For visible light, 700nm(red) > λ > 400 nm(indigo) See page 631 for wavelengths of colours

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1 1 Light Section 20.5, Electromagnetic waves Light is an electromagnetic which travels at m/s in a vacuum. It obeys the relationship: λf = c just like other traveling waves. For visible light, 700nm(red) > λ > 400 nm(indigo) See page 631 for wavelengths of colours Index of refraction = n = Speed of light in vacuum = c/v See table 23.1, page 722 Speed of light in medium Recall that the velocity of light is different in different media. In Diamond it is about 40% as fast as in a vacuum. When light moves from one medium to another, its velocity changes and its wavelength changes, but its frequency does not change. This is like having a transverse wave travel from a thin rope to a thick rope. When a light wave travels from one medium to another, its wavelength changes. Let λ be the wavelength in vacuum. In a medium with index of refraction, n, the wavelength is λ n. λ = c/f; λ n = v n /f ; then λ n /λ = v n /c = 1/n and λ n = λ/n Section 20.7 Doppler effect for light For a receding source, λ = (1 + v s /c)/(1 - v s /c) λ o For an approaching source, λ = (1 - v s /c)/(1 + v s /c) λ o Section 21.6 Thin film optical coatings Red shift Blue shift Light originating from other galaxies is red shifted. This provides evidence for the Big Bang- the time when the universe started expanding from a hot dense source. This happened about 14 billion years ago. An electromagnetic wave undergoes a phase change of 180 o upon reflection from a medium that has a higher index of refraction than the one in which the wave is travelling. When light is reflected from a mirror it undergoes a phase change of 180 o. Phase change of 180 o No phase change Air medium(n>n air ) air medium (n>n air ) air (Hard Reflection) (Soft Reflection)

2 2 In Physics 101, we are not going to discuss the Physics of this phase change. The mechanical wave analogy is the reflection of waves on a string which is rigidly attached to the wall as compared to one which can move freely up and down. Vs Consider a thin film of uniform thickness t and index of refraction n with light rays incident from nearly normal to the two surfaces of the film. 180 phase No phase change change This can be a model for soap bubbles or peacock feathers. n air air t Note that here we are drawing the light waves at an angle nearly normal to the surface. Otherwise the path length would not be 2t and the path lengths of the two rays outside the surface would be different When a light wave travels from one medium to another, its frequency does not change but its wavelength changes: λ n = λ/n where λ is the wavelength in free space. (or air where n= ) Applying the above rules, the condition for constructive interference of the two waves is that the phase difference between the two outgoing waves is 2π Phase difference between the two waves = Phase change in film plus phase change upon reflection = φ = 2π r/λ + φ o For constructive interference, φ = m2π; for destructive interference φ = (m + ½ )2π For the above film: φ = (2π) 2t/λ n + π = 2π(2t/λ n + ½) This is all you need to know to solve this kind of problem. Then for constructive interference here: 2π(2t/λ n + ½) = m2π, m =0, 1, 2, divide by 2π to get: 2t/λ n = (m + ½), m =0, 1, 2, 3,.. These values make the phase change a multiple of 2π. Note that m is always positive (or 0) since t > 0. Note we could also have used 2nt = (m - ½)λ

3 3 Or 2nt =(m + ½)λ And for destructive interference: 2t/λ n = m, m = 1, 2, 3.. Or 2nt = mλ These values make the phase change an odd multiple of π. For these problems, it is most important to understand the principles. You need not try to memorise the formulae for each situation but rather you should work out each situation using λ n = λ/n and the π phase change law. Problem: A thin film of oil (n=1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film appears red (640)nm and has no blue (512nm) color. How thick is the oil film? 1 2 t Oil Water We assume that the light doesn t reflect from the dark surface of the pavement Note that the refractive index of water is from table 23.1, hence the light wave undergoes a 180 phase change at both surfaces, hence the phase shift is entirely determined by the path length difference. φ = 2π r/λ + φ o φ o = 0 Then, the phase difference between path 1 and path 2 for constructive interference (red light) is 2π(2t/(λ(red)/n oil )); for constructive interference, this phase difference is m2π so: 2t = mλ(red)/n oil ; t = mλ(red)/2n oil = m(640nm)/2.5 = m(256nm) = 256nm, 512nm, 768nm,.. and for destructive interference (blue light) 2t = (m+ ½)λ(blue)/n oil ; t = (m+½) (512nm)/2.50 = (m+½) (204.8nm) = 102.4nm, 307.2nm, 512nm Then t = 512 nm

4 4 Problem: An air wedge is formed between two glass plates separated at one edge by a very fine wire. When the wedge is illuminated from above by 600nm light, 30 dark fringes are observed across the plates. Calculate the radius of the wire. 1 2 y t Note that since the top and bottom plates have flat parallel surfaces, they result in uniform phase shifts of the light waves. Only the air between the plates produces a phase shift which varies along the horizontal direction. n glass > n air, therefore there is a π phase shift when light reflects from the bottom surface of the air cavity but no phase shift when light reflects from the top surface of the air cavity. Recall φ = 2π r/λ + φ o Beam 1 undergoes no phase shift relative to the incident beam due to reflection, Beam 2 has a π phase shift. Phase shift between the two beams from the air gap: φ = 2y/(λ/n)*2π + π = [2ny/λ](2π) + π = (m + ½)2π for destructive interference (dark fringe). This simplifies to 2ny/λ = m Note that y varies from 0 to t, the thickness of the wire, since the phase shift is π for y=0, the pattern starts with a dark fringe at the left. As we move from left to right, 30 fringes, or 29 times 2π phase shift have occurred. Then 2nt/λ = 29 and t = 29λ/(2n) = 29(600x10-9 m)/(2 1.00) = 8.7x10-6 m. Then the radius of the wire is 4.35 µ A typical human hair is 50µ.

5 5 Section 22-6: The Michelson Interferometer The Michelson interferometer is used to measure light wavelengths or other lengths with great precision. M 1 M S Beam splitter L 1 M 1 - fixed mirror M S L 2 M 2 M 2 - Adjustable mirror Light source (coherent, monochromatic) Observer If M 2 is not exactly perpendicular to M 1, the observer observes a fringe pattern. If they were exactly perpendicular, the image would be uniform. At any point in the image at the observer, the intensity undergoes one complete cycle as M 2 is moved by λ/2. (since light travels a distance 2L 2 ). Problem: Light of wavelength nm is used to calibrate a Michelson interferometer. How many dark fringes are counted as the mirror is moved 0.180mm? Note that the Michelson Answer: # fringes = L/λ/2 = 2(0.180x10-3 m)/(550.2x10-9 interferometer can be m) = = 654 used to measure index of refraction (Mastering Chapter 22: Wave Optics Physics) Huygens' Principle: Every point in a wavefront can be considered as a source of tiny wavelets that spread out spherically at the speed of the wave itself. At a later time the shape of the wave front is the line tangent to all the wavelets Examples: Diffraction: A simple application of Huygens' law. At each point in the wave front, tiny wavelets move forward at the wave velocity with a spherical wavefront. This is a wave property, particles. The property of light 'bending' around corners leads to interference and diffraction.

6 6 Section 22-2 Interference of light Young's Double Slit Experiment Consider a system consisting of light source, a double slit and a viewing screen r 1 Light Source d y L r 2 dsin If L >>d, the angles from each ray are equal to. Conditions for Interference (or light waves) For observation of interference effects with light waves, the light source should be coherent (i.e. constant phase relationship) and monochromatic (i.e. same wavelength). Interference fringes are seen on viewing screen at a distance y from the centre line. If we assume that r 1 and r 2 are parallel (note L>> d), then when r = r 2 - r 1 = dsin = mλ m = 0, ±1, ±2,.. there is constructive interference. when r = r 1 - r 2 = dsin = (m+½)λ m = 0, ±1, ±2,.. there is destructive interference. tan = y/l m is the order of the fringe. The intensity of the fringes is greatest for the central fringe (m=0) and decreases for higher orders. This is due to diffraction, which we discuss later. Problem: A Young's double slit experiment uses 589 nm light and a slit to screen distance of 2.00m. The 10 th interference minimum is observed 7.26 mm from the central maximum. What is the spacing of the slits? y

7 7 Interference minima occur when dsin = (m+½)λ, m= 0,±1, ±2,.. The 10 th interference minimum occurs when m = +9; tan = y/l; Note that the peaks are quite Then = tan -1 (y/l) = tan -1 [(7.26x10-3 close to each other- you m)/(2.00m)] = rad might need glasses Then d = (9+½)λ/sin = 9.5(589x10-9 m)/sin(0.0036) = m = 1.55mm Intensity in the Double Slit Interference Pattern Consider two light beams passing through the double slit. They have electric fields given by P D 1 = D M sin(ωt) and D 2 = D M sin(ωt + φ) D 1 We are covering the spatial phase dependence by inserting the relative phase difference φ. D 2 dsin The phase difference φ is 2π times the difference in path lengths (d sin) divided by λ i. e. φ = 2πdsin/λ The phase difference passes through 2π radians along each wavelength. This is exactly analogous to the speaker problem At the screen, we measure the superposition of the two beams Recall the identity sin (a) + sin (b) = 2cos[(a-b)/2] sin[(a+b)/2] Then at P, D = D 1 + D 2 = 2D M cos( φ/2)sin(ωt + φ/2) The intensity of a light wave is proportional to the square of the wave amplitude (recall that I = ½ρω 2 D M 2 v for a single traveling wave) So the light intensity, I, at point P is proportional to D 2 = 4D M 2 cos 2 ( φ/2)sin 2 (ωt + φ/2) The time-average of sin 2 (ωt + φ/2) is ½. This variation of intensity with time at ~ Hz (= c/λ= 3x10 8 m/s/600x10-9 m = 5x10 14 ) is not visible. Note that I 1 is the intensity arising from a single slit. The factor ½ from I the sin 2 = 4I 1 cos 2 ( φ/2) or 4I 1 cos 2 (πdsin/λ) using φ = 2πdsin/λ. average is included in I 1. Going back to the diagram for the Young's Double slit experiment, we see that for large L and small y, we can write sin = y/l then φ = 2πdy/λL and I = 4I 1 cos 2 (πdy/λl). This expression enables us to calculate the intensity anywhere in the double slit interference pattern; before we could only find the maxima and the minima. We will learn soon that due to diffraction that the central maximum is the most intense and the others fall off with distance from the central maximum. This is due to interference effects between light rays going through the same slit.

8 8 φ -4π -3π -2π -π 0 π 2π 3π 4π Intensity -2λL/d -3/2λL/d -λl/d -λl/2d 0 λl/2d λl/d 3/2λL/d 2λL/d PHI(rad) Plot of theoretical intensity in the Young s double slit measurement Section 22-3 The diffraction grating A diffraction grating contains a very large number of slits produced by a large number of parallel lines on a glass plate. Gratings can contain 10,000 lines per cm. Coherent monochromatic light source d dsin On the screen there are well defined peaks where dsin = mλ, m=0,1,2,3,4.. In contrast to the maxima observed for single slits and Young's two slit experiment, the maxima observed for light passing through a diffraction grating are sharper and narrower.

9 9 Two slits: Many slits: Recall that for the two slit experiment, the maximum intensity is 4I 1. For the diffraction grating, the intensity of the peaks is N 2 I 1. N is a very large number! Because the diffraction grating gives rise to much narrower lines, we can separately observe maxima from light rays of different wavelengths. Recall sin = mλ/d All wavelengths contribute to the same m= 0 line which occurs at the centre of the viewing screen. However, for the higher order peaks, light rays of different wavelength produce different maxima. Problem: Light of wavelength 440 nm produces a first order peak displaced 10 cm above the central maximum on the viewing screen which is located 25 cm away from the diffraction grating. (a) How many lines/cm are on the grating? (b) How many orders of peak can we observe from this light ray? 25 cm 10cm (a) Use dsin = mλ for a maximum. Here tan = 10cm/25cm = 0.4 = 21.8 o. Then d = mλ/sin = 440nm/sin(21.8 o ) = 440x10-9 m/sin (21.8 o ) = m This corresponds to the line spacing, then the number of lines/m is 1/d = lines/m = 8,440 lines/cm. (b) for the second order peak, sin = 2λ/d = 2( m)/( m) = 0.745, for the third order peak, sin = 3λ/d = 1.1 Not possible, therefore, only two orders can be observed from this light ray. The diffraction grating can be used as a spectrometer to measure the wavelength of light from the angle at which the light ray is diffracted. Source A diffraction spectrometer

10 10 Section 22.4 Single - Slit Diffraction 3λ/a 2λ/a Single slit of thickness, a. λ/a 0 -λ/a -2λ/a -3λ/a sin The pattern of intensity on the viewing screen has a central bright fringe flanked by much weaker maxima alternating with dark fringes. Consider a single slit. Previously, we treated each slit as a point (or line) source of light. Each portion of the slit acts as a source of light waves. a/2 There is a viewing screen far off in the distance. a a/2 The path difference of a/2sin holds for the comparison of any light ray from the bottom half with its counterpart in the upper half. a/2sin

11 11 Divide the slit into two halves. The path difference between light rays in the upper half and the lower half is a/2sin. We increase from zero until we find the first minimum due to π phase shift between parallel beams. Destructive interference occurs when a/2sin = λ/2 or sin = λ/a Divide the screen into four equal parts: Then the path difference is a/4sin and destructive interference occurs when a/4sin = λ/2 which yields sin = 2λ/a The general condition for destructive interference is sin = mλ/a. This determines the positions of the fringes. The complete intensity pattern for single slit diffraction follows the following relationship which will not be derived here: I = I max [sin(πasin/λ)] 2 [πasin/λ] 2 Intensity PHI (rad) Theoretical plot of intensity from a single slit due to diffraction

12 12 Problem A screen is placed 50.0 cm from a single slit, which is illuminated with 690 nm light. If the distance between the first and third minima in the diffraction pattern is 3.00mm what is the width of the slit? Third minimum L First minimum Then the first minimum occurs at y = 1.5mm. The first minimum occurs at sin = λ/a and the third occurs at sin = 3λ/a. So, 2λ/a = 3.00 mm. using the small angle approximation since L>>a The displacement along y is Ltan and the spacing between minima is 1.5mm Then tan = 1.50x10-3 m/50.0x10-2 m = rad. Then we can use the first minimum position to derive a = λ/sin = 690x10-9 m/ = 2.30x10-4 m or 0.230mm Diffraction in the Double Slit Experiment In the double slit experiment, we calculated the intensity of fringes as a function of. I = I max cos 2 [πdsin/λ] d a a Maxima occur at sin = mλ/d Minima occur at sin = (m + ½ )λ/d In fact, this was not rigorously correct. It turns out that due to the finite thickness of each slit this intensity function must be multiplied by the diffusion correction. [Sin(πasin/λ)] 2 [πasin/λ] 2 where a is the thickness of the slit.

13 13 Minima occur at sin = mλ/a, Hence the combined pattern looks like: Intensity PHI (rad) Theoretical intensity for light passing through two slits. The higher frequency oscillations are determined by the distance between the two slits and peaks occur at sin = mλ/d where d is the spacing between the two slits. The diffraction minima occur where sin = mλ/a where a is the thickness of each slit. Note that d is always larger than a.

14 14 Problem from Exam Final 2006: A screen is placed 200 cm behind two narrow slits that are separated by 0.480mm. The figure shows the two slit interference pattern not including the effects of diffraction. (a) What is the wavelength of the light? (b) Consider point A on the screen. What is the path difference between the distance from A to one slit and the distance to the other slit? (c) A film of index of refraction n= 1.5 is now placed in front of one of the slits in the above setup. What should be the thickness, t, of the film so that point A moves to the central axis? 1 cm A y=0 (d) Light of wavelength 600nm is emerging from a single rectangular slit of width 0.1mm. A screen of width.1m is placed at L=2.0m from the slit. How many diffraction minima will appear on the screen? (e) 0.1 mm 0.10m (a) dsin = mλ; then λ = dsin/m; sin = tan = y/l; then λ = dy/l/m = 0.480x10-3 m*.0025m/2.00m = 600nm (b) phase difference = 2π r/λ = 4π; r= 2λ = 1200nm (c) We need a phase change of 4π. The phase change with path length is φ in air = 2π(t/λ); φ in film = 2π(nt/λ) so the phase difference is: 2π(nt/λ- t/λ) = 4π; then t/λ(n-1) = 2; t = 2λ/(n-1) = 2400nm (d) diffraction minima occur at asin = mλ or ay/l = mλ; then: y = mλl/a = m(.012) then we have minima at ±0.012m, ±0.024m, ±0.036m, ±0.048m. or 8 minima.

15 15 Section 22-5, 23-8 Circular Aperture diffraction and resolution The ability of optical systems to distinguish between closely spaced objects is limited because of the wave nature of light. The opening represents the lens of a camera or our eye. As the two light sources move closer together, the diffraction patterns overlap and we can no longer separate the two images. When the central maximum of one diffraction pattern falls on the first minimum of the other diffraction pattern, the images are said to be just resolved. This limiting condition of resolution is known as the Rayleigh criterion. The first minimum occurs when sin = λ/a as discussed above. For λ<<a which holds in most instances sin ~ and min = λ/a

16 16 For a circular aperture, it can be shown that min = 1.22λ/D where D is the diameter of the aperture. Note also, from your text, page 700, the width of the central maximum of the diffraction pattern from a circular aperture is: = 2*1.22λ/D = 2.44λ/D. or width on screen, w = L = 2.44λL/D. Resolution of the eye: At a distance of 25 cm, what is the minimum separation of light sources that the eye can resolve? The pupil diameter is approximately 2mm. The centre of the spectrum of visible light is 500 nm. min = 1.22λ/D = 1.22(500x10-9 m)/2x10-3 m = 3. x10-4 rad 25 cm is about as close as we can focus our eyes. s Pupil diameters vary from 1mm to 8mm. The eye is most sensitive to 550nm light. L Minimum separation s = L = (0.25m)(3x10-4 ) = 8x10-5 m = 0.08 mm - the thickness of a human hair. Problem: When Mars is nearest the earth, the distance separating the two planets is 88.6x10 6 m. Mars is viewed by a telescope whose mirror has a diameter of 30.0cm. (a) for light of wavelength 590nm, what is the angular resolution of the telescope? (b) What is the smallest distance that can be resolved on Mars? The mirror is equivalent to the aperture of the telescope. Then min = 1.22λ/D = 2.40x10-6 rad. The smallest distance that can be resolved on Mars is then s = L min = (88.6x10 6 m)(2.40x10-6 ) = 213m. Could you see life on Mars with this resolution?