Question

Parallel rays of monochromatic light with wavelength 582 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm 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 4.40×10^−4 W/m^2.

Required:
What is the intensity at a point on the screen that is 0.710 mm from the center of the central maximum?

Answer 1

I = 0.56 10⁻⁴ W / m²

Step-by-step explanation:

Double-slit interference is described by the expression

d sin θ = m λ

if we use trigonometry we can find the angle

tan θ= y / L

how the angles are small

tan θ = sin θ / cos θ = sin θ

sin θ = y / L

we substitute

d y / L = m λ

if we take into account that each slit also produces a diffraction phenomenon, the intensity distribution is the product of the intensity of the slits by the intensity of the diffraction process

I = I₀ cos² (d a) [(sin ba) / ba]²

a = π sin θ /λ

where the separation of the slits and b is the width of the slits

we substitute

I = I₀ [cos (d a)]² [sin ba /ba]²

a = π y / (L λ)

let's reduce the magnitudes to the SI system

λ = 582 nm = 582 10⁻⁹ m

L = 75.0 cm = 75.0 10⁻² m

d = 0.640 mm = 0.640 10⁻³ m

b = 0.434 mm = 0.434 10⁻³ m

Io = 4.40 10⁻⁴ W / m²

let's calculate for y = 0.710 mm = 0.710 10⁻³ m

a = π 0.710 10⁻³ / (75 10⁻² 582 10⁻⁹)

a = 5.11004 10³

I = 4.40 10⁻⁴ [cos (0.640 10⁻³ 5.11004 10³)]² (sin (0.434 10⁻³ 5.11004 10³) / (0.434 10⁻³ 5.11 10³)²

I = 4.40 10⁻⁴ cos² (3.2704) (sin 2.2178 / 2.2178)²

remember that the angles are in radians

I = 4.40 10⁻⁴ 0.9835 (0.79789 / 2.2178)²

I = 4.40 10⁻⁴ 0.9835 0.1294

I = 0.56 10⁻⁴ W / m²