9.6 Diffraction from an array of apertures

Waves and optics Page 1 / 1

We derive the diffraction pattern from an array of rectangular apertures.

An array of rectangular apertures

Say we have an array of rectangular apertures sitting in the $x - y$ plane and light hits this aperture traveling in the positive z direction. There are $N$ apertures arranged vertically (in the $y$ direction). Each aperture has a width in the $x$ direction of $a$ and a height in the $y$ direction of $b$ . For convenience, the apertures are aligned with their centers at $x = 0$ . The apertures are equally spaced by a distance $d$ .

The electric field at some point

P

away from the array is

E = \sum_{n_{y} = 1}^{N} E_{n_{y}} (a, b)

where

E_{n_{y}} (a, b)

is the field from the

n_{y}

slit at that point. The

y

position of the center of the aperture is

(n_{y} - 1) d

so we write

y = (n_{y} - 1) d + y^{'}

and

(x, y^{'})

is the position of a point in the aperture with respect to the center of the aperture. We can write

E_{n_{y}} (a, b) = \int_{- b / 2}^{b / 2} ⅆ y^{'} \int_{- a / 2}^{a / 2} ⅆ x \frac{ɛ_{A}}{R} e^{- i (k r_{n_{y}} - ω t)}

If the point of observation is

(x_{P}, y_{P}, z_{P})

then

r_{n_{y}} = {[(x_{P} - x)}^{2} + {(y_{P} - y)}^{2} + {{(z_{P} - z)}^{2}]}^{1 / 2}

but we take z to be zero at the aperture so

\begin{matrix} r_{n_{y}} = {[(x_{P} - x)}^{2} + {(y_{P} - y)}^{2} + {z_{P}^{2}]}^{1 / 2} \\ = {[x_{P}^{2} - 2 x x_{P} + x^{2} + y_{P}^{2} - 2 y y_{P} + y^{2} + z_{P}^{2}]}^{1 / 2} \\ = R [1 - 2 x x_{P} / R^{2} - 2 y y_{P} / R^{2} + (x^{2} + y^{2}) / {R^{2}]}^{1 / 2} \end{matrix}

where

R^{2} = x_{P}^{2} + y_{P}^{2} + z_{P}^{2}

, the distance from the origin. In the far field approximation

(x^{2} + y^{2}) / R^{2} = 0

and we can write:

r_{n_{y}} \approx R {[1 - 2 x x_{P} / R^{2} - 2 y y_{P} / R^{2}]}^{1 / 2} .

We use the first two terms in the binomial expansion and get

\begin{matrix} r_{n_{y}} \approx R [1 - x x_{P} / R^{2} - y y_{P} / R^{2}] \\ = R - x x_{P} / R - y y_{P} / R \\ = R - x x_{P} / R - [(n_{y} - 1) d + y^{'}] y_{P} / R \\ = R - x x_{P} / R - (n_{y} - 1) ⅆ y_{P} / R - y^{'} y_{P} / R \end{matrix}

so now we have

E_{n_{y}} (a, b) = \int_{- b / 2}^{b / 2} ⅆ y^{'} \int_{- a / 2}^{a / 2} ⅆ x \frac{ɛ_{A}}{R} e^{- i (k (R - x x_{P} / R - (n_{y} - 1) ⅆ y_{P} / R - y^{'} y_{P} / R) - ω t)} .

We rearrange to get

E_{n_{y}} (a, b) = \frac{ɛ_{A}}{R} e^{- i (k R - ω t)} e^{i k (n_{y} - 1) ⅆ y_{P} / R} \int_{- b / 2}^{b / 2} ⅆ y^{'} e^{i k y^{'} y_{P} / R} \int_{- a / 2}^{a / 2} ⅆ x e^{i k x x_{P} / R} .

We define

k x_{P} / R = k_{x}

and

k y_{P} / R = k_{y}

so that

E_{n_{y}} (a, b) = \frac{ɛ_{A}}{R} e^{- i (k R - ω t)} e^{i k_{y} (n_{y} - 1) d} \int_{- b / 2}^{b / 2} ⅆ y^{'} e^{i k_{y} y^{'}} \int_{- a / 2}^{a / 2} ⅆ x^{'} e^{i k_{x} x^{'}}

E = \frac{ɛ_{A}}{R} e^{- i (k R - ω t)} \sum_{n_{y} = 1}^{N} e^{i k_{y} (n_{y} - 1) d} \int_{- b / 2}^{b / 2} ⅆ y^{'} e^{i k_{y} y^{'}} \int_{- a / 2}^{a / 2} ⅆ x e^{i k_{x} x} .

We see that each piece of this is something we did before

E = \frac{ɛ_{A}}{R} e^{- i (k R - ω t)} e^{i k_{y} (N - 1) d / 2} \frac{\sin (N k_{y} d / 2)}{\sin (k_{y} d / 2)} b s i n c (k_{y} b / 2) a s i n c (k_{x} a / 2)

or if we define

k R_{c} = k R - (N - 1) d k_{y} / 2

$E = \frac{ɛ_{A} a b}{R} e^{- i (k R_{c} - ω t)} \frac{\sin (N k_{y} d / 2)}{\sin (k_{y} d / 2)} s i n c (k_{y} b / 2) s i n c (k_{x} a / 2)$

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Source: OpenStax, Waves and optics. OpenStax CNX. Nov 17, 2005 Download for free at http://cnx.org/content/col10279/1.33

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