Wave Optics PDF

# Wave Optics ## Doppler Effect You have already encountered Doppler effect for sound waves in Chapter 15 of Class XI textbook. For velocities small compared to the speed of light, we can use the same formulae which we use for sound waves. The fractional change in frequency Δv/v is given by -v_radial/c, where v_radial is the component of the source velocity along the line joining the observer to the source relative to the observer; v_radial is considered positive when the source moves away from the observer. Thus, the Doppler shift can be expressed as: Δν/ν = v_radial/c The formula given above is valid only when the speed of the source is small compared to that of light. A more accurate formula for the Doppler effect which is valid even when the speeds are close to that of light, requires the use of Einstein's special theory of relativity. The Doppler effect for light is very important in astronomy. It is the basis for the measurements of the radial velocities of distant galaxies. ### Example 10.1 What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm? **Solution** Since νλ = c, Δν/ν = Δλ/λ (for small changes in ν and λ). For we get [using Eq. (10.9)] - Δλ = 589.6 – 589.0 = + 0.6 nm - Δν/ν = v_radial/c = + Δλ/λ = 0.6/589.0 - v_radial = 0.6/589.0 * c = +3.06 x 105 ms-1 - v_radial = 306 km/s Therefore, the galaxy is moving away from us. ## Reflection and Refraction of Light - When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. - Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light. - When light travels from a rarer to a denser medium, the speed decreases. - No. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation. - In the wave picture of light, the intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light. - For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per unit time. ## Coherent and Incoherent Addition of Waves In this section, we will discuss the interference pattern produced by the superposition of two waves. You may recall that we had discussed the superposition principle in Chapter 15 of your Class XI textbook. Indeed the entire field of interference is based on the superposition principle according to which at a particular point in the medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves. ### Interference of Waves from Two Coherent Sources Consider two needles S₁ and S₂ moving periodically up and down in an identical fashion in a trough of water. They produce two water waves, and at a particular point, the phase difference between the displacements produced by each of the waves does not change with time; when this happens the two sources are said to be coherent. Since the distances S₁ P and S₂ P are equal, waves from S₁ and S₂ will take the same time to travel to the point P and waves that emanate from S₁ and S₂ in phase will also arrive, at the point P, in phase. Thus, if the displacement produced by the source S₁ at the point P is given by - y₁ = acos wt then, the displacement produced by the source S₂ (at the point P) will also be given by - Y₂ = acos wt Thus, the resultant of displacement at P would be given by - y = y₁ + y₂ = 2acos wt Since the intensity is the proportional to the square of the amplitude, the resultant intensity will be given by - I = 4I where I represents the intensity produced by each one of the individual sources; I is proportional to a². In fact at any point on the perpendicular bisector of S₁S₂, the intensity will be 4I. The two sources are said to interfere constructively and we have what is referred to as constructive interference. #### Constructive Interference The waves emanating from S₁ will arrive exactly two cycles earlier than the waves from S₂ and will again be in phase. Thus, if the displacement produced by S₁ is given by - y₁ = acos wt then the displacement produced by S₂ will be given by - Y₂ = acos (wt – 4π) = acos wt where we have used the fact that a path difference of 2λ corresponds to a phase difference of 4π. The two displacements are once again in phase and the intensity will again be 4I giving rise to constructive interference. #### Destructive Interference The waves emanating from S₁ will arrive exactly two and a half cycles later than the waves from S₂. Thus if the displacement produced by S₁ is given by - y₁ = acos wt then the displacement produced by S₂ will be given by - y₂ = acos (wt + 5π) = -acos wt where we have used the fact that a path difference of 2.5λ corresponds to a phase difference of 5π. The two displacements are now out of phase and the two displacements will cancel out to give zero intensity. This is referred to as destructive interference. **To summarise:** If we have two coherent sources S₁ and S₂ vibrating in phase, then for an arbitrary point P whenever the path difference, - S₁P ~ S₂P = ηλ (n = 0, 1, 2, 3,…) we will have constructive interference and the resultant intensity will be 4I; the sign ~ between S₁P and S₂ P represents the difference between S₁P and S₂ P. On the other hand, if the point P is such that the path difference, - S₁P ~ S₂P = (n + ½)λ (n = 0, 1, 2, 3,…) we will have destructive interference and the resultant intensity will be zero. #### Intensity at an Arbitrary Point Let the phase difference between the two displacements be φ. Thus, if the displacement produced by S₁ is given by - Y₁ = acos wt then, the displacement produced by S₂ would be - y₂ = acos (wt + φ) and the resultant displacement will be given by - Y = Y₁+Y₂ = a[cos wt + cos (wt + φ)] = 2acos (φ/2)cos (wt + φ/2) The amplitude of the resultant displacement is 2acos (φ/2) and therefore the intensity at that point will be - I = 4Icos²(φ/2) #### The Case of Incoherent Sources If ¢ = 0, ±2 π, ± 4 π,... which corresponds to the condition given by Eq. (10.10) we will have constructive interference leading to maximum intensity. On the other hand, if ¢ = ± π, ± 3π, ± 5π [which corresponds to the condition given by Eq. (10.11)] we will have destructive interference leading to zero intensity. Now if the two sources are coherent (i.e., if the two needles are going up and down regularly) then the phase difference φ at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time. However, if the two needles do not maintain a constant phase difference, then the interference pattern will also change with time and, if the phase difference changes very rapidly with time, the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. When this happens, we will observe an average intensity that will be given by - = 4I < cos²(φ/2) > where angular brackets represent time averaging. Indeed it is shown in Section 7.2 that if φ(t) varies randomly with time, the time-averaged quantity < cos²(φ/2) > will be 1/2. This is also intuitively obvious because the function cos²(φ/2) will randomly vary between 0 and 1 and the average value will be 1/2. The resultant intensity will be given by - I = 2I at all points. When the phase difference between the two vibrating sources changes rapidly with time, we say that the two sources are incoherent and when this happens the intensities just add up. This is indeed what happens when two separate light sources illuminate a wall. ## Interference of Light Waves and Young's Experiment We will now discuss interference using light waves. If we use two sodium lamps illuminating two pinholes, we will not observe any interference fringes. This is because of the fact that the light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase changes in times of the order of 10-10 seconds. Thus the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent, when this happens, as discussed in the previous section, the intensities on the screen will add up. ### Young's Experiment The British physicist Thomas Young used an ingenious technique to "lock" the phases of the waves emanating from S₁ and S₂. He made two pinholes S₁ and S₂ (very close to each other) on an opaque screen. S₁ and S₂ then behave like two coherent sources because light waves coming out from S₁ and S₂ are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S₁ and S2. Thus, the two sources S₁ and S₂ will be locked in phase; i.e., they will be coherent like the two vibrating needle in our water wave example. #### Interference Fringes Thus spherical waves emanating from S₁ and S₂ will produce interference fringes on the screen GG', as shown in Fig. 10.12(b). The positions of maximum and minimum intensities can be calculated by using the analysis given in Section 10.4 where we had shown that for an arbitrary point P on the line GG' to correspond to a maximum, we must have - S₂P – S₁P = ηλ; η = 0, 1, 2 ... Now, - d(S₂P)² - d(S₁P)² = [D² + (x + d/2)²] - [D² + (x - d/2)²] = 2xd where S₁S₂ = d and OP = x. Thus, - S₂P – S₁P ≈ 2xd/2D If x, d<<D then negligible error will be introduced if S₂P + S₁P (in the denominator) is replaced by 2D. For example, for d = 0.1 cm, D = 100 cm, OP = 1 cm (which correspond to typical values for an interference experiment using light waves), we have - S₂P + S₁P ≈ [(100)² + (1.05)²]1/2 + [(100)² + (0.95)²]1/2 ≈ 200.01 cm Thus if we replace S₂P + S₁P by 2 D, the error involved is about 0.005%. In this approximation, Eq. (10.16) becomes - S₂P - S₁P ≈ xd/D Hence we will have constructive interference resulting in a bright region when - x = x_n = ηλD/d ; n = 0, ±1, ± 2, ... On the other hand, we will have a dark region near - x = x_n = (n + ½)λD/d; n = 0, ±1, ± 2 Thus dark and bright bands appear on the screen, as shown in Fig. 10.13. Such bands are called fringes. Equations (10.18) and (10.19) show that dark and bright fringes are equally spaced and the distance between two consecutive bright and dark fringes is given by - β = x_n+1 - x_n = λD/d which is the expression for the fringe width. Obviously, the central point O (in Fig. 10.12) will be bright because S₁O = S₂O and it will correspond to n = 0. If we consider the line perpendicular to the plane of the paper and passing through O [i.e., along the y-axis] then all points on this line will be equidistant from S₁ and S₂ and we will have a bright central fringe which is a straight line as shown in Fig. 10.13. In order to determine the shape of the interference pattern on the screen we note that a particular fringe would correspond to the locus of points with a constant value of S₂P-S₁P. Whenever this constant is an integral multiple of λ, the fringe will be bright and whenever it is an odd integral multiple of λ/2 it will be a dark fringe. Now, the locus of the point P lying in the x-y plane such that S₂P - S₁P (= Δ) is a constant, is a hyperbola. Thus the fringe pattern will strictly be a hyperbola; however, if the distance Dis very large compared to the fringe width, the fringes will be very nearly straight lines as shown in Fig. 10.13. ## Diffraction If we look clearly at the shadow cast by an opaque object, close to the region of geometrical shadow, there are alternate dark and bright regions just like in interference. This happens due to the phenomenon of diffraction. Diffraction is a general characteristic exhibited by all types of waves, be it sound waves, light waves, water waves or matter waves. Since the wavelength of light is much smaller than the dimensions of most obstacles; we do not encounter diffraction effects of light in everyday observations. However, the finite resolution of our eye or of optical instruments such as telescopes or microscopes is limited due to the phenomenon of diffraction. We will now discuss the phenomenon of diffraction. ### The Single Slit In the discussion of Young's experiment, we stated that a single narrow slit acts as a new source from which light spreads out. Even before Young, early experimenters – including Newton – had noticed that light spreads out from narrow holes and slits. It seems to turn around corners and enter regions where we would expect a shadow. These effects, known as diffraction, can only be properly understood using wave ideas. After all, you are hardly surprised to hear sound waves from someone talking around a corner! When the double slit in Young's experiment is replaced by a single narrow slit (illuminated by a monochromatic source), a broad pattern with a central bright region is seen. On both sides, there are alternate dark and bright regions, the intensity becoming weaker away from the centre. To understand this, go Fig 10.15, which shows a parallel beam of light falling normally on a single slit LN of width a. The diffracted light goes on to meet a screen. The midpoint of the slit is M. A straight line through M perpendicular to the slit plane meets the screen at C. We want the intensity at any point P on the screen. As before, straight lines joining P to the different points L,M,N, etc., can be treated as parallel, making an angle θ with the normal MC. #### Calculation of Intensity at a Point P on the Screen The basic idea is to divide the slit into much smaller parts, and add their contributions at P with the proper phase differences. We are treating different parts of the wavefront at the slit as secondary sources. Because the incoming wavefront is parallel to the plane of the slit, these sources are in phase. The path difference NP – LP between the two edges of the slit can be calculated exactly as for Young's experiment. From Fig. 10.15, - NP - LP = NG = asin θ ≈ aθ. Similarly, if two points M₁ and M₂ in the slit plane are separated by y, the path difference M₂P – M₁P ≈ yd. We now have to sum up equal, coherent contributions from a large number of sources, each with a different phase. This calculation was made by Fresnel using integral calculus, so we omit it here. The main features of the diffraction pattern can be understood by simple arguments. At the central point C on the screen, the angle θ is zero. All path differences are zero and hence all the parts of the slit contribute in phase. This gives maximum intensity at C. Experimental observation shown in Fig. 10.15 indicates that the intensity has a central maximum at θ = 0 and other secondary maxima at θ ≈ (n + 1/2)λ/α, and has minima (zero intensity) at θ ≈ ηλ/α, n = ±1, ±2, 3, .... It is easy to see why it has minima at these values of angle. Consider first the angle θ where the path difference aθ is λ. Then, - αθ = λ/α. Now, divide the slit into two equal halves LM and MN each of size α/2. For every point M₁ in LM, there is a point M₂ in MN such that M₁M₂ = α/2. The path difference between M₁ and M₂ at P = M₂P – M₁P = αθ/2 = λ/2 for the angle chosen. This means that the contributions from M₁ and M₂ are 180° out of phase and cancel in the direction θ = λ/α. Contributions from the two halves of the slit LM and MN, therefore, cancel each other. Equation (10.22) gives the angle at which the intensity falls to zero. One can similarly show that the intensity is zero for θ = ηλ/α, with n being any integer (except zero!). Notice that the angular size of the central maximum increases when the slit width α decreases. It is also easy to see why there are maxima at θ = (n + 1/2)λ/α and why they go on becoming weaker and weaker with increasing n. Consider an angle θ = 3λ/2a which is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two thirds of the slit, the path difference between the two ends would be - 2α × θ = 2α × 3λ/2α = λ. The first two-thirds of the slit can therefore be divided into two halves which have a λ/2 path difference. The contributions of these two halves cancel in the same manner as described earlier. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima. Clearly, this will be much weaker than the central maximum (where the entire slit contributes in phase). One can similarly show that there are maxima at (n + 1/2) θ/α with n = 2, 3, etc. These become weaker with increasing n, since only one-fifth, one-seventh, etc., of the slit contributes in these cases. The photograph and intensity pattern corresponding to it is shown in Fig. 10.16. There has been prolonged discussion about difference between intereference and diffraction among scientists since the discovery of these phenomena. In this context, it is interesting to note what Richard Feynman has said in his famous Feynman Lectures on Physics. No one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between them. The best we can do is, roughly speaking, is to say that when there are only a few sources, say two interfering sources, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often used. In the double-slit experiment, we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole, and the double-slit interference pattern. This is shown in Fig. 10.17. It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference. The number of interference fringes occuring in the broad diffraction peak depends on the ratio d/a, that is the ratio of the distance between the two slits to the width of a slit. In the limit of a becoming very small, the diffraction pattern will become very flat and we will obsrve the two-slit interference pattern [see Fig. 10.13(b)]. ## Resolving Power of Optical Instruments In Chapter 9 we had discussed about telescopes. The angular resolution of the telescope is determined by the objective of the telescope. The stars which are not resolved in the image produced by the objective cannot be resolved by any further magnification produced by the eyepiece. The primary purpose of the eyepiece is to provide magnification of the image produced by the objective. Consider a parallel beam of light falling on a convex lens. If the lens is well corrected for aberrations, then geometrical optics tells us that the beam will get focused to a point. However, because of diffraction, the beam instead of getting focused to a point gets focused to a spot of finite area. In this case the effects due to diffraction can be taken into account by considering a plane wave incident on a circular aperture followed by a convex lens. The analysis of the corresponding diffraction pattern is quite involved; however, in principle, it is similar to the analysis carried out to obtain the single-slit diffraction pattern. Taking into account the effects due to diffraction, the pattern on the focal plane would consist of a central bright region surrounded by concentric dark and bright rings. A detailed analysis shows that the radius of the central bright region is approximately given by - r_0 ≈ 1.22λf/2a = 0.61λf/a where f is the focal length of the lens and 2a is the diameter of the circular aperture or the diameter of the lens, whichever is smaller. Typically if λ≈ 0.5 µm, f≈ 20 cm and a≈ 5cm we have - r_0 ~ 1.2 μm Although the size of the spot is very small, it plays an important role in determining the limit of resolution of optical instruments like a telescope or a microscope. For the two stars to be just resolved - fΔθ ≈ r_0 ≈ 0.61λf/a implying -_ Δθ ~ 0.61λ/a Thus Δθ will be small if the diameter of the objective is large. This implies that the telescope will have better resolving power if a is large. It is for this reason that for better resolution, a telescope must have a large diameter objective. **Example 10.6** Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch? **Solution** A 100 inch telescope implies that 2a = 100 inch = 254 cm. Thus if, - λ = 6000Å = 6 × 10-5 cm then - _ Δθ ~ (0.61 × 6 × 10-5)/127 ≈ 2.9 × 10-7 radians We can apply a similar argument to the objective lens of a microscope. In this case, the object is placed slightly beyond f, so that a real image is formed at a distance v. The magnification ratio of image size to object size is given by m v/f. It can be seen from Fig 10.20 that D/f≈ 2 tan β where 2β is the angle subtended by the diameter of the objective lens at the focus of the microscope. You can estimate the resolving power of your eye with a simple experiment. Make black stripes of equal width separated by white stripes; see figure here. All the black stripes should be of equal width, while the width of the intermediate white stripes should increase as you go from the left to the right. For example, let all black stripes have a width of 5 mm. Let the width of the first two white stripes be 0.5 mm each, the next two white stripes be 1 mm each, the next two 1.5 mm each, etc. Paste this pattern on a wall in a room or laboratory, at the height of your eye. Now watch the pattern, preferably with one eye. By moving away or closer to the wall, find the position where you can just see some two black stripes as separate stripes. All the black stripes to the left of this stripe would merge into one another and would not be distinguishable. On the other hand, the black stripes to the right of this would be more and more clearly visible. Note the width d of the white stripe which separates the two regions, and measure the distance D of the wall from your eye. Then d/D is the resolution of your eye. You have watched specks of dust floating in air in a sunbeam entering through your window. Find the distance (of a speck) which you can clearly see and distinguish from a neighboring speck. Knowing the resolution of your eye and the distance of the speck, estimate the size of the speck of dust. When the separation between two points in a microscopic specimen is comparable to the wavelength λ of the light, the diffraction effects become important. The image of a point object will again be a diffraction pattern whose size in the image plane will be - υ θ = υ (1.22λ/D) Two objects whose images are closer than this distance will not be resolved, they will be seen as one. The corresponding minimum separation, d_min, in the object plane is given by - d_min = [(1.22λ)]/(D/v) = 1.22 λv/D = 1.22 fλ/D Now, combining Eqs. (10.26) and (10.28), we get - d_min = 1.22λ/2 tan β. ## Diffraction Pattern of a Single Slit It is surprisingly easy to see the single-slit diffraction pattern for oneself. The equipment needed can be found in most homes - two razor blades and one clear glass electric bulb preferably with a straight filament. One has to hold the two blades so that the edges are parallel and have a narrow slit in between. This is easily done with the thumb and forefingers. Keep the slit parallel to the filament, right in front of the eye. Use spectacles if you normally do. With slight adjustment of the width of the slit and the parallelism of the edges, the pattern should be seen with its bright and dark bands. Since the position of all the bands (except the central one) depends on wavelength, they will show some colours. Using a filter for red or blue will make the fringes clearer. With both filters available, the wider fringes for red compared to blue can be seen. In this experiment, the filament plays the role of the first slit S in Fig 10.16. The lens of the eye focuses the pattern on the screen (the retina of the eye). With some effort, one can cut a double slit in an aluminium foil with a blade. The bulb filament can be viewed as before to repeat Young's experiment. In daytime, there is another suitable bright source subtending a small angle at the eye. This is the reflection of the Sun in any shiny convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle of (1/2)°. In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy. ## Resolving Power of Optical Instruments In Chapter 9 we had discussed about telescopes. The angular resolution of the telescope is determined by the objective of the telescope. The stars which are not resolved in the image produced by the objective cannot be resolved by any further magnification produced by the eyepiece. The primary purpose of the eyepiece is to provide magnification of the image produced by the objective. Consider a parallel beam of light falling on a convex lens. If the lens is well corrected for aberrations, then geometrical optics tells us that the beam will get focused to a point. However, because of diffraction, the beam instead of getting focused to a point gets focused to a spot of finite area. In this case the effects due to diffraction can be taken into account by considering a plane wave incident on a circular aperture followed by a convex lens. The analysis of the corresponding diffraction pattern is quite involved; however, in principle, it is similar to the analysis carried out to obtain the single-slit diffraction pattern. Taking into account the effects due to diffraction, the pattern on the focal plane would consist of a central bright region surrounded by concentric dark and bright rings. A detailed analysis shows that the radius of the central bright region is approximately given by - r_0 ≈ 1.22λf/2a = 0.61λf/a where f is the focal length of the lens and 2a is the diameter of the circular aperture or the diameter of the lens, whichever is smaller. Typically if λ≈ 0.5 µm, f≈ 20 cm and a≈ 5cm we have - r_0 ~ 1.2 μm Although the size of the spot is very small, it plays an important role in determining the limit of resolution of optical instruments like a telescope or a microscope. For the two stars to be just resolved - fΔθ ≈ r_0 ≈ 0.61λf/a implying -_ Δθ ~ 0.61λ/a Thus Δθ will be small if the diameter of the objective is large. This implies that the telescope will have better resolving power if a is large. It is for this reason that for better resolution, a telescope must have a large diameter objective. **Example 10.6** Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch? **Solution** A 100 inch telescope implies that 2a = 100 inch = 254 cm. Thus if, - λ = 6000Å = 6 × 10-5 cm then - _ Δθ ~ (0.61 × 6 × 10-5)/127 ≈ 2.9 × 10-7 radians We can apply a similar argument to the objective lens of a microscope. In this case, the object is placed slightly beyond f, so that a real image is formed at a distance v. The magnification ratio of image size to object size is given by m v/f. It can be seen from Fig 10.20 that D/f≈ 2 tan β where 2β is the angle subtended by the diameter of the objective lens at the focus of the microscope. You can estimate the resolving power of your eye with a simple experiment. Make black stripes of equal width separated by white stripes; see figure here. All the black stripes should be of equal width, while the width of the intermediate white stripes should increase as you go from the left to the right. For example, let all black stripes have a width of 5 mm. Let the width of the first two white stripes be 0.5 mm each, the next two white stripes be 1 mm each, the next two 1.5 mm each, etc. Paste this pattern on a wall in a room or laboratory, at the height of your eye. Now watch the pattern, preferably with one eye. By moving away or closer to the wall, find the position where you can just see some two black stripes as separate stripes. All the black stripes to the left of this stripe would merge into one another and would not be distinguishable. On the other hand, the black stripes to the right of this would be more and more clearly visible. Note the width d of the white stripe which separates the two regions, and measure the distance D of the wall from your eye. Then d/D is the resolution of your eye. You have watched specks of dust floating in air in a sunbeam entering through your window. Find the distance (of a speck) which you can clearly see and distinguish from a neighboring speck. Knowing the resolution of your eye and the distance of the speck, estimate the size of the speck of dust. When the separation between two points in a microscopic specimen is comparable to the wavelength λ of the light, the diffraction effects become important. The image of a point object will again be a diffraction pattern whose size in the image plane will be - υ θ = υ (1.22λ/D) Two objects whose images are closer than this distance will not be resolved, they will be seen as one. The corresponding minimum separation, d_min, in the object plane is given by - d_min = [(1.22λ)]/(D/v) = 1.22 λv/D = 1.22 fλ/D Now, combining Eqs. (10.26) and (10.28), we get - d_min = 1.22λ/2 tan β. ## Diffraction Pattern of a Single Slit It is surprisingly easy to see the single-slit diffraction pattern for oneself. The equipment needed can be found in most homes - two razor blades and one clear glass electric bulb preferably with a straight filament. One has to hold the two blades so that the edges are parallel and have a narrow slit in between. This is easily done with the thumb and forefingers. Keep the slit parallel to the filament, right in front of the eye. Use spectacles if you normally do. With slight adjustment of the width of the slit and the parallelism of the edges, the pattern should be seen with its bright and dark bands. Since the position of all the bands (except the central one) depends on wavelength, they will show some colours. Using a filter for red or blue will make the fringes clearer. With both filters available, the wider fringes for red compared to blue can be seen. In this experiment, the filament plays the role of the first slit S in Fig 10.16. The lens of the eye focuses the pattern on the screen (the retina of the eye). With some effort, one can cut a double slit in an aluminium foil with a blade. The bulb filament can be viewed as before to repeat Young's experiment. In daytime, there is another suitable bright source subtending a small angle at the eye. This is the reflection of the Sun in any shiny convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle of (1/2)°. In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.

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