Question

1. A screen is located 1.5 m from a slit of width 0.7 mm.

a. Find the width of the central maximum produced when illuminated with light wavelength 480 nm and;

b. the angular width of the central maximum.

Answer 1

a. The width of the central maximum is 2.04 mm

b. The angular width of the central maximum is 0.00137°

Given:

Distance between screen and slit = 1.5 m = L

Width of slit = 0.7 mm = d

Wavelength of light = 480 nm = λ

Explanation:

a. Angular positions of destructive interference is given as:

d × sin θ₁ = λ

On substituting the values, we get,

0.7 × 10⁻³ × sin θ₁ = 480 × 10⁻⁹

sin θ₁ = (480 × 10⁻⁹)/(0.7 × 10⁻³) = 0.000685

θ₁ = sin⁻¹ (0.000685)

∴ θ₁ = 0.039°

The spatial position of angular width is given as:

y₁ = L × tan θ₁

y₁ = 1.5 × tan (0.039)

∴ y₁ = 0.00102

The width of the central maximum is given as:

⇒ 2y₁ = 2 × 0.00102

∴ 2y₁ = 0.00204 m = 2.04 mm

b. The angular width of the central maximum is given as:

β = 2λ/d

On substituting the values, we get,

β = (2 × 480 × 10⁻⁹)/(0.7 × 10⁻³)

∴ β = 0.00137°