Question

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.)

Answer 1

`\theta = 0.195^0`

Step-by-step explanation:

wavelength
`\lambda = 648 nm \ = 648*10^(-9)m`

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

`dsin \theta = n \lambda\\\\\theta = sin^((-1))((n \lambda)/(d))\\\\\\\theta = sin^((-1))((1*648*10^(-9))/(0.190*10^(-3)))`

`\theta = 0.195^0`

The first maximum will appear at an angle
`\theta = 0.195^0` from the beam axis