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4.3: Double-Slit Diffraction

When we studied interference in Young’s double-slit experiment, we ignored the diffraction effect in each slit. We assumed that the slits were so narrow that on the screen you saw only the interference of light from just two point sources. If the slit is smaller than the wavelength, then [link](a) shows that there is just a spreading of light and no peaks or troughs on the screen. Therefore, it was reasonable to leave out the diffraction effect in that chapter. However, if you make the slit wider, [link](b) and (c) show that you cannot ignore diffraction. In this section, we study the complications to the double-slit experiment that arise when you also need to take into account the diffraction effect of each slit.

To calculate the diffraction pattern for two (or any number of) slits, we need to generalize the method we just used for a single slit. That is, across each slit, we place a uniform distribution of point sources that radiate Huygens wavelets, and then we sum the wavelets from all the slits. This gives the intensity at any point on the screen. Although the details of that calculation can be complicated, the final result is quite simple:


The diffraction pattern of two slits of width D that are separated by a distance d is the interference pattern of two point sources separated by d multiplied by the diffraction pattern of a slit of width D.

In other words, the locations of the interference fringes are given by the equation \(d \, sin \, \theta = m \lambda\), the same as when we considered the slits to be point sources, but the intensities of the fringes are now reduced by diffraction effects, according to [link]. [Note that in the chapter on interference, we wrote \(d \, sin \, \theta = m \lambda\) and used the integer m to refer to interference fringes. [link] also uses m, but this time to refer to diffraction minima. If both equations are used simultaneously, it is good practice to use a different variable (such as n) for one of these integers in order to keep them distinct.]

Interference and diffraction effects operate simultaneously and generally produce minima at different angles. This gives rise to a complicated pattern on the screen, in which some of the maxima of interference from the two slits are missing if the maximum of the interference is in the same direction as the minimum of the diffraction. We refer to such a missing peak as a missing order. One example of a diffraction pattern on the screen is shown in Figure \(\PageIndex{1}\). The solid line with multiple peaks of various heights is the intensity observed on the screen. It is a product of the interference pattern of waves from separate slits and the diffraction of waves from within one slit.

Figure shows a graph of I by I0 versus theta. Three curves are shown on the graph. Interference curve has a smaller wavelength. Diffraction curve has a larger wavelength and a y value of 1 at x equal to 0. Resultant curve has the same wavelength as the interference and its amplitude is modified according to the diffraction curve. Each wave crest of the interference wave is labeled m equal to 1, m equal to 2 and so on. The diffraction wave has a zero at m equal to 3, and theta equal to 30 degrees. Hence, the resultant wave, too, has a zero. This is labeled missing order m equal to 3.

Figure \(\PageIndex{1}\): Diffraction from a double slit. The purple line with peaks of the same height are from the interference of the waves from two slits; the blue line with one big hump in the middle is the diffraction of waves from within one slit; and the thick red line is the product of the two, which is the pattern observed on the screen. The plot shows the expected result for a slit width \(D = 2\lambda\) and slit separation \(d = 6\lambda\). The maximum of \(m = \pm 3\) order for the interference is missing because the minimum of the diffraction occurs in the same direction.

Example \(\PageIndex{1}\): Intensity of the Fringes

Figure \(\PageIndex{1}\) shows that the intensity of the fringe for m=3 is zero, but what about the other fringes? Calculate the intensity for the fringe at m=1 relative to \(I_0\), the intensity of the central peak.


Determine the angle for the double-slit interference fringe, using the equation from Interference, then determine the relative intensity in that direction due to diffraction by using [link].


From the chapter on interference, we know that the bright interference fringes occur at \(d \, sin \, \theta = m \lambda\), or

\[sin \, \theta = \dfrac{m\lambda}{d}.\]

From [link],

\[I = I_0 \left(\dfrac{sin \, \beta}{\beta}\right)^2, \, where \, \beta = \dfrac{phi}{2} = \dfrac{\pi D sin \, \theta}{\lambda}.\]

Substituting from above,

\[\beta = \dfrac{\pi D sin \, \theta}{\lambda} = \dfrac{\pi D}{\lambda} \cdot \dfrac{m\lambda}{d} = \dfrac{m\pi D}{d}.\]

For \(D = 2\lambda, \, d = 6 \lambda\), and \(m = 1\),

\[\beta = \dfrac{(1)\pi (2\lambda)}{(6 \lambda)} = \dfrac{\pi}{3}.\]

Then, the intensity is

\[I = I_0 \left(\dfrac{sin \, \beta}{\beta}\right)^2 = I_0 \left(\dfrac{sin \, (\pi/3)}{\pi/3}\right)^2 = 0.684 I_0.\]


Note that this approach is relatively straightforward and gives a result that is almost exactly the same as the more complicated analysis using phasors to work out the intensity values of the double-slit interference (thin line in Figure \(\PageIndex{1}\)). The phasor approach accounts for the downward slope in the diffraction intensity (blue line) so that the peak near m=1 occurs at a value of θ ever so slightly smaller than we have shown here.

Two-Slit Diffraction

Suppose that in Young’s experiment, slits of width 0.020 mm are separated by 0.20 mm. If the slits are illuminated by monochromatic light of wavelength 500 nm, how many bright fringes are observed in the central peak of the diffraction pattern?


From [link], the angular position of the first diffraction minimum is \(\theta \approx sin \, \theta = \dfrac{\lambda}{D} = \dfrac{5.0 \times 10^{-7}m}{2.0 \times 10^{-5}m} = 2.5 \times 10^{-2} rad\).

Using \(sin \, \theta = m \lambda\) for \(\theta = 2.5 \times 10^{-2} rad\), we find

\[m = \dfrac{d \, sin \, \theta}{\lambda} = \dfrac{(0.20 \, mm)(2.5 \times 10^{-2} rad)}{(5.0 \times 10^{-7}m)} = 10,\]

which is the maximum interference order that fits inside the central peak. We note that \(m=±10\) are missing orders as \(θ\) matches exactly. Accordingly, we observe bright fringes for

m = −9, −8, −7, −6, −5, −4, −3, −2, −1, 0, +1, +2, +3, +4, +5, +6, +7, +8, and +9

for a total of 19 bright fringes.

Exercise \(\PageIndex{1}\)

For the experiment in Example, show that m=20m=20 is also a missing order.


From \(d \, sin \, \theta = m\lambda\), the interference maximum occurs at \(2.87^o\) for \(m = 20\). From [link], this is also the angle for the second diffraction minimum. (Note: Both equations use the index m but they refer to separate phenomena.)

Explore the effects of double-slit diffraction. In this simulation written by Fu-Kwun Hwang, select N=2 using the slider and see what happens when you control the slit width, slit separation and the wavelength. Can you make an order go “missing?”


Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).