Aslit 0.380 mm wide is illuminated by parallel rays of light that have a wavelength of 600 nm . the diffraction pattern is observed on a screen that is 1.40 m from the slit. the intensity at the center of the central maximum (θ=0∘) is i0.

x = 2.19 mm

I = 3.03E-9

Explanation:

Sin θ = λ/a = 0.600µm / 380µm = 0.00158

θ = sin¯¹ (0.00158)

θ = 0.090°

x = Rtanθ = 1.4 tan(0.090)

x = 2.19 mm

Approximating small amgle will be fine in this situation. By using small angle approximation, θ here will be 0.045°

Therefore sinθ = 0.785×10¯³

Using the formula

β = 2πa sin θ/λ

β = 2×3.142×380µm×0.785×10¯³/0.600µm

β = 3.124

So,

β/2 = 1.562

Utilizing above value in the formula

I = Io (sin(β/2)/(β/2))²

I = (10×10¯6) (0.0272/1.562)²

I = 3.03E-9

x = 2.19 mm

I = 3.03*10^-9

Explanation:

step by step Explanation:

diffraction is the overlapping of a wave particle when its made to pass through an aperture

Sin θ = λ/a = 0.600nm / 380nm = 0.00158

taking the sine inverse of bot sides

θ = sin¯¹ (0.00158)

angle of diffraction

θ = 0.090°

x = Rtanθ = 1.4 tan(0.090)

x = 2.19 mm

using small angle approximation,θ here will be half the previous value θ/2 = 0.090°/2

θ = 0.0.45°

so sinθ = 0.785×10¯³

Using the formula

β = 2πa sin θ/λ

β = 2×3.142×380m×0.785×10¯³/0.600nm

β = 3.124

therefore we can write

β/2 = 1.562

substituting the parameters in to the equation

I = Io (sin(β/2)/(β/2))²

I = (10×10¯6) (0.0272/1.562)²

I = 3.03*10^-9