# Fourier optics (diffraction from pinholes)

A plane wave of wavelength $\lambda$ and unit amplitude is normally incident on a mask placed in the xy-plane at $z=0$. The mask contains two infinitesimally small pinholes, located on the x-axis ($y=0$) at $x=-d/2$ and $x=d/2.$ Transmitted light is viewed on a screen at a distance $z$ from the mask. Use the paraxial diffraction integral to show that the intensity of light viewed at screen is$$|f_z(x,y)|^2 = \frac{4}{\lambda^2 z^2} \cos^2(\frac{kdx}{2z})$$

Attempt:

Here (page 111 of this book) is an explanation of the method we are meant to be using. The diffraction integral is given by:

$$f_z(x,y)= \frac{1}{j\lambda z} \int^\infty_{-\infty} \int^\infty_{-\infty} f_0(x_0, y_0) \exp \Big[ \frac{jk}{2z} \left( (x-x_0)^2 + (y-y_0)^2 \right) \Big] dx_0 dy_0 \tag{1}$$

This gives the field amplitudes. We know the amplitude is $1,$ so I guess the expression for the original plane wave would be $f_0(x_0, y_0) = e^{j (k(x,y)- \omega t)}.$

How do we proceed from here? What steps are exactly involved? I am unable to follow the textbook due to the lack of worked examples.

To solve this first you take two spherical waves emanating from two pin holes $\frac{1}{\lambda z} exp(ikr)$. The intensity from individual wave at any point (x,z) will be mod squared sum of the two waves. Here $r$ will be $\sqrt{z^2+(x\pm d/2)^2}$ expand the bracket and take out $z$, expand binomially keep only first term. Now add the two expressions with +d and -d you will get the cosine term. Take the mod square of the expression you will get your answer. I have solved it on paper (the complex exponent will vanish).