Physics II Chapter L Solutions

Chapter L - Problems Blinn College - Physics 2426 - Terry Honan Problem L.1 Young's double slit experiment is performed by shooting a He-Ne laser beam (l = 632.8 nm) through two slits separated by 0.15 mm onto a screen 8 m away. (a) What is the distance between bright fringes on the screen? (b) What is the distance between dark fringes? (c) What is the distance from the center of the central maximum to the third dark fringe? (d) What is the smallest distance from the central maximum where the intensity is 3/4 its maximum? Solution to L.1 l = 632.8 µ 10-9 m , L = 8 m and d = 0.15 µ 10-3 m (a) d sin q = m l Assuming the small angle approximation sin q = y we get d L y L = m l. (The small angle approximation will be justified if the y value is much less than L.) The bright fringes are at integer values of m so between bright fringes we have Dm = 1. y=m lL lL ï Dy = Dm = 0.0337 = 3.37 cm d d Note that 0.0337 m ` 5 m. This justifies the small angle approximation. (b) The distance between dark bands is the same as the distance between bright fringes. 1 2 DJm + N = Dm = 1 ï Dy = 3.37 cm (c) Still assume the small angle approximation. For destructive interference (interference minimum) we get: 1 2 d sin q = Jm + N l ï d y L 1 2 = Jm + N l. 1 2 1 2 Note that the values of Jm + N are 0.5, 1.5, 2.5, ... , so the third minimum is Jm + N = 2.5 . 1 lL lL = 2.5 = 8.43 cm d 2 d y = Jm + N (d) The intensity formula is I = I0 cos2 K p d sin q O l Using the small angle approximation we get I = I0 cos2 K 3/4 the maximum intensity means that pdy lL O I = 0.75 . I0 cos-1 pdy I = ï y = 5.62 mm I0 lL 2 Chapter L - Problems cos-1 pdy I = ï y = 5.62 mm I0 lL Make sure your calculator is in the radians mode for this calculation. Problem L.2 700 nm light shines through a single slit of width 1.2 µ 10-5 m and shines on a screen 55 cm away. (a) What is the distance from the center to the third maximum? (Do not use the small angle approximation here? (b) Repeat part (a) using the small angle approximation. Solution to L.2 d = 1.2 µ 10-5 m, l = 700 µ 10-9 m and L = 55 cm (a) The central maximum is at y = 0. For constructive interference, using the small angle approximation, we get: d sin q = m l and m = 3 ï q = sin-1 tan q = y L ml = 10.079 ° d ï y = L tan q = 0.55 tan q = 0.0978 m (b) With the small angle approximation d y L = m l and m = 3 ï y = mlL = 0.963 m d Problem L.3 When 632.8 nm light shines through a diffraction grating the first bright line is observed at a deflection angle of 15°. What is the spacing between the slits in the grating? Solution to L.3 For a diffraction grating the condition for constructive interference is the same as for the single slit, d sin q = m l but there is destructive interference everywhere else. The first bright fringe corresponds to m = 1. d sin q = m l, m = 1, q = 15 ° and l = 632.8 µ 10-9 m ï d = 2.44 µ 10-6 m Problem L.4 The visible spectrum varies between lviolet = 400 nm and lred = 700 nm. Consider a diffraction grating with an arbitrary slit separation d. Show that for white light the m = 1 maximum gives pattern that doesn't overlap the m = 2 maximum. Also show that the m = 2 and m = 3 patterns always overlap. Solution to L.4 First consider the m = 1 and m = 2 maxima. d sin q = m l ï sin q = ml d Chapter L - Problems d sin q = m l ï sin q = 3 ml d It is sufficient to consider the extremes only. sin q1,red = 1 lred d = 2 lviolet 700 nm 800 nm and sin q2,violet = = d d d Thus for any d we have sin q1,red < sin q2,violet . Since sine is an increasing function (in the first quadrant) we can conclude that q1,red < q2,violet . Now consider the m = 2 and m = 3 maxima. sin q2,red = 2 lred d = 3 lviolet 1400 nm 1200 nm and sin q3,violet = = d d d Thus for any d we have sin q3,violet < sin q2,red , giving q3,violet < q2,red . Problem L.5 A diffraction grating has 300 lines ê mm. What is the highest order violet (400 nm) line seen? What is the highest order red line (700 nm). Solution to L.5 If there are 250 lines/mm, then the slit spacing is d= 1 mm. 300 1 µ 10-3 = 3.3333 µ 10-6 m = 3333.3 nm 300 Here the small angle approximation cannot be used. The largest angle is 90° so we have sin q § 1. The condition for seeing the mth bright fringe is: d sin q = m l and sin q § 1 ï sin q = ml d §1 ï m§ d l The highest order fringe is the largest m (an integer) that satisfies this inequality. l = 400 nm ï m § 3333.3 nm = 8.333 ï mmax = 8 400 nm (Note that since d ê l is a real number the distinction between ' § ' and '<' in m § d should not be important. In this problem it l matters; it changes the answer. Whether the answer should be 10 or 9 is a matter of interpretation and the problem is ambiguous.) l = 700 nm ï m § 3333.3 nm = 4.76 ï mmax = 4 700 nm Problem L.6 550 nm light passes through a single vertical slit of width 0.16 mm. What is the width of the central maximum observed on a screen 1.2 m away? 4 Chapter L - Problems Solution to L.6 l = 550 µ 10-9 m , L = 1.2 m and a = 0.16 µ 10-3 m The condition for destructive interference (dark fringes) for a single slit is a sin q = m l, where m = ± 1, ± 2, ± 3, .... Note that m = 0 is not a dark fringe; it is the center of the central bright fringe. The the central bright fringe is from m = -1 to m = +1. Use the small angle approximation. a sin q = m l ï a y L =ml ï y= mlL 1µ550µ10-7 µ1.2 = = 4.125 µ 10-3 a 0.16µ10-3 The width of the central maximum is 2 y = 8.25 mm . Problem L.7 When a He-Ne laser beam (l = 632.8 nm) passes through a narrow vertical slit a diffraction pattern is observed on a screen 80 cm away. If the distance from the first dark fringe to the third is 2.5 mm then what is the width of the slit? Solution to L.7 The condition for destructive interference with the small angle approximation becomes a sin q = m l ï a y=m y L =ml lL lL 632.8µ10-9 µ0.80 ï Dy = Dm ï 0.0025 = H3 - 1L ï a = 0.405 mm a a a Problem L.8 When light of wavelength l passes through a narrow vertical slit the highest order dark fringe observed (over all possible angles) is 5. (The fifth fringe can be seen and the sixth cannot.) What can one conclude about the width of the slit? Solution to L.8 The small angle approximation cannot be used. a sin q = m l. To do this use the fact that sin q § 1 for m = 5 and sin q ¥ 1 for m = 6. sin q = ml 5l 6l ï § 1 and ¥1 ï 5l§a§6l a a a Problem L.9 A soap film in air has a thickness of 120 nm. What visible wavelengths are strongly reflected by this film? Take the index of the soap film to be the same as for water, 1.33. The visible spectrum is between 400 nm and 700 nm. Solution to L.9 Strongly reflected implies constructive interference. The thickness of the film is t = 120 nm. The soap is between air layers on both sides so n = 1.333 > n£ = 1. Chapter L - Problems 5 Strongly reflected implies constructive interference. The thickness of the film is t = 120 nm. The soap is between air layers on both sides so n = 1.333 > n£ = 1. 1 l 2nt 2µ1.33µ120 nm 319.23 nm ï l= = = 1 1 1 2 n Jm+ N Jm+ N Jm+ N n > n£ and const. int. ï 2 t = Jm + N 2 2 2 1 2 For values of Jm + N = 0.5, 1.5, 2.5, ... we get a list of possible wavelengths. l = 638 nm, 213 nm, 128 nm, ... Since the visible spectrum is between 400 nm and 700 nm, the only visible wavelength in the list is 638 nm. Problem L.10 Red light (l = 650 nm) is normally incident on a thin oil layer (n = 1.25) that sits on a puddle of water. (a) What is the smallest nonzero thickness that will strongly reflect the red light? (b) What is the smallest nonzero thickness that will minimally reflect the red light? Solution to L.10 l = 650 nm. The index of the film is n = 1.25, which is larger than that of water In£ = 1.33M, so n < n£ (a) Strongly reflect implies constructive interference 2nt=ml ï t=m l = m µ 260 nm 2n The values of m are 0, 1, 2,... so the smallest nonzero thickness is when m = 1. tmin = 260 nm (b) Minimally reflect implies destructive interference 1 l 1 = Jm + N µ 260 nm 2 2n 2 1 2 2 n t = Jm + N l ï t = Jm + N Since the values of m + 1 1 are 0.5, 1.5, 2.5, ... . It follows the the minimum is when m + = 0.5. 2 2 tmin = 130 nm Problem L.11 You are asked to design an antiglare coating for a computer monitor. Sitting on the glass (n = 1.50) screen is a thin coating of magnesium fluoride with an index of refraction of 1.38. To be an antiglare coating it must satisfy two conditions. It must give destructive interference for light in the middle of the visible spectrum, at 550 nm. Secondly, there must be no constructive interference for any wavelength in the visible spectrum, from 400 nm to 700 nm. First find the different thicknesses that give the destructive interference and then use the lack of consructive interference to select the thickness uniquely. Solution to L.11 Here we have n = 1.38 for the coating and the glass is n£ = 1.50. For destructive interference at 550 nm: 1 2 1 l 1 = Jm + N 199.275 nm 2 2n 2 n < n£ and dest. int. ï 2 n t = Jm + N l. ï t = Jm + N 6 Chapter L - Problems 1 2 1 l 1 = Jm + N 199.275 nm 2 2n 2 n < n£ and dest. int. ï 2 n t = Jm + N l. ï t = Jm + N 1 2 1 3 5 7 , , , ... ï t = 99.637 nm, 298.91 nm, 498.19 nm, ... 2 2 2 2 Jm + N = , Thus the minimum thickness is t0 = 99.637 nm and the others are 3 t0 , 5 t0 , 7 t0 , .... Now consider the constructive interference condition for the smallest thickness t0 = 99.637 nm n < n£ and dest. int. ï 2 n t0 = m l ï l = 2 n t0 m = 275 nm m m = 1, 2, 3, ... .. ï l = 276 nm, 138 nm, 92 nm, ... . The visible spectrum is between 400 nm and 700 nm, so there are no visible wavelengths that are intensified. It should also be clear that if we choose other thicknesses that give destructive interference 3 t0 , 5 t0 , 7 t0 , ..., then we will multiply the above wavelength list by 3, 5, 7, 9, .. . This will always give constructive interference at some visible wavelength. 99.6 nm The unique thickness is thus Problem L.12 Unpolarized light with an intensity of 4000 W ë m2 passes through three polarizing filters, the first is at a 15° angle from vertical, the second is 40° from vertical and the third is horizontal. What are the intensities between the first and second filters, between the second and third filters and after the third filter? Solution to L.12 When polarized light with intensity I0 passes through a polarizing filter it leaves with half the intensity I = I0 ë 2 and is polarized along the axis of the filter. It follows that the intensity between the first and second filters is: I= 1 1 W I0 = 4000 = 2000 , 2 2 m2 and this is polarized at q1 = 15 °. When polarized light with intensity I0 passes through a filter it leaves polarized along the axis of the filter and with intensity I = I0 cos2 q, where q is the angle between the polarization of the light and the axis of the filter. After the second filter at q2 = 40 ° the intensity is I £ = I cos2 Hq2 - q1 L = 2000 cos2 H40 ° - 15 °L = 1644.8 = 1645 W m2 , The same analysis gives the intensity after the third filter. I ″ = I £ cos2 Iq3 - q2 M = 1644.8 cos2 H90 ° - 40 °L = 679 W m2 , Problem L.13 When the sun is what angle above the horizon will sunlight reflected off a still pond be totally polarized? Solution to L.13 The polarization angle for light from air reflecting off water is q p . Chapter L - Problems The polarization angle for light from air reflecting off water is q p . n 1.33 tan q p = 2 = ï q p = 53.1 ° n1 1 This angle is the incident angle q1 which is is measured from the normal. The angle above the horizon is its complement. 90 ° - 53.1 ° = 36.9 ° 7