Question

Parallel rays of monochromatic light with wavelength 587 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cmcm 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 5.00×10^−4 W/m^2 , what is the intensity at a point on the screen that is 0.890 mm from the center of the central maximum?

Answer 1

I = 1.82 10⁻⁴ W / m²

Step-by-step explanation:

This is a problem of light interference by a double slit, the expression that describes the process is

I = I₀ cos² (πd /λ sin θ)

Where te is the interference angle, as in these experiments the screen is far from the slits we can use trigonometry

tan θ = y / L

Since the angles are very small

tan θ = sin θ/cos θ = sin θ =θ y / L

We substitute

I = I₀ cos² (πd/λ y/L)

Let's apply this equation to our case

λ = 587 10⁻⁹ m

L = 0.750 m

d = 0.640 10⁻³ m

I₀ = 5.00 10⁻⁴ W / m²

y = 0.890 10⁻³ m

Let's calculate

I = 5 10⁻⁴ cos² (π 0.640 10⁻³ / 587 10⁻⁹ 0.890 10⁻³ /0.750)

I = 5 10⁻⁴ cos² (4.0646)

Be careful that the angles are in radians

I = 1.82 10⁻⁴ W / m²