 0.00481 m

so we substitute

d = λD / y₁

d = ( 5.8 × 10⁻⁷ m × 3.00 m ) / 0.00481 m

d = 0.00000174 m² / 0.00481 m

d = 3.6117 × 10⁻⁴ m

Now, position of dark fringe  [tex]y_m[/tex] on screen in double slit experiment is expressed as;

[tex]y_m[/tex] = ( m + 1/2 )λD / d

we substitute 0 for m

y₀ = ( 0 + 1/2 )λD / d

y₀ = λD / 2d

2y₀d = λD

λ =  2y₀d  / D

we substitute

λ =  ( 2(0.00481 m) ( 3.6117 × 10⁻⁴ m) )  / 3.0 m

λ = 1.15815 × 10⁻⁶ m

λ = 1.15815 μm

Therefore, the required wavelength is 1.15815 μm

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