Question

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.107 mm wide. At the point in the pattern which is an angular distance of 3.09° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad.
a) What is the wavelenght of the radiation?
b) What is the intensity at this point, if the intensity at the center of the central maximum is I/O?

Answer 1

Answer:

a. λ = 647.2 nm

b. I₀  9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

a.

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m  ⇒  λ = 647.2 nm

b.

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀  9.36 x 10⁻⁵