Interference, Diffraction, and Polarization of Light PDF

# Interference, Diffraction and Polarization of Light ## Introduction Light is a transverse electromagnetic wave that can be seen by the typical human. The wave nature of light was first illustrated through experiments on diffraction and interference. Like electromagnetic waves, light can travel t...

# Interference, Diffraction and Polarization of Light ## Introduction Light is a transverse electromagnetic wave that can be seen by the typical human. The wave nature of light was first illustrated through experiments on diffraction and interference. Like electromagnetic waves, light can travel through a vacuum. The transverse nature of light can be demonstrated through polarization. ## Interference of Light When two beams of light cross each other, the resultant amplitude and intensity may be different from the two beams acting separately. This modification of intensity obtained by the superposition of two or more beams of light is called interference. * If the resultant intensity is zero or in general less than the separate intensities, we have destructive interference. * If the resultant intensity is greater, it is called constructive interference. The first man successfully to demonstrate the interference of light was Thomas Young. ### Constructive Interference of Light  ### Destructive Interference of Light  ## Young's experiment Young's experiment, that provided classical investigation into the nature of light and the basic element in the development of the wave theory, was first performed by the English physicist and physician, Thomas Young in 1801. In this experiment, Young identified the phenomenon called interference. Observing that when light from a single source is split into two beams, and the two beams are then recombined, the combined beam shows a pattern of light and dark fringes. Young concluded that the fringes result from the fact that when the beams recombine their peaks and troughs may not be in phase. * When two peaks coincide they reinforce each other, and a line of light results; * when a peak and a trough coincide they cancel each other, and a dark line results. Young's experiment with double slits is shown in figure 7.2.  **Figure 7.2 Young's experiment Double Slits** If a white screen is placed in the region beyond the slits, a pattern of bright and dark interference bands (fringes) can be seen. The key to experiment is the use of a single pinhole 'S' to illuminate the aperture. This provides the necessary mutual, coherence between the light coming from the two slits 'S_' and 'S_'. An equation for the intensity at any point 'P' on the screen. The phase difference between the two waves arriving at P, having transverse different distances S_1_P and S_2_P given by **Phase difference** $= \frac{2\pi}{\lambda}(path difference) \newline \delta = \frac{2\pi}{\lambda}(S_2P - S_1P) = k(d_2 - d_1)$ It is assumed that the waves start out from 'S_' and 'S_2_' in the same phase. Furthermore, the amplitudes are practically the same. The intensity at 'P' was given by equation: $P = I_{A}^{2} = 4a^{2}cos^{2} ( \frac{\delta}{2})$ Where 'a' is the amplitude of the separate waves and 'A' that of their resultant. In Young's experiment 'D' is very practically much larger than 'd' or 'x'. Hence' θ and O' are very small and ≃ θ'. And the path difference is equal to an integral number of wavelengths. $d_{2} - d_{1}$ = mλ (m = 0, 1, 2, 3, .....) Also the path difference = d sin θ = d sin θ'. For small angle 0, sin e equals tan θ Hence the path difference = $ d_{2}-d_{1} = d\ sin \theta = d\frac{x}{D}$. $\delta = \frac{2\pi }{\lambda} (d_{2}-d_{1}) = \frac{2\pi }{\lambda} d\sin \theta = \frac{2\pi }{\lambda} \frac{xd}{D} $ Using equation 7.2 the intensity has maximum values equal to 4a² whenever 8 is an integral multiple of 2π. This will occur when the path difference is an integral multiple of λ. Therefore $\frac{xd}{D} =0, \lambda, 2\lambda, 3\lambda, .... = m\lambda$ x = distance of a bright fringe from the center, PO. The minimum value of intensity is zero, and this occurs when $\delta = \pi, 3\pi, 5\pi, ... \newline \frac{xd}{D} = \frac{ \lambda} {2}, \frac{3 \lambda} {2}, ...= ( m + \frac{1}{2}) \lambda $ x = distance of a dark fringe from the center, PO. The whole number 'm', which characterizes a particular bright fringe, is called the order of interference. Thus, the fringes with m = 0, 1, 2, 3, ....are called the zero, first, second ....etc, order. The distance on the screen between two successive fringes is obtained by changing 'm' by unity in equation 7.5 or 7.6. It is equal to $\frac{ \lambda D} {d}$. ## Example 7.1 Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10.95° relative to the incident beam. What is the wavelength of the light? **Solution** For the third bright line m = 3 d = 0.0100 mm and θ=10.95°.  d sinθ = mλ λ = $\frac{dsinθ} {m}$ Substituting known values yields λ =$ \frac{0.0100 (sin 10.95°)}{3} = \frac{6.33 × 10^-4}{3} mm = 633\ nm$. ## Interference patterns do not have an infinite number of lines Since there is a limit to how big m can be. What is the highest-order constructive interference possible with the system described in example 7.1? **Solution** For the third bright line m=3 d = 0.0100 mm and θ-10.95°. d sinθ = mλ m =$ \frac{d\ sin θ} {λ}$ Taking sin θ = 1 , Substituting the values of d and λ gives m = $\frac{0.0100 × 10^{-3}}{633 × 10^{-9}} ≈ 15.8$ Therefore, the largest integer m can be is 15, m=15. ## Example 7.3 In Young's experiment, the distance between the two slits is 0.8 mm and the distance of the screen from the slits is 1.2m. If the fringe width is 0.75 mm, calculate the wavelength of light. **Solution** Distance between slits = d = 0.8 mm = 0.8 x 10^-3 m = 8 x 10^-4 m. Distance between slit and screen = D = 1.2 m, Fringe width = x = 0.75 mm = 0.75×10^-3 m = 7.5 x 10^-4 m. Wavelength of light used, λ =? The fringe width is given by....... x= $\frac{λD} {d}$ λ = $\frac{xd}{D} = \frac{(7.5 × 10^-4)(8 × 10^{-4})}{1.2} = 5 × 10^{-7} m = 5000 × 10^{-10} m = 5000 Å$ **Ans: Wavelength of light used is 5000 Å** ## Example 7.4 The distance between two consecutive bright bands in Young's experiment is 0.32 mm when the red light of wavelength 6400 Å is used. By how much will this distance change if the light is substituted by the blue light of wavelength 4800 Å with the same setting? **Solution** For red light: fringe width = x_r = 0.32 mm, wavelength = λ_r = 6400 Å, For blue light: wavelength = λ_b = 4800 Å Change in fringe width = Δx = ? The fringe width is given by x_r = $\frac{λ_rD} {d}$ ..... (1) x_b = $\frac{λ_bD} {d}$ ..... (2) Dividing Equation (2) by (1) $\frac{x_b}{x_r} = \frac{λ_bD}{d} ÷ \frac{λ_rD}{d} = \frac{λ_b}{λ_r}$ $\frac{x_b}{x_r} = \frac{λ_b}{λ_r}$ x_b = $\frac{λ_b}{λ_r}$ × x_r = ($\frac{4800}{6400}$) × 0.32 = 0.24 mm :. Δx = x_b - x_r = 0.32 - 0.24 = 0.08 mm **Ans: The distance between two consecutive bright bands will change by 0.08 mm.** ## Thin film interference The multicolored patterns on an oil or gasoline slick on a wet pavement, or in soap bubbles are common examples of thin film interference. **Thin-film interference** is a natural phenomenon in which light waves reflected by the upper and lower boundaries of a thin film interfere with one another, either enhancing or reducing the reflected light. When the thickness of the film is an odd multiple of one quarter-wavelength of the light on it, the reflected waves from both surfaces interfere to cancel each other (assuming the refractive index of the thin film has a value between the refractive indices of the materials above and below it). Since the wave cannot be reflected, it is completely transmitted instead. When the thickness is a multiple of a half-wavelength of the light, the two reflected waves reinforce each other, increasing the reflection and reducing the transmission. Thus when white light, which consists of a range of wavelengths, is incident on the film, certain wavelengths (colors) are intensified while others are attenuated. Thin-film interference explains the multiple colors seen in light reflected from soap bubbles and oil films on water. It is also the mechanism behind the action of antireflection coatings used on glasses and camera lenses. If the thickness of the film is much larger than the coherence length of the incident light, then the interference pattern will be washed out due to the line width of the light source. Figure 7.4 (a) and (b) shows thin film interference of light in oil and glasses.  **Figure 7.4 (a) and (b) Thin film interference of light in oil and glasses.** ## Diffraction of Light When a beam of light passes through a narrow slit, it spreads out to a certain extent into the region of the geometrical shadow. This effect is one of the simplest examples of diffraction, i.e, the failure of the light to travel in straight lines. This phenomenon can be satisfactorily explained only by assuming a wave character of light as shown in figure 7.5.  **Figure 7.5 Diffraction** It is diffraction that gives the slits in Young's experiment the character of point sources. This depends on the slit width and wavelength of the light. To have appreciable diffraction, the slit width must be on the same order or less than the wavelength in figure 7.6. When the slit width is much smaller than the wavelength, spherical waves spread out from the slit as though they were from a point source. Notice in the figure the bending or diffraction around the edge of the slit. If the slit width is greater than the wavelength, little diffraction occurs.  **Figure 7.6 Diffraction depends on the slit width and wavelength of the light** ## Huygens' principle and diffraction When light passes through an aperture, every point on the light wave within the aperture can be viewed as a source creating a circular wave that propagates outward from the aperture. The aperture thus creates a new wave source that propagates in the form of a circular wavefront is shown in figure 7. 7 (a) and (b).  **Figure 7.7 Huygens' principle states that every point on a wave front may be considered as a source of secondary waves.** The centre of the wavefront has greater intensity while the edges have a lesser intensity is as shown in figure 7.8. This explains the observed diffraction pattern and why a perfect image of the aperture on a screen is not created.  **Figure 7.8 Huygens' principle of diffraction pattern of light** ## The single slit diffraction When light passes through a single slit whose width w is on the order of the wavelength of the light, then can be observed a single slit diffraction pattern on a screen that is a distance L >> w away from the slit. The intensity is a function of angle. Huygens' principle tells us that each part of the slit can be thought of as an emitter of waves. All these waves interfere to produce the diffraction pattern. Constructive interference can be occured when two crests meet each other and destructive interference can be occured when crest meets trough. The single slit diffraction pattern of light is shown in figure 7.9.  **Figure 7.9 Single slit diffraction pattern of light** In order to study the diffraction pattern on a screen (figure 7.10), the single-slit experiment is employed. Consider a monochromatic source of light that passes through a slit width "a" as shown in the figure. At point P on the screen, the secondary waves interfere destructively and produce a dark fringe. Let "D" be the distance between the slit and the screen, and "y" be the distance between point P and point O, the center of the screen. AC is perpendicular to BP. Let be the angle of diffraction, and | is the angle BAC.  **Figure 7.10 Single slit diffraction** We assume that the screen is at a considerable distance from the slit, i.e., D >> a. Hence, 0 = 0' and, sin 0 ≈ tan 0 ≈ 0 = $\frac{y}{D}$. The path difference between the two rays AP and BP is given by, Δ = ΒΡ - ΑP = BC In the right-angled triangle BСА, sin 0' = sin 0 = $\frac{BC}{BA}$ BC = BA sin 0 = a sin 0 Therefore, A = a sin 0 ## Diffraction Minima The condition for minima or dark fringe is, Path difference = integral multiple of wavelength Δ = nλ a sin 0 = nλ $\frac{a\ sin θ} {λ} =n $ (n = ±1, 2, 3, ..., etc.) y_n = $\frac{nλD}{a}$ This equation gives the distance of the nth dark fringe from the center. The fringe width is given by, β = y_n+1 -y_n =(n+1)$\frac{λD}{a}$ -n$\frac{λD}{a}$ or, β = $\frac{λD}{a}$ ## Diffraction Maxima The condition for maxima or bright fringe is, Path difference = non-integral multiple of wavelength Δ = (n+$\frac{1}{2}$)λ (n = ±1, 2, 3, ..., etc.) a sin 0 = $\frac{(n + \frac{1}{2})λ}{a}$ y = $\frac{(n + \frac{1}{2})λD}{a}$ The intensity of single-slit diffraction is given by, I = I_0 \[\frac{sin (a \ sin \frac{θ}{λ})}{(a \ sin \frac{θ}{λ})}\]^2 The differences between single slit and double-slits diffraction is listed in table 7.1. The comparison of diffraction intensities of light for single slit and double slits is also shown in figure 7. 11. | Single-slit diffraction | Double-slits diffraction | | --- | --- | | Consist of one slit | Consist of two slits | | Waves that originate within the same slit interfere | Slits are so small that each one is considered as a single light source, and the interference of waves originating within the same slit can be neglected. | | The fringes are broad and not as sharp as double-slit | The fringes are narrow and sharp | | All bright fringes are visible | Some bright fringes are missing since they suppressed by the minima of a single-slit interference pattern. | | Intensity: I = I_0 \[\frac{sin (a \ sin \frac{θ}{λ})}{(a \ sin \frac{θ}{λ})}\]^2 | Intensity: I= I_0 cos² [$\frac{nd \ sin θ}{λ}$] \[\frac{sin (π a \ sin \frac{θ}{λ})}{(a \ sin \frac{θ}{λ})}\]^2 |  **Figure 7.11 Comparison of diffraction intensities of light for single slit and double slits** ## Example 7.5 Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0° relative to the incident direction of the light. (a) What is the width of the slit? (b) At what angle is the first minimum produced? **Solution** λ = 550 nm, m=2 and θ_2 = 45.0°. For minimum diffraction, a sin θ = mλ a = $\frac{mλ}{sin θ_2}$ = $\frac{2(550 × 10^{-9})}{sin 45.0°}$ = $\frac{1100 × 10^{-9}}{0.707}$ = 1.56 × 10^-6 m sin θ_1 = $\frac{λ} {a}$ = $\frac{550 × 10^{-9}}{1.56 × 10^{-6}}$ = 0.354 θ_1 = sin^-1 0.354 = 20.7°. ## Polarization Polarization is the attribute that a wave's oscillations have a definite direction relative to the direction of propagation of the wave. Waves having such a direction are said to be polarized. For an EM wave, we define the direction of polarization to be the direction parallel to the electric field.  **Figure 7.12 Polarization of light.** ## Types of Polarization There are three types of polarization depending on how the electric field is oriented: linear polarization, circular polarization and elliptical polarization. These are shown in figure 7.13.  **Figure 7.13 Different types of polarization of light.** ## Uses of Polarization The lenses of the sunglasses have polarizing filters that are oriented vertically with respect to the frames. Polarizing sunglasses are very useful when driving in the sun or at the beach where sunlight is reflected from the surface of the road or water leading to glare that can be almost blinding. One of the most common uses of polarization today is the liquid crystal display (LCD) used in numerous applications including wrist watches, computer screens, timer-clocks and many others.