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Two slits separated by a distance of d = 0.190 mm are located at a distance of D = 1.91 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 ? = 648 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.

At what angle from the beam axis will the first (m=1) maximum appear? (You can safely use the small angle approximation.)

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Answer:

[tex]\theta = 0.195^0[/tex]

Explanation:

wavelength [tex]\lambda = 648 nm \ = 648*10^{-9}m[/tex]

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

[tex]dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})[/tex]

[tex]\theta = 0.195^0[/tex]

The first maximum will appear at an angle [tex]\theta = 0.195^0[/tex] from the beam axis