n the experiment on diffraction due to a single slit, show that:(i) The intensity of diffraction fringes decreases as the other (n) increases. (ii) Angular width of the central maximum is twice that of the first order secondary maximum.

In a diffraction experiment due to a single slit, the intensity and angular width of diffraction fringes can be derived using the following principles:

Part (i): The intensity of diffraction fringes decreases as the order (n) increases
For single-slit diffraction, the condition for minima is given by:

a * sin(θ) = n * λ

where:

a is the width of the slit,
θ is the angle at which the minima occurs,
λ is the wavelength of light,
n is the order of the minima (n = 1, 2, 3,...).
The intensity I at an angle θ in single-slit diffraction is given by the formula:

I(θ) = I₀ * (sin(β) / β)²

where:

I₀ is the maximum intensity (at the central maximum),
β = (π * a * sin(θ)) / λ.
For the first order minimum (n = 1), the angle θ₁ is such that:

a * sin(θ₁) = λ.

For higher-order minima, say the second minimum (n = 2), the angle θ₂ satisfies:

a * sin(θ₂) = 2λ.

Thus, as n increases, sin(θ) increases, meaning the angle θ increases, and the intensity of the maxima between the minima decreases because the intensity is inversely related to the square of β. Therefore, the intensity of the diffraction fringes decreases as the order (n) increases.

Part (ii): Angular width of the central maximum is twice that of the first-order secondary maximum
The angular width of the central maximum is defined as the angular distance between the first minima on either side of the central maximum. The angular position of the first minimum is given by:

a * sin(θ₁) = λ (for the first minimum).

Therefore, θ₁ = sin⁻¹(λ / a).

The angular width of the central maximum is the distance between the first minima on both sides, which is 2 * θ₁.

For the first-order secondary maximum (n = 1), the angle at which this maximum occurs can be derived using the condition for maxima. For single-slit diffraction, the maxima occur at angles given by:

a * sin(θ) = (m + 1/2) * λ, where m = 0, 1, 2, 3,... (for secondary maxima).

For the first-order secondary maximum (m = 0), the angle θ₁' is:

a * sin(θ₁') = λ / 2.

So, θ₁' = sin⁻¹(λ / 2a).

The angular width of the first-order secondary maximum is twice this angle, which gives 2 * θ₁'.

Finally, it can be shown that the angular width of the central maximum (2 * θ₁) is indeed twice the angular width of the first-order secondary maximum (2 * θ₁'), because θ₁ = 2 * θ₁'.

Therefore, the angular width of the central maximum is twice that of the first-order secondary maximum.