Two slits separated by a distance of d = 0.130 mm are located at a distance of D

Question

Two slits separated by a distance of d = 0.130 mm are located at a distance of D = 1.65 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a monochromatic and coherent light source with a wavelength of lambda = 626 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima. What is the path length difference between the waves at the first maximum (m=l) on the screen? At what angle from the beam axis will the first (m = l) maximum appear? (You can safely use the small angle approximation.)

Explanation / Answer

Hi,

To solve this problem we use the following equations:

dsin() = m (1); where is the wavelength of the incident light, d is the distance between the slits, m is the order number and is the angle between from the beam axis to a certain position.

If the distance from the slits to the screen is big compared to the distance between the slits, then we can use the next equation:

= dsin() (2); where is known as the path lenght difference

The small angle approximation implies that:

sin() = ; as long as is in radians.

1. The path length difference between the waves at the first maximum is equal to:

= dsin() = m ::::::: m = 1 ; = 626 nm :::::::::: = 626 nm

2. The angle we are looking for is equal to:

dsin() = m :::::::: m = 1; = 626 nm ; d = 0.130 mm; sin() =

= (1)(626*10-9 m)/(0.130*10-3 m) = 4.84*10-3 rad = 0.277°

I hope it helps.

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