Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm.

If the intensity at the center of the central maximum is 5.0010?4W/m2 , what is the intensity at a point on the screen that is 0.830mm from the center of the central maximum?

The intensity at a point on the screen that is 0.830mm from the center of the central maximum is; I =  8.64 × 10⁻⁶ W/m²

We are given;

Wavelength; λ = 571 nm = 571 × 10⁻⁹ m

Distance of interference from screen; R = 75 cm = 0.75 m

center to center distance between slits; d = 0.640 mm = 0.00064 m

width of each slit; a = 0.434 mm = 0.000434 m

Intensity at center; I₀ = 5 × 10⁻⁴ W/m²

Distance of intensity from center of the central maximum; y = 0.83 mm = 0.00083 m

Let us first find the angle θ from the formula;

tan θ = y/R

tan θ = 0.00083/0.75

tan θ = 0.001107

θ = tan⁻¹0.001107

θ = 0.06343

θ is very small and as such we use the formula;

∅ = (2πd sinθ)/λ

∅ = (2π * 0.00064 * sin 0.06343)/(571 × 10⁻⁹)

∅ = 7.7964 rad

Similarly, for distance between the centers of slit;

β = (2πa sinθ)/λ

β =  (2π * 0.000434 * sin 0.06343)/(571 × 10⁻⁹)

β = 5.287 rad

Formula for the Intensity at a point on the screen that is 0.830mm from the center of the central maximum is;

I =  8.64 × 10⁻⁶ W/m²

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Explanation:

We are given that

Wavelength,

R=75 cm=

1 m=100 cm

Because is small.

,

Therefore

snell's law

refractve index = sin incidence / sin refraction