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$$|f_z(x,y)|^2 = \frac{4}{\lambda^2 z^2} \cos^2(\frac{kdx}{2z})$$


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.


1 Answer 1


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).

the expression in your question is for exact integral you may find the book by Goodman "Introduction to Fourier optics" useful in solving this expression by Fresnel approximation. However for paraxial approximation the approach presented by me is sufficient.

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    – garyp
    Commented Jun 4, 2016 at 11:36
  • $\begingroup$ @garyp I have removed the straight forward steps. I hope it is now in accordance with the group policy. $\endgroup$
    – hsinghal
    Commented Jun 4, 2016 at 14:07

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