Diffraction

Section 14.13 Diffraction

Huygens’ Principle and Diffraction.

You might have noticed that a siren can be heard from around the corner even when siren is not directed towards you. If you go to a beach or a lake, you will notice that waves tend to go around rocks with waves even behind the rock. These are examples of a fundamental tendency of waves to spread out and around obstacles. This property is called diffraction.

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We understand diffraction by using a model of wave propagation enunciated in 1690 by the Dutch physicist Christiaan Huygens, called Huygens principle . It states that:

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Every point on a primary wavefront serves as a point source for spherical secondary wavelets of the same frequency and speed as the original primary wave such that the primary wavefront at a later time is the envelop of these secondary spherical wavelets as illustrated in Figure 14.42.

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Figure 14.42. Traveling of waves according to Huygens secondary spherical wavelets.
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Despite the successes of Huygens’ principle, it suffers from several drawbacks. It does not tell us what to do with the backward traveling part of the secondary spherical wavelet. Furthermore, it makes use of only the envelop of the secondary wavelets, and does not tell us what to do with the rest of the wavelet. More importantly, it gives rise to the same wavefront regardless of the value of the wavelength. Thus, if we follow the prescription outlined above for waves of two different wavelengths, we will get exactly the same result. But we know from experiments that different diffraction patterns result for the same aperture depending on the wavelength.

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Augustin Jean Fresnel used the ideas of interference to modify the Huygens’ principle so that one can understand the diffraction phenomena better. Fresnel hypothesized that:

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Every point of a primary wave could be thought of as producing secondary wavelet whose overlap and interference forms the primary wave at an instant later.

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The modified version is called the Huygens-Fresnel principle. The main difference comes from the introduction of interference in Fresnel’s modification in place of the envelop that determines the primary wave later. We will now apply the Huygens-Fresnel principle to simple cases of diffraction from single-slit, double-slit, circular aperture, double-slit and diffraction grating to gain qualitative and quantitative understanding of the diffraction phenomena.

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The diffraction of wave by an obstacle or a slit leads to spread of wave. Near the obstacle, the wave shows complicated wave pattern, called Fresnel diffraction pattern. But, far away from the obstacle, the compicating miniwaves die out and a simpler patter emerges. This is called Fraunhoffer diffraction pattern. As an example, we will nest look at Fraunhoffer diffraction pattern from a single slit.

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Subsection 14.13.1 Diffraction Through a Single Slit

Suppose you block a wave except for a single slit through which wave can leak to the other side and produce diffraction pattern illustrated in Figure 14.43. For example, you may place a small opening in an opaque board and shine light through it. If the opening is of the a few times more than the wavelength of wave, you will notice the diffraction effect in a pronounced way.

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Figure 14.43. Diffraction pattern of diffraction through an obstacle that has a narrow single slit opening only.
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According to the Huygens-Fresnel principle, every point of the wavefront at the slit will be point source of new waves on the other side of the slit. These waves will interfere with each other when they superimpose on each other. These interferences produce a central bulge and side bulges in specific directions as illustrated in Figure 14.43. The full calculations are beyond the scope of this text, but a simple argument can give us the direction in which waves will interfere destructively and hence will give zero intensity.

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Let \(\lambda\) be the wavelength of the wave and \(b\) the width of the slit. Let us place a \(y\) axis at the slit with origin at the center. Then, the slit occupies \(- b/2 \le y \le b/2 \) as shown in Figure 14.44. Let us look at a spot on a screen far away, in the direction \(\theta_1\) where wave from \(y=-b/2\) arrives at peak and \(y=0\) arrives at trough, and hence interfering destructively. Then, by pairing any point on the upper half of the slit with a corresponding lower point, we can say that in the direction \(\theta_1\text{,}\) we will have minima of diffraction. The calculation is similar to that of interference of waves and gives the following relation. Equating difference in path, \(\frac{b}{2}\sin\theta_1\text{,}\) to half a wave length for destructive interference, we get

\begin{equation*} \frac{b}{2}\sin\theta_1 = \frac{\lambda}{2}. \end{equation*}

Therefore, we have

\begin{equation} b\sin\theta_1 = \lambda.\ \ \ (m= 1\text{ diffraction minimum})\tag{14.97} \end{equation}

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Figure 14.44. The direction of a diffraction minimum can be obtained by pairing points on the wavefront at the slit so that the net interference of all waves from the secondary wavelets from the slit is a destructive interference. In this figure, point at the top is paired with point at the center with the difference in path equalling \(b/2\, \sin\theta_1\text{.}\) If this path equals \(\lambda/2\text{,}\) these two waves will interfere destructively. Other points from the upper part can be similarly paired with similar conclusion.
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Similarly, if we split the wavefront in four parts, \(-b/4\le y\lt -b/2\text{,}\) \(-b/4\le y\lt 0\text{,}\) \(0\le y \lt b/4\text{,}\) \(b/4\le y \le b/2\text{.}\) We pair points in \(-b/4\le y\lt -b/2\) with points in \(-b/4\le y\lt 0\) as we did above for direction \(\theta_1\) above, to obtain a direction \(\theta_2\) in which these waves will interfere destructively. This gives

\begin{equation*} \frac{b}{4}\sin\theta_2 = \frac{\lambda}{2}. \end{equation*}

Therefore, we have

\begin{equation} b\sin\theta_2 = 2\lambda.\ \ \ (m= 2\text{ diffraction minimum})\tag{14.98} \end{equation}

Clearly, we can continue this process of splitting the wavefront at the slit and pairing waves to find directions in which neighboring parts will interfere destructively. This gives a general formula for order \(n\) diffraction minimum.

\begin{equation} b\sin\theta_n = \pm n \lambda,\ \ \ n = 1, 2, 3, \cdots,\tag{14.99} \end{equation}

with angles below horizontal line being taken to be negative. Note that since \(-1 \le \sin\theta \le 1\text{,}\) maximum value of \(n\) is limited to the case when multiple of wavelength exceeds the width of the slit. Thus, if the slit width was less than one wavelength, we will just see a central bulge and no minimum.

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The central bulge is spread over a angle of \(\theta_1 + |\theta_{-1}|\text{.}\) central most intense part of the wave will be within an angle \(\theta\) about the center of the slit.

\begin{equation} \Delta\theta_\text{central} = 2\sin\left(\frac{\lambda}{b} \right).\tag{14.100} \end{equation}

In case of \(\lambda \le\le b\text{,}\) we can use small angle approximation of sine and simplify this to

\begin{equation} \Delta\theta_\text{central} = \frac{2\lambda}{b}.\tag{14.101} \end{equation}

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Example 14.45. Single-slit Fraunhofer Diffraction of Light.

Light of wavelength \(550\,\text{nm}\) passes through a slit of width \(2\, \mu\, \text{m}\text{.}\) Find the location of four minima on a screen about the central bright location (a) in terms of the angle subtendend with the horizontal direction from the center of the slit and (b) the positions of the dark bands if the screen is \(50\text{ cm}\) away.

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Answer.

(a) \(\pm 16^{\circ},\ \pm 33^{\circ}\text{,}\) (b) \(\pm 14\ \text{cm},\ \pm 33\ \text{cm}\text{.}\)

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Solution 1. a

The minima are given by the following condition as long as it makes sense mathematically due to the limitation posed by the value of \(\sin\theta_n\)

\begin{equation*} b\sin\theta = n \lambda \ \ (n = \pm 1, \pm 2, \pm 3, \cdots). \end{equation*}

For the four minima around the central maximum we will try \(n = \pm 1, \pm 2\text{.}\) The corresponding directions from the slit are

\begin{align*} \amp \theta_{\pm 1} = \pm\sin\left( \frac{\lambda}{b}\right) = \pm 16^{\circ},\\ \amp \theta_{\pm 2} = \pm\sin\left( \frac{2\lambda}{b}\right) = \pm 33^{\circ}. \end{align*}

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Solution 2. b

The location of the diffraction minima on the screen can be deduced from the right-angled triangles. Let positive \(y\) axis be pointed up on the screen, then we will have

\begin{equation*} y = (50\ \text{cm})\ \tan\theta. \end{equation*}

Denote the positions of the four minima by \(y_{+1}\text{,}\) \(y_{-1}\text{,}\) \(y_{+2}\) and \(y_{-2}\text{.}\)

\begin{align*} \amp y_{\pm 1} = (50\ \text{cm})\ \tan\theta_{\pm 1} = \pm 14\ \text{cm}\\ \amp y_{\pm 2} =(50\ \text{cm})\ \tan\theta_{\pm 2} = \pm 33\ \text{cm} \end{align*}

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