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?

What is the intensity is 1.3349 × 10⁻⁷ w/m²

Explanation:

Given that;

λ = 582 nm = 582 × 10⁻⁹

R = 75.0 cm = 0.75 m

d = 0.640 mm = 0.000640 m

a = 0.434 mm = 0.000434 m

I₀ =  4.40×10⁻⁴ W/m²

y = 0.710 mm  = 0.00071 m

Now to get our tanФ we say

tanФ = y/R =  0.00071 / 0.75 = 0.0009466  

Ф is so small

∴ tanФ ≈ sinФ

So

∅ = 2πdsinФ / λ

we substitute

∅ = ( 2π × 0.000640 × 0.0009466  ) /  582 × 10⁻⁹

=  6.54 rad

Now

β = 2πasinФ / λ

we substitute

β = ( 2π × 0.000434 × 0.0009466  ) /  582 × 10⁻⁹

β = 4.435 rad

I = I₀ cos²(∅/2) [(sin(β/2))/(β/2)]²

we substitute

I = 4.40×10⁻⁴ cos(3.27)² [ (sin(2.2175)) / (2.2175) ]²  

= 4.40×10⁻⁴ × 0.9967 × 0.0003044

= 1.3349 × 10⁻⁷ w/m²

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.