In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 527 nm. An interference pattern is observed on the screen. The distance between the screen and the double-slit is 1.54 m and the distance between the two slits is 0.102 mm. A light wave propogates from each slit to the screen. What is the path length difference between the distance traveled by the waves for the fifth-order maximum (bright fringe) on the screen

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Moya1316go Moya1316go · Answer 1

Answer:

Λ = 5.14 10⁻⁴ m

Explanation:

This is a double slit experiment, which for the case of constructive interference

d sin θ = m λ

let's use trigonometry

tan θ = y / L

as the angles are very small

tan θ = [tex]\frac{sin \theta}{cos \theta}[/tex] = sin θ

sin θ = y / L

we substitute

d y / L = m λ

y = m λ L / d

we calculate for the interference of order m = 5

y = 5 527 10⁻⁹ 1.54/0.102 10⁻³

y = 3.978 10⁻² m

Now we can find the difference in length between the two rays, that of the central maximum and this

let's use the Pythagorean theorem

L’= [tex]\sqrt{L^2 +y^2}[/tex]

L ’= [tex]\sqrt{1.54^2 +(3.978 \ 10^{-2})^2 }[/tex]

L ’= 1.54051 m

optical path difference

Λ = L’- L

Λ = 1.54051 - 1.54

Λ = 5.14 10⁻⁴ m