Question

Calculate the Fraunhofer spectrum produced by a diffraction grating, uniformly illuminated by light of wavelength $2\pi /k$, as follows. Consider a grating with $4N$ equal strips each of width $a$ and alternately opaque and transparent. The aperture function is then $$ f(y)= \begin{cases}A & \text { for }(2 n+1) a \leq y \leq(2 n+2) a, \quad-N \leq n

Short answer

The intensity distribution $I(\theta )$ is proportional to $\frac{{\sin }^2(2qaN)}{q^2{\cos }^2(qa/2)}$. Significant peaks occur when $\theta =0$ or the cosine term vanishes.

Step-by-step solution

- Define the Problem

Given a diffraction grating with 4N equal strips each of width a, alternating between opaque and transparent, determine the Fraunhofer spectrum produced. The wavelength of the light is given as $\frac{2\pi }{k}$, and the aperture function is provided.

- Express Aperture Function

The aperture function is defined as: $f(y)=\{\begin{array}{lll}A& \text{ for }(2n+1)a\le y\le (2n+2)a,-N\le n<N0& \text{ otherwise }\end{array}$

- Define Diffraction Angle and Fourier Transform

For diffraction at angle $\theta$, the Fourier transform can be expressed in terms of q, where $q=k\sin \theta$, as: $\tilde{f}(q)=(2\pi {)}^{-1/2}\sum _{r=-N}^{N-1}\exp (-2iarq){\int }_a^{2a}A\exp (-iqu)du$

- Evaluate the Integral Part

The integral to evaluate is: ${\int }_a^{2a}A\exp (-iqu)du$. This evaluates to: $A{\left[\frac{\text{exp}(-iqu)}{-iq}\right]}_a^{2a}=\frac{A}{-iq}\text{[}\exp (-iq(2a))-\exp (-iqa)\text{]}=\frac{A}{-iq}\exp (-iqa)\text{[}\exp (-iqa)\exp (-iqa)\text{]}=\frac{A(\exp (-iqa)-\exp (-i2qa))}{-iq}$

- Simplify the Sum Expression

Evaluating the sum: $\sum _{r=-N}^{N-1}\exp (-2iaraq)$, substitute the integral result. The entire expression now reads: $\tilde{f}(q)=(2\pi {)}^{-1/2}\frac{A\exp (-iqa)}{-iq}\sum _{r=-N}^{N-1}\exp (-iqar)\text{[}\exp (-iqa)-\exp (-i2qa)\text{]}$

- Combine and Simplify Further

Simplify the complete expression to derive: $\tilde{f}(q)=(2\pi {)}^{-1/2}\exp (-iqa/2)\frac{A\sin (2qaN)}{q\cos (qa/2)}$

- Obtain Intensity Distribution

The intensity distribution $I(\theta )$ is proportional to the modulus squared of the transform: $I(\theta )\propto \frac{{\sin }^2(2qaN)}{q^2{\cos }^2(qa/2)}$

- Analyze for Large N

For large N, the maxima and minima in the numerator become close and average to $\frac{1}{2}$. Significant peaks occur when cosine term vanishes or when $\theta =0$. Hence, the corresponding values of $\tilde{f}(q)=\frac{2aNA}{(2\pi {)}^{1/2}}\text{ and }\frac{4aNA}{(2\pi {)}^{1/2}(2m+1)\pi }$.

Key concepts

Diffraction Grating

A diffraction grating is an optical component with a pattern of regular, closely spaced lines or slits that diffract light. When light passes through this grating, it bends and creates an interference pattern.
The grating in this exercise has 4N strips, alternately opaque and transparent, each with a width of 'a'. This pattern creates complex interference effects.
The configuration of the grating can be described through its aperture function as given:
$$f(y)=\{\begin{array}{lll}A& \text{for }(2n+1)a\le y\le (2n+2)a,-N\le n<N0& \text{otherwise}\end{array}$$
Here, 'A' is the amplitude of the light wave passing through the transparent regions,
and 'y' is the position across the grating. This function alternates value between 'A' and '0' based on whether the region is transparent or opaque.
This kind of structure allows us to observe diffraction patterns which are explained further using concepts like Fourier Transforms and intensity distribution.

Fourier Transform

The Fourier transform is a mathematical tool that transforms a function of time (or space) into a function of frequency. In the context of diffraction, it helps analyze how different spatial frequencies (patterns) contribute to the overall diffraction pattern.
For this grating, the relevant Fourier Transform is expressed as:
$$\tilde{f}(q)=(2\pi {)}^{-1/2}\sum _{r=-N}^{N-1}\exp (-2iarq){\int }_a^{2a}A\exp (-iqu)du$$
Where:

• q = k * sin(\theta)
• k is the wavenumber of light, related to the wavelength by k = 2\pi/λ
• \theta is the diffraction angle from the normal

This formula breaks down the grating's aperture function into a sum of sinusoidal functions of different frequencies (q values), weighted by their amplitudes.
The integral part evaluates to:
$$\tilde{f}(q)=(2\pi {)}^{-1/2}\exp (-iqa/2)\frac{A\sin (2qaN)}{q\cos (qa/2)}$$
This transform tells us how each spatial frequency component (q) contributes to the overall diffraction pattern.

Intensity Distribution

Intensity distribution describes how bright different parts of the diffraction pattern are. The intensity is proportional to the square of the amplitude of the Fourier transform.
In this case, the intensity distribution I(θ) is given by:
$$I(\theta )\propto \frac{{\sin }^2(2qaN)}{q^2{\cos }^2(qa/2)}$$
The numerator ${\sin }^2(2qaN)$ produces closely spaced maxima and minima, especially for large N. These oscillations result in bright and dark fringes in the diffraction pattern.
For large N, the contributions average out to about 1/2, but significant peaks in intensity occur when the cosine term in the denominator vanishes, leading to constructive interference.

• The main peaks occur at θ = 0 and where $\cos (qa/2)=0$.
• These are where intense, bright spots will appear.

The calculation shows that these peaks, related to the spatial frequency, result in observable maxima of the diffraction pattern.
This means that the constructive interference leads to sharp, bright fringes, and their position and intensity can be predicted from the equations derived above.