Diffraction

1.0Diffraction of Light

The phenomenon of diffraction was first discovered by Grimaldi. Its experimental study was done by Newton and Young. The theoretical explanation was first given by Fresnel.

(1) The phenomenon of bending of light around the corners of an obstacle/aperture of the size of the wavelength of light is called diffraction.

(2) The phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wave front is defined as diffraction of light.

(3) Diffraction is the characteristic of all types of waves.

(4) Greater the wave length of wave higher will be its degree of diffraction.

2.0Conditions for Diffraction

For significant diffraction to be observed, the size of the obstacle or aperture must be of the same order of magnitude as the wavelength of the wave.

• If the size is much larger than the wavelength, the wave will cast a sharp shadow, and diffraction will be negligible.
• If the size is much smaller than the wavelength, the wave will spread out almost completely.
• For light, this means diffraction is most noticeable with very narrow slits, since the wavelength of visible light is very small (on the order of hundreds of nanometers).

3.0Diffraction from a Single Slit

In this section we’ll discuss the diffraction pattern formed by plane-wave (parallel ray) monochromatic light when it emerges from a long, narrow slit, as shown in Fig. We call the narrow dimension the width, even though in this figure it is a vertical dimension. According to geometric optics, the transmitted beam should have the same cross section as the slit, as in Fig. a.What is actually observed is the pattern shown in Fig. b. The beam spreads out vertically after passing through the slit. The diffraction pattern consists of a central bright band, which may be much broader than the width of the slit, bordered by alternating dark and bright bands with rapidly decreasing intensity. About 85% of the power in the transmitted beam is in the central bright band, whose width is inversely proportional to the width of the slit. In general, the smaller the width of the slit, the broader the entire diffraction pattern. (The horizontal spreading of the beam in Fig. b is negligible because the horizontal dimension of the slit is relatively large.)You can observe a similar diffraction pattern by looking at a point source, such as a distant street light, through a narrow slit formed between your two thumbs held in front of your eye; the retina of your eye corresponds to the screen.

In figure below (a) The “shadow” of a horizontal slit as incorrectly predicted by geometric optics. (b) A horizontal slit actually produces a diffraction pattern. The slit width has been greatly exaggerated.

Diffraction by a single rectangular slit. The long sides of the slit are perpendicular to the figure.

4.0Intensity in the Single-Slit Pattern

We can derive an expression for the intensity distribution for the single-slit diffraction pattern by the phasor-addition method. We imagine a plane wave front at the slit subdivided into many strips. We superpose the contributions of the Huygens wavelets from all the strips at a point P on a distant screen at an angle from the normal to the slit plane (Fig. a). To do this, we use a phasor to represent the sinusoidally varying field from the E field from each individual strip. The magnitude of the vector sum of the phasors at each point P is the amplitude $E_P$ of the total E field at the point. The intensity at P is proportional to $E_P^2$. At the point O shown in fig. a corresponding to the center of the pattern where $\theta =0$, there are negligible path differences for $x>>a$. The phasors are all essentially in phase (that is, have the same direction). In fig. b we draw the phasors at time t = 0 and denote the resultant amplitude at O by $E_0$. In this illustration, we have divided the slit into 15 strips.

Now consider wavelets arriving from different strips at point P in fig. a at an angle $\theta$ from O. Because of the differences in path length, there are now phase differences between wavelets coming from adjacent strips; the corresponding phasor diagram is shown in fig. c. The vector sum of the phasors is now part of the perimeter of a many sided polygon, and $E_P$, the amplitude of the resultant electric field at P , is the chord. The angle is the total phase difference between waves from the top strip of fig. a and the wave from the bottom strip; that is $\beta$ is the phase of the wave received at P from the top strip with respect to the wave received at P from the bottom strip.

Using phasor diagrams to find the amplitude of the E field in single-slit diffraction. Each phasor represents the E fields from a single strip within the

We may imagine dividing the slit into narrower and narrower strips. In the limit that there is an infinite number of infinitesimally narrow strips, the curved trails of phasors become an arc of a circle (fig. d). which are length equal to the length $E_0$ in fig. b. The center C of this arc is found by constructing perpendiculars at A and B. From the relationship among arc length, radius, and angle, the radius of the arc is $\frac{E_0}{\beta }$ The amplitude $E_p$ of the resultant electric field at P is equal to the chord AB, which is $2(E_0)\sin (\beta /2)$.

(Note that $\beta$ must be in radians)

$E_P=E_0\frac{\sin (\varphi /2)}{\varphi /2}$ (amplitude in single slit diffraction)........(1)

The intensity at each point on the screen is proportional to the square of the amplitude given by Eq. (i), if $I_0$ is the intensity in the straight-ahead direction where $\theta =0$ and

$\beta =0$, then the intensity I at any point is

$I=I_0{\left(\frac{\sin (\varphi /2)}{\varphi /2}\right)}^2\text{Intensity in single slit diffraction}$........(2)

We can express the phase difference in terms of geometric quantities, as we did for the two-slit pattern. The phase difference is $2\frac{2\pi }{\lambda }$ times the path difference. The path difference between the ray from the top of the slit and the ray from the middle of the slit is $(a/2)\sin \theta$. The path difference between the rays from the top of the slit and the bottom of the slit is twice this, so

$\beta =\frac{2\pi }{\lambda }a\sin \theta$.......(3)

And equation (2) becomes

$I=I_0{\left(\frac{\sin (\frac{\pi a\sin \theta }{\lambda })}{\frac{\pi a\sin \theta }{\lambda }}\right)}^2$(Intensity in single slit diffraction)........(3)

This equation expresses the intensity directly in terms of the angle . In many calculations it is easier first to calculate the phase angle $\beta$, using equation (3) and then to use equation (2).

Equation (4) is plotted in figure below. Note that the central intensity peak is much larger than any of the others. This means that most of the power in the wave remains within an angle from the perpendicular to the slit, where $\sin \theta \approx \lambda /a$(the first diffraction minimum).

Note:

The peak intensities in figure below decrease rapidly as we go away from the center of the pattern. (Compare the figure which shows a single slit diffraction pattern for light.)

The dark fringes in the pattern are the places where I = 0. These occur at a point for which the numerator of the equation is zero so that $\beta$ is a multiple of $2\pi$. From equation this corresponds to

$a\sin \theta =m\lambda (m=1,2,\dots ).......(5)$

$\sin \theta =\frac{m\lambda }{a}(m=1,2,\dots )$

Note again that $\beta =0$ (corresponding to \theta = 0) is not a minimum. The equation is indeterminate at $\beta =0$, but we can evaluate the limit as $0\beta =0$ using L’Hopital’s rule. We find that at \beta = 0,I = I_0, as we should expect.

Intensity Maxima in the Single-Slit Pattern

We can also use equation (2) from above to calculate the positions of the peaks, or intensity maxima, and intensities at these peaks. This is not quite as simple as it may appear. We might expect the peaks to occur where the sine function reachecs the value $\pm 1$ -- namely, where $\beta =\pm \pi ,\pm 3\pi ,\pm 5\pi$ or in general,

$\beta =\pm (2m+1)m+0,1,2$,……..(6)

This is approximately correct, but because of the factor ( $\frac{\beta }{2}{)}^2$ in the denominator of eq. (2), the maxima don’t occur precisely at these points. When we take the derivative of eq. (2) with respect to   and set it equal to zero to try to find the maxima and minima, we get a transcendental equation that has to be solved numerically. In fact there is no maximum near $\beta =\pm \pi$. The first maxima on either side of the central maximum, near $\beta =\pm 3\pi$, actually occur at $\pm 2.860\pi$. The second side maxima, near $\beta =\pm 5\pi$, are actually at $\pm 4.918\pi$ and so on. The error in equation (6) vanishes in the limit of large m — that is, for intensity maxima far from the center of the pattern.

To find the intensities at the side maxima, we substitute these values of back into eq. (2). Using the approximate expression in equation (6), we get

$I_m\approx I_0{\left(\frac{2}{(2m+1)\pi }\right)}^2$ .......(7)

Where Im is the intensity of the mth side maximum and $I_0$ is the intensity of the central maximum. Equation (7) gives the series of intensities

$0.0450I_{0}0.0162I_{0}0.0083I_0$

and so on. As we have pointed out, this equation is only approximately correct. The actual intensities of the side maxima turn out to be

$0.0472I_{0}0.0165I_{0}0.0083I_0$

Note that the intensities of the side maxima decrease very rapidly, as fig. below also shows. Even the first side maxima has less than 5% of the intensity of the central maximum.

Figure: Intensity versus angle in single slit diffraction. The values of m label intensity minima given by equation (5). Most of the wave power goes into the central intensity peak (between the m = 1 and m = -1 intensity minima). Only the diffracted waves within the central intensity peak are visible; the waves at larger angles are too faint to see.

Width of the Single-Slit Pattern

For small angles the angular spread of the diffraction pattern is inversely proportional to the slit width a or, more precisely, to the ratio of a to the wavelength $\lambda$. Figure shows graphs of intensity I as a function of the angle for three values of the ratio $a/\lambda$

The single-slit diffraction pattern depends on the ratio of the slit width a to the wavelength $\lambda$

$(a)a=\lambda (b)a=5\lambda (c)a=8\lambda$

If the slit width is equal to or narrower than the wavelength, only one broad maximum forms. The wider the slit (or the shorter the wavelength), the narrower and sharper the central peak.

With light waves, the wavelength $\lambda$ is often much smaller than the slit width a, and the values of $\theta$ in equations (5) and (6) are so small that the approximation sin  = is very good. With this approximation the position ${\theta }_1$ of the first minimum beside the central maximum, corresponding to $\beta /2=\pi$, is from

${\theta }_1\approx \frac{\lambda }{a}$……. (6)

This characterizes the width (angular spread) of the central maximum, and we see that it is inversely proportional to the slit width a. When the small-angle approximation is value, the central maximum is exactly twice as wide as each side maximum. When a is of the order of a centimetre or more, ${\theta }_1$ is so small that we can consider practically all the light to be concentrated at the geometrical focus. But when a is less than $\lambda$, the central maximum spreads over 180°, and the fringe pattern is not seen at all.

It’s important to keep in mind that diffraction occurs for all kinds of waves, not just light. Sound waves undergo diffraction when they pass through a slit or aperture such as an ordinary doorway. The sound waves used in speech have wavelengths of about a meter or greater, and a typical doorway is less that 1 m wide; in this situation, a is less than and the central intensity maximum extends over 180°. This is why the sounds coming through an open doorway can easily be heard by an eavesdropper hiding out of sight around the corner. In the same way, sound waves can bend around the head of an instructor who faces the blackboard while lecturing. By c contrast, there is essentially no diffraction of visible light through a doorway because the width a is very much greater than the wavelength $\lambda (\text{of order 5}\times 10^{-7}m)$. You can hear around corners because typical sound waves have relatively long wavelengths you cannot see around corners because the wavelength of visible light is very short.

5.0Circular Apertures and Resolving Power

We have studied in detail the diffraction patterns formed by long, thin slits or arrays of slits. But an aperture of any shape forms a diffraction pattern. The diffraction pattern formed by a circular aperture is of special interest because of its role in limiting how well an optical instrument can resolve fine details. In principle, we could compute the intensity at any point P in the diffraction pattern by dividing the area of the aperture into small elements, finding the resulting wave amplitude and intensity at P. In practice, the integration cannot be carried out in terms of elementary functions. We will simply describe the pattern and quote a few relevant numbers

The diffraction pattern formed by a circular aperture consists of a central bright spot surrounded by a series of bright and dark rings, as Fig. below shows.Diffraction pattern formed by a circular aperture of diameter D. The pattern consists of a central bright spot and alternating dark and bright rings. The angular radius ${\theta }_1$ of the first dark ring is shown.

We can describe the pattern in terms of the angle, representing the angular radius of each ring. If the aperture diameter is D and the wavelength is $\lambda$, the angular radius ${\theta }_1$ of the first dark ring is given by

$\sin {\theta }_1=1.22\frac{\lambda }{d}$ (diffraction by a circular aperture)

Diffraction pattern formed by a circular aperture.

The discussion shows that a converging lens can never form a point image of a distant point source. In the best conditions, it produces a bright disc surrounded by dark and bright rings. If we assume that most of the light is concentrated within the central bright disc, we can say that the lens produces a disc image for a distant point source. This is not only true for a distant point source but also for any point source. The radius of the image disc is

$R=1.22\frac{\lambda }{d}$

Where d is the distance between image plane and lens.                     

6.0Limit of Resolution

The fact that a lens forms a disc image of a point source, puts a limit on resolving two neighbouring points imaged by a lens.

Suppose $S_{1}and{S}_2$ are two-point sources placed before a converging lens (as shown in figure). If the separation between the centres of the image-discs is small in comparison to the radii of the discs, the discs will largely overlap on one another and it will appear like a single disc. The two points are then not resolved. If $S_{1}and{S}_2$ are moved apart, the centres of their image-discs also move apart. For a sufficient separation, one can distinguish the presence of two discs in the pattern. In this case, we say that the points are just resolved.

The angular radius of the diffraction disc is given by sin=1.22 b , where b is the radius of the lens. Thus, increasing the radius of the lens improves the resolution. This is the reason why objective lenses of powerful microscopes and telescopes are kept large in size.

The human eye is itself a converging lens which forms the image of the points (we see) on the retina. The above discussion then showed that two points very close to each other cannot be seen as two distinct points by the human eye.

7.0Rayleigh Criterion

Whether two-disc images of nearby points are resolved or not may depend on the person viewing the images. Rayleigh suggested a quantitative criterion for resolution. Two images are called just resolved in this criterion if the centre of one bright disc falls on the periphery of the second. This means, the radius of each bright disc should be equal to the separation between them. In this case, the resultant intensity has a minimum between the centres of the images. The figure shows the variation of intensity when the two images are just resolved.

Illustration-1:The Fraunhofer diffraction pattern of a single slit is formed at the focal plane of a lens of focal length 1m. The width of the slit is 0.3 mm. If the third minimum is formed at a distance of 5mm from the central maximum then calculate the wavelength of light.

Solution:

$x_n=\frac{nf\lambda }{a}$

$\lambda =\frac{a{x}_n}{nf}=\frac{3\times 10^{-4}\times 5\times 10^{-3}}{3\times 1}=5000\text{A˚}[\because n=3]$

Illustration-2:Light of wavelength $5000\mathring{A}$ is incident normally on a slit of width 0.1mm. Find out the width of the central bright line on a screen distance 2 m from the slit?

Solution:

$W_x=\frac{2D\lambda }{a}=\frac{2\times 2\times 5\times 10^{-7}}{10^{-4}}=20mm$

Illustration-3:Light of wavelength $6000\mathring{A}$ is incident normally on a slit of width $24\times 10^{-5}cm$. Find out the angular position of the second minimum from the central maximum?

Solution:

$\lambda =6\times 10^{-7}m,a=24\times 10^{-5}\times 10^{-2}m$

$a\sin \theta =2\lambda$

$\sin \theta =\frac{2\lambda }{a}=\frac{2\times 6\times 10^{-7}}{24\times 10^{-7}}=\frac{1}{2}\Rightarrow \theta =30^{\circ }$

Illustration-4:The intensity at the center of the pattern is $I_0$ for a light of wave length $6.33\times 10^{-7}m$. What is the intensity at a point on the screen 3.0 mm from the center of the pattern. The screen is placed at a distance of 6 m.

Solution:We have y = 3 mm and x = 6m, so $\tan \theta =y/x=(3\times 10^{-3})/(6\text{ m})=5\times 10^{-4}$.This is so small that the values of $\tan \theta ,\sin \theta ,$ and $\theta$ (in radians) are all nearly the same. Then, using equation

$\frac{\pi a\sin \theta }{\lambda }=\frac{\pi (2.4\times 10^{-4}\text{ m})(5\times 10^{-4})}{6.33\times 10^{-7}\text{ m}}=0.60$

Illustration-5:

(a) The intensity at the center of a single-slit diffraction pattern is I0 . What is the intensity at a point in the pattern where there is a 66-radian phase difference between wavelets from the two edges of the slit?

(b) If this point is $7.0^{\circ }$ away from the central maximum, how many wavelengths wide is the slit?

Solution:

1. We have $\frac{\beta }{2}=33\text{ rad}$

$I=I_0{\left[\frac{\sin (33\text{ rad})}{33\text{ rad}}\right]}^2=(9.2\times 10^{-4})I_0$

2. $I=I_0{\left[\frac{\sin (33\text{ rad})}{33\text{ rad}}\right]}^2=(9.2\times 10^{-4})I_0$ 

For example for 550 nm light the slit width is

$a=(86)(550\text{ nm})=4.7\times 10^{-5}\text{ m}=0.047\text{ mm or roughly }\frac{1}{20}\text{ mm}$