White Light Diffraction Grating

In this paper, 10 monochromatic lights uniformly selected from the white light spectrum are subjected to grating diffraction. One convert each RGB value and use the Fraunhofer diffraction light intensity formula to obtain the relative intensity at a specific diffraction angle position, taking θ = 2° as an example. Intense white light is incident on a diffraction grating that has 584 lines/mm. (a) What is the highest order in which the complete visible spectrum can be seen with this grating?

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  • Q- White light ( = 400 – 700 nm) is incident on a 600 line/mm diffraction grating. What is the width of the first order rainbow on the screen 2.0 m behind the grating? The angle of deviation of the first order maximum is given by d sin Here d is the line spacing and is the wavelength of the light.

Learning Objectives

By the end of this section, you will be able to:

  • Discuss the pattern obtained from diffraction grating.
  • Explain diffraction grating effects.

Figure 1. The colors reflected by this compact disc vary with angle and are not caused by pigments. Colors such as these are direct evidence of the wave character of light. (credit: Infopro, Wikimedia Commons)

An interesting thing happens if you pass light through a large number of evenly spaced parallel slits, called a diffraction grating. An interference pattern is created that is very similar to the one formed by a double slit (see Figure 2). A diffraction grating can be manufactured by scratching glass with a sharp tool in a number of precisely positioned parallel lines, with the untouched regions acting like slits. These can be photographically mass produced rather cheaply. Diffraction gratings work both for transmission of light, as in Figure 2, and for reflection of light, as on butterfly wings and the Australian opal in Figure 3 or the CD in Figure 1. In addition to their use as novelty items, diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than double slits do. That is, their bright regions are narrower and brighter, while their dark regions are darker. Figure 4 shows idealized graphs demonstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, finger-like structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their pigmentation. This is called iridescence.

Figure 1. A diffraction grating is a large number of evenly spaced parallel slits. (a) Light passing through is diffracted in a pattern similar to a double slit, with bright regions at various angles. (b) The pattern obtained for white light incident on a grating. The central maximum is white, and the higher-order maxima disperse white light into a rainbow of colors.

Figure 2. (a) This Australian opal and (b) the butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. (credits: (a) Opals-On-Black.com, via Flickr (b) whologwhy, Flickr)

Figure 4. Idealized graphs of the intensity of light passing through a double slit (a) and a diffraction grating (b) for monochromatic light. Maxima can be produced at the same angles, but those for the diffraction grating are narrower and hence sharper. The maxima become narrower and the regions between darker as the number of slits is increased.

The analysis of a diffraction grating is very similar to that for a double slit (see Figure 5). As we know from our discussion of double slits in Young’s Double Slit Experiment, light is diffracted by each slit and spreads out after passing through. Rays traveling in the same direction (at an angle θ relative to the incident direction) are shown in Figure 5. Each of these rays travels a different distance to a common point on a screen far away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled.

As seen in Figure 5, each ray travels a distance d sin θ different from that of its neighbor, where d is the distance between slits. If this distance equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition necessary to obtain constructive interference for a diffraction grating is d sin θ = mλ, for m = 0, 1, −1, 2, −2, . . . (constructive) where d is the distance between slits in the grating, λ is the wavelength of light, and m is the order of the maximum. Note that this is exactly the same equation as for double slits separated by d. However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles.

In Figure 5, we see a diffraction grating showing light rays from each slit traveling in the same direction. Each ray travels a different distance to reach a common point on a screen (not shown). Each ray travels a distance d sin θ different from that of its neighbor.

Where are diffraction gratings used? Diffraction gratings are key components of monochromators used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength emitted by molecules in diseased cells in a biopsy sample or to help excite strategic molecules in the sample with a selected frequency of light. Another vital use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings are available for selecting specific wavelengths for such use.

Take-Home Experiment: Rainbows on a CD

The spacing d of the grooves in a CD or DVD can be well determined by using a laser and the equation d sin θ = mλ, form = 0, 1, −1, 2, −2, . . . . However, we can still make a good estimate of this spacing by using white light and the rainbow of colors that comes from the interference. Reflect sunlight from a CD onto a wall and use your best judgment of the location of a strongly diffracted color to find the separation d.

Example 1. Calculating Typical Diffraction Grating Effects

Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it to a screen 2.00 m away.

  1. Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760 nm).
  2. What is the distance between the ends of the rainbow of visible light produced on the screen for first-order interference? (See Figure 6.)
Light

Figure 6. The diffraction grating considered in this example produces a rainbow of colors on a screen a distance from the grating. The distances along the screen are measured perpendicular to the x-direction. In other words, the rainbow pattern extends out of the page.

Strategy

The angles can be found using the equation d sin θ = mλ (form = 0, 1, −1, 2, −2, . . . ) once a value for the slit spacing d has been determined. Since there are 10,000 lines per centimeter, each line is separated by 1/10,000 of a centimeter. Once the angles are found, the distances along the screen can be found using simple trigonometry.

Solution for Part 1

The distance between slits is [latex]d=frac{1text{ cm}}{10,000}=1.00times10^{-4}text{ cm}[/latex] or 1.00 × 10−6 m. Let us call the two angles θV for violet (380 nm) and θR for red (760 nm). Solving the equation d sin θV = mλ for sin θV, [latex]sintheta_{text{V}}=frac{mlambda_{text{V}}}{d}[/latex], where m = 1 for first order and λV = 380 nm = 3.80 × 10−7 m. Substituting these values gives

[latex]displaystylesintheta_{text{V}}=frac{3.80times10^{-7}text{ m}}{1.00times10^{-6}text{ m}}=0.380[/latex]

Thus the angle θV is θV = sin−1 0.380 = 22.33º.

Similarly,

[latex]displaystylesintheta_{text{R}}=frac{7.60times10^{-7}text{ m}}{1.00times10^{-6}text{ m}}[/latex]

Thus the angle θR is θR = sin−1 0.760 = 49.46º.

Notice that in both equations, we reported the results of these intermediate calculations to four significant figures to use with the calculation in Part 2.

Solution for Part 2

The distances on the screen are labeled yV and yR in Figure 6. Noting that [latex]tantheta=frac{y}{x}[/latex], we can solve for yV and yR. That is, yV = x tan θV = (2.00 m)(tan 22.33º) = 0.815 m and yR = x tan θR = (2.00 m)(tan 49.46º) = 2.338 m.

The distance between them is therefore yRyV = 1.52 m.

Discussion

The large distance between the red and violet ends of the rainbow produced from the white light indicates the potential this diffraction grating has as a spectroscopic tool. The more it can spread out the wavelengths (greater dispersion), the more detail can be seen in a spectrum. This depends on the quality of the diffraction grating—it must be very precisely made in addition to having closely spaced lines.

Section Summary

A diffraction grating is a large collection of evenly spaced parallel slits that produces an interference pattern similar to but sharper than that of a double slit.
There is constructive interference for a diffraction grating when d sin θ = (for m = 0 , 1, –1, 2, –2, …), where d is the distance between slits in the grating, λ is the wavelength of light, and m is the order of the maximum.

Conceptual Questions

  1. What is the advantage of a diffraction grating over a double slit in dispersing light into a spectrum?
  2. What are the advantages of a diffraction grating over a prism in dispersing light for spectral analysis?
  3. Can the lines in a diffraction grating be too close together to be useful as a spectroscopic tool for visible light? If so, what type of EM radiation would the grating be suitable for? Explain.
  4. If a beam of white light passes through a diffraction grating with vertical lines, the light is dispersed into rainbow colors on the right and left. If a glass prism disperses white light to the right into a rainbow, how does the sequence of colors compare with that produced on the right by a diffraction grating?
  5. Suppose pure-wavelength light falls on a diffraction grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? What happens to the interference pattern if a longer-wavelength light falls on the same grating? Explain how these two effects are consistent in terms of the relationship of wavelength to the distance between slits.
  6. Suppose a feather appears green but has no green pigment. Explain in terms of diffraction.
  7. It is possible that there is no minimum in the interference pattern of a single slit. Explain why. Is the same true of double slits and diffraction gratings?

Problems & Exercises

  1. A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum be for 520-nm-wavelength green light?
  2. Find the angle for the third-order maximum for 580-nm-wavelength yellow light falling on a diffraction grating having 1500 lines per centimeter.
  3. How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 470-nm blue light at an angle of 25.0º?
  4. What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of 60.0º?
  5. Calculate the wavelength of light that has its second-order maximum at 45.0º when falling on a diffraction grating that has 5000 lines per centimeter.
  6. An electric current through hydrogen gas produces several distinct wavelengths of visible light. What are the wavelengths of the hydrogen spectrum, if they form first-order maxima at angles of 24.2º, 25.7º, 29.1º, and 41.0º when projected on a diffraction grating having 10,000 lines per centimeter?
  7. (a) What do the four angles in the above problem become if a 5000-line-per-centimeter diffraction grating is used? (b) Using this grating, what would the angles be for the second-order maxima? (c) Discuss the relationship between integral reductions in lines per centimeter and the new angles of various order maxima.
  8. What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light?
  9. The yellow light from a sodium vapor lamp seems to be of pure wavelength, but it produces two first-order maxima at 36.093º and 36.129º when projected on a 10,000 line per centimeter diffraction grating. What are the two wavelengths to an accuracy of 0.1 nm?
  10. What is the spacing between structures in a feather that acts as a reflection grating, given that they produce a first-order maximum for 525-nm light at a 30.0º angle?
  11. Structures on a bird feather act like a reflection grating having 8000 lines per centimeter. What is the angle of the first-order maximum for 600-nm light?
  12. An opal such as that shown in Figure 2 acts like a reflection grating with rows separated by about 8 μm. If the opal is illuminated normally, (a) at what angle will red light be seen and (b) at what angle will blue light be seen?
  13. At what angle does a diffraction grating produces a second-order maximum for light having a first-order maximum at 20.0º?
  14. Show that a diffraction grating cannot produce a second-order maximum for a given wavelength of light unless the first-order maximum is at an angle less than 30.0º.
  15. If a diffraction grating produces a first-order maximum for the shortest wavelength of visible light at 30.0º, at what angle will the first-order maximum be for the longest wavelength of visible light?
  16. (a) Find the maximum number of lines per centimeter a diffraction grating can have and produce a maximum for the smallest wavelength of visible light. (b) Would such a grating be useful for ultraviolet spectra? (c) For infrared spectra?
  17. (a) Show that a 30,000-line-per-centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a first-order maximum? (c) What is the greatest number of lines per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light?
  18. A He–Ne laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.50 m from the CD, and the first fringe is 0.600 m from the central maximum, what is the spacing of grooves on the CD?
  19. The analysis shown in the figure below also applies to diffraction gratings with lines separated by a distance d. What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 600-nm light, if the screen is 1.50 m away?

    Figure 6. The distance between adjacent fringes is [latex]Delta y=frac{xlambda}{d}[/latex], assuming the slit separation d is large compared with λ.

  20. Unreasonable Results. Red light of wavelength of 700 nm falls on a double slit separated by 400 nm. (a) At what angle is the first-order maximum in the diffraction pattern? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
  21. Unreasonable Results. (a) What visible wavelength has its fourth-order maximum at an angle of 25.0º when projected on a 25,000-line-per-centimeter diffraction grating? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
  22. Construct Your Own Problem. Consider a spectrometer based on a diffraction grating. Construct a problem in which you calculate the distance between two wavelengths of electromagnetic radiation in your spectrometer. Among the things to be considered are the wavelengths you wish to be able to distinguish, the number of lines per meter on the diffraction grating, and the distance from the grating to the screen or detector. Discuss the practicality of the device in terms of being able to discern between wavelengths of interest.

White Light Incident On Diffraction Grating

Glossary

constructive interference for a diffraction grating: occurs when the condition d sin θ = (form = 0,1,–1,2,–2, . . .) is satisfied, where d is the distance between slits in the grating, λ is the wavelength of light, and m is the order of the maximum

diffraction grating: a large number of evenly spaced parallel slits

Selected Solution to Problems & Exercises

1. 5.97º

3. 8.99 × 103

5. 707 nm

7. (a) 11.8º,12.5º,14.1º,19.2º; (b) 24.2º,25.7º,29.1º,41.0º; (c) Decreasing the number of lines per centimeter by a factor of x means that the angle for the x­-order maximum is the same as the original angle for the first-­order maximum.

9. 589.1 nm and 589.6 nm

11. 28.7º

13. 43.2º

15. 90.0º

Light Diffraction Glasses

17. (a) The longest wavelength is 333.3 nm, which is not visible; (b) 333 nm (UV); (c) 6.58 × 103 cm

19. 1.13 × 10−2 m

21. (a) 42.3 nm; (b) Not a visible wavelength. The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light.

We classically think of light as always traveling in straight lines, but when light waves pass near a barrier they tend to bend around that barrier and become spread out. Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength.

A very simple demonstration of diffraction can be conducted by holding your hand in front of a light source and slowly closing two fingers while observing the light transmitted between them. As the fingers approach each other and come very close together, you begin to see a series of dark lines parallel to the fingers. The parallel lines are actually diffraction patterns. This phenomenon can also occur when light is 'bent' around particles that are on the same order of magnitude as the wavelength of the light. A good example of this is the diffraction of sunlight by clouds that we often refer to as a silver lining, illustrated in Figure 1 with a beautiful sunset over the ocean.

We can often observe pastel shades of blue, pink, purple, and green in clouds that are generated when light is diffracted from water droplets in the clouds. The amount of diffraction depends on the wavelength of light, with shorter wavelengths being diffracted at a greater angle than longer ones (in effect, blue and violet light are diffracted at a higher angle than is red light). As a light wave traveling through the atmosphere encounters a droplet of water, as illustrated below, it is first refracted at the water:air interface, then it is reflected as it again encounters the interface. The beam, still traveling inside the water droplet, is once again refracted as it strikes the interface for a third time. This last interaction with the interface refracts the light back into the atmosphere, but it also diffracts a portion of the light as illustrated below. This diffraction element leads to a phenomenon known as Cellini's halo (also known as the Heiligenschein effect) where a bright ring of light surrounds the shadow of the observer's head.

The terms diffraction and scattering are often used interchangeably and are considered to be almost synonymous. Diffraction describes a specialized case of light scattering in which an object with regularly repeating features (such as a diffraction grating) produces an orderly diffraction of light in a diffraction pattern. In the real world most objects are very complex in shape and should be considered to be composed of many individual diffraction features that can collectively produce a random scattering of light.

One of the classic and most fundamental concepts involving diffraction is the single-slit optical diffraction experiment, first conducted in the early nineteenth century. When a light wave propagates through a slit (or aperture) the result depends upon the physical size of the aperture with respect to the wavelength of the incident beam. This is illustrated in Figure 3 assuming a coherent, monochromatic wave emitted from point source S, similar to light that would be produced by a laser, passes through aperture d and is diffracted, with the primary incident light beam landing at point P and the first secondary maxima occurring at point Q.

As shown in the left side of the figure, when the wavelength (λ) is much smaller than the aperture width (d), the wave simply travels onward in a straight line, just as it would if it were a particle or no aperture were present. However, when the wavelength exceeds the size of the aperture, we experience diffraction of the light according to the equation:

White Light Strikes A Diffraction Grating At Normal Incidence

sinθ = λ/d

Where θ is the angle between the incident central propagation direction and the first minimum of the diffraction pattern. The experiment produces a bright central maximum which is flanked on both sides by secondary maxima, with the intensity of each succeeding secondary maximum decreasing as the distance from the center increases. Figure 4 illustrates this point with a plot of beam intensity versus diffraction radius. Note that the minima occurring between secondary maxima are located in multiples of π.

This experiment was first explained by Augustin Fresnel who, along with Thomas Young, produced important evidence confirming that light travels in waves. From the figures above, we see how a coherent, monochromatic light (in this example, laser illumination) emitted from point L is diffracted by aperture d. Fresnel assumed that the amplitude of the first order maxima at point Q (defined as εQ) would be given by the equation:

dεQ = α(A/r)f(χ)d

where A is the amplitude of the incident wave, r is the distance between d and Q, and f(χ) is a function of χ, an inclination factor introduced by Fresnel.

Diffraction of Light

Explore how a beam of light is diffracted when it passes through a narrow slit or aperture. Adjust the wavelength and aperture size and observe how this affects the diffraction intensity pattern.

Diffraction of light plays a paramount role in limiting the resolving power of any optical instrument (for example: cameras, binoculars, telescopes, microscopes, and the eye). The resolving power is the optical instrument's ability to produce separate images of two adjacent points. This is often determined by the quality of the lenses and mirrors in the instrument as well as the properties of the surrounding medium (usually air). The wave-like nature of light forces an ultimate limit to the resolving power of all optical instruments.

White Light Diffraction Grating

Our discussions of diffraction have used a slit as the aperture through which light is diffracted. However, all optical instruments have circular apertures, for example the pupil of an eye or the circular diaphragm and lenses of a microscope. Circular apertures produce diffraction patterns similar to those described above, except the pattern naturally exhibits a circular symmetry. Mathematical analysis of the diffraction patterns produced by a circular aperture is described by the equation:

sinθ(1) = 1.22(λ/d)

where θ(1) is the angular position of the first order diffraction minima (the first dark ring), λ is the wavelength of the incident light, d is the diameter of the aperture, and 1.22 is a constant. Under most circumstances, the angle θ(1) is very small so the approximation that the sin and tan of the angle are almost equal yields:

From these equations it becomes apparent that the central maximum is directly proportional to λ/d making this maximum more spread out for longer wavelengths and for smaller apertures. The secondary mimina of diffraction set a limit to the useful magnification of objective lenses in optical microscopy, due to inherent diffraction of light by these lenses. No matter how perfect the lens may be, the image of a point source of light produced by the lens is accompanied by secondary and higher order maxima. This could be eliminated only if the lens had an infinite diameter. Two objects separated by a distance less than θ(1) can not be resolved, no matter how high the power of magnification. While these equations were derived for the image of a point source of light an infinite distance from the aperture, it is a reasonable approximation of the resolving power of a microscope when d is substituted for the diameter of the objective lens.

Thus, if two objects reside a distance D apart from each other and are at a distance L from an observer, the angle (expressed in radians) between them is:

White Light Through A Diffraction Grating

θ = D / L

which leads us to be able to condense the last two equations to yield:

Where D(0) is the minimum separation distance between the objects that will allow them to be resolved. Using this equation, the human eye can resolve objects separated by a distance of 0.056 millimeters, however the photoreceptors in the retina are not quite close enough together to permit this degree of resolution, and 0.1 millimeters is a more realistic number under normal circumstances.

The resolving power of optical microscopes is determined by a number of factors including those discussed, but in the most ideal circumstances, this number is about 0.2 micrometers. This number must take into account optical alignment of the microscope, quality of the lenses, as well as the predominant wavelengths of light used to image the specimen. While it is often not necessary to calculate the exact resolving power of each objective (and would be a waste of time in most instances), it is important to understand the capabilities of the microscope lenses as they apply to the real world.

Contributing Authors

Mortimer Abramowitz - Olympus America, Inc., Two Corporate Center Drive., Melville, New York, 11747.

Michael W. Davidson - National High Magnetic Field Laboratory, 1800 East Paul Dirac Dr., The Florida State University, Tallahassee, Florida, 32310.