Chapter 27 - Interference and the Wave Nature of Light PDF

Physics Cutnell & Johnson Chapter 27 Interference and Diffraction Copyright ©2022 John Wiley & Sons, Inc. Is light a particle or a wave? A little review… Copyright ©2022 John Wiley & Sons, Inc. 2 27.1 The Principle of Linear Superposition (1 of 3) We kn...

Physics Cutnell & Johnson Chapter 27 Interference and Diffraction Copyright ©2022 John Wiley & Sons, Inc. Is light a particle or a wave? A little review… Copyright ©2022 John Wiley & Sons, Inc. 2 27.1 The Principle of Linear Superposition (1 of 3) We know that light bends when light enters another medium… but what happens when light enters more light? If light is a particle, we expect two beams to bounce off each other But this does not happen… light is probably not a particle Copyright ©2022 John Wiley & Sons, Inc. 3 27.1 The Principle of Linear Superposition (1 of 3) We know that light bends when light enters another medium… but what happens when light enters more light? More light No light More light No light More light Light is a wave! Like sound, light will interfere with itself! We should see constructive and destructive interference Copyright ©2022 John Wiley & Sons, Inc. 4 27.1 The Principle of Linear Superposition (1 of 3) But why do waves have constructive or destructive interference? Path length difference! Like sound… in Chapter 17 More light No light More light No light More light Light is a wave! Like sound, light will interfere with itself! We should see constructive and destructive interference Copyright ©2022 John Wiley & Sons, Inc. 5  Review of Chapter 17 17.1 The Principle of Linear Superposition (3 of 3) The Principle of Linear Superposition When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves. Copyright ©2022 John Wiley & Sons, Inc. 6  Review of Chapter 17 17.2 Constructive and Destructive Interference of Sound Waves (8 of 8) constructive interference: destructive interference: Copyright ©2022 John Wiley & Sons, Inc. 7  Review of Chapter 17 17.2 Constructive and Destructive Interference of Sound Waves (8 of 8) constructive interference: destructive interference: Copyright ©2022 John Wiley & Sons, Inc. 8  Review of Chapter 17 17.2 Constructive and Destructive Interference of Sound Waves (6 of 8) Calculate the path length difference. 2 2 = 3.20 m   2.40 m   2.40 m 1.60 m Calculate the wavelength. −1 =343 𝑚. 𝑠 v 343m s −1  f 214 Hz 1.60 m = =214 𝑠 For two wave sources vibrating in phase, a difference in path lengths that is zero or an integer number (1, 2, 3,.. ) of wavelengths leads to constructive interference; LOUD SOUND Copyright ©2022 John Wiley & Sons, Inc. 9 27.1 The Principle of Linear Superposition (1 of 3) When two or more light waves pass through a given point, their electric fields combine according to the principle of superposition. 𝐿2 − 𝐿1 𝐿2 𝐿1 Bright band The waves emitted by the sources start out in phase and arrive at point P in phase, leading to constructive interference. 𝐿2 − 𝐿1 =m 𝜆 ;𝑚=0 , 1 , 2 , 3 ,… Copyright ©2022 John Wiley & Sons, Inc. 10 27.1 The Principle of Linear Superposition (2 of 3) The waves emitted by the sources start out in phase and arrive at point P out of phase, leading to destructive interference. 𝐿2 − 𝐿1 𝐿2 𝐿1 (1 ) 𝐿2 − 𝐿1 = m+ 𝜆 ;𝑚=0 , 1 , 2 , 3 ,… 2 dark band Copyright ©2022 John Wiley & Sons, Inc. 11 27.1 The Principle of Linear Superposition (3 of 3) If constructive or destructive interference is to continue ocurring at a point, the sources of the waves must be coherent sources. Two sources are coherent if the waves they emit maintain a constant phase relation (Frequency and wavelength must be the same). Copyright ©2022 John Wiley & Sons, Inc. 12 27.2 Young’s Double Slit Experiment (1 of 7) In Young’s experiment, two slits acts as coherent sources of light. Light waves from these slits interfere constructively and destructively on the screen. Copyright ©2022 John Wiley & Sons, Inc. 13  27.2 Young’s Double Slit Experiment (2 of 7) 𝐿2 − 𝐿1 =m 𝜆 ;𝑚=0, 1, 2 , 3 ,… ( ) 𝐿2 − 𝐿1 = m+ 1 2 𝜆 ;𝑚=0 , 1 , 2 , 3 ,… The waves interfere constructively or destructively, depending on the difference in distances between the slits and the screen. Copyright ©2022 John Wiley & Sons, Inc. 14 27.2 Young’s Double Slit Experiment (3 of 7) We only observe the interference pattern on the screen Copyright ©2022 John Wiley & Sons, Inc. 15  27.2 Young’s Double Slit Experiment (4 of 7) Path length difference triangle () d = distance between two slits (slit seperation) d sin 𝜃=(𝐿¿ ¿2− 𝐿1 )/ d¿ Bright fringes of a double-slit Dark fringes of a double-slit 𝐿2 − 𝐿1 =m 𝜆 ;𝑚=0, 1, 2 , 3 ,… ( ) 𝐿2 − 𝐿1 = m+ 1 2 𝜆 ;𝑚=0 , 1 , 2 , 3 ,… d sin 𝜃=( m+ ) 𝜆 ;𝑚=0 , 1 , 2 ,3 ,… 1 dsin 𝜃=m 𝜆 ;𝑚=0,1,2 ,3 ,… 2 Copyright ©2022 John Wiley & Sons, Inc. 16  Example 1 Young’s Double-Slit Experiment Red light (664 nm) is used in Young’s experiment with slits separated by 120. The screen is located a distance 2.75 m from the slits. (a) Find the angle of the central and the third-order bright fringe. (b) Find the distance between the two bright fringes 𝜆=664×10 𝑚 (a)sin 𝜃= m 𝜆 ;𝑚=0 ,1 ,2 , 3 , … −9 𝑑=120×10 𝑚 −6 d 𝑚𝑐𝑒𝑛𝑡𝑟𝑒 =0 𝑚3 𝑟𝑑 =3 𝜃=sin −1 m 𝜆 d ( ) ( ) ( ) −9 (0)(664 × 10 −9 𝑚 ) (3)(664 × 10 𝑚 ) 𝜃 0=? 𝜃0=sin− 1 𝜃 3=sin − 1 120× 10 −6 𝑚 120× 10 − 6 𝑚 𝜃 3=? ∘ ∘ 𝜃 0= 0 𝜃 3= 0.951 Copyright ©2022 John Wiley & Sons, Inc. 17  Example 1 Young’s Must be able to Double-Slit Experiment calculate distance between any two fringes! Red light (664 nm) is used in Young’s experiment with slits separated by 120. The screen is located a distance 2.75 m from the slits. (a) Find the angle of the central and the third-order bright fringe. (b) Find the distance between the two bright fringes 𝜆=664×10 𝑚 (a)sin 𝜃= m 𝜆 ;𝑚=0 ,1 ,2 , 3 , … −9 𝑑=120×10 𝑚 −6 d 𝑚𝑐𝑒𝑛𝑡𝑟𝑒 =0 𝑚3 𝑟𝑑 =3 𝜃=sin −1 m 𝜆 d ( ) ( ) ( ) −9 (0)(664 × 10 −9 𝑚 ) (3)(664 × 10 𝑚 ) 𝜃 0=? 𝜃0=sin− 1 𝜃 0=sin − 1 120× 10 −6 𝑚 120× 10 − 6 𝑚 𝜃 3=? ∘ ∘ 𝜃 0= 0 𝜃 3= 0.951 (b) 𝑦 3 − 𝑦 0¿0.04 =? 6m−0m¿0.04 6m 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒¿ 𝑦 𝑡𝑎𝑛 𝜃= 𝑦3 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐿 𝜃3 𝑦 0=𝐿𝑡𝑎𝑛 𝜃¿0(2.75𝑚)𝑡𝑎𝑛(0¿∘ )0 m 𝐿 𝑦 3=𝐿𝑡𝑎𝑛 𝜃¿3(2.75𝑚)𝑡𝑎𝑛(0.951∘) m Copyright ©2022 John Wiley & Sons, Inc. 18 Conceptual Example 2 White Light and Young’s Experiment The figure shows a photograph that illustrates the kind of interference fringes that can result when white light is used in Young’s experiment. (a) Why does Young’s experiment split white light into its constituent colors? (b) In any group of colored fringes, such as the two singled out, why is red farther out from the central fringe than green is? (c)Why is the central fringe white? m𝜆 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 sin 𝜃= ;𝑚=0 ,1 ,2 , 3 , … (b) 𝑡𝑎𝑛 𝜃= (a) 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 d 𝜃=sin −1 ( ) m𝜆 d larger (red) larger distance Larger (red) larger smaller (green) smaller smaller (green) smaller distance (c) At centre, , so for 𝑦 wavelengths (all colours) 𝜃 𝐿 All colours together = white Copyright ©2022 John Wiley & Sons, Inc. 19 27.3 Thin Film Interference (1 of 7) Light can make a similar interference pattern when it light is both reflected and refracted through a thin film, ex. Soap bubble or petrol stain… Copyright ©2022 John Wiley & Sons, Inc. 20 27.3 Thin Film Interference (1 of 7) Interference = Path length difference! Copyright ©2022 John Wiley & Sons, Inc. 21 27.3 Thin Film Interference (1 of 7) Copyright ©2022 John Wiley & Sons, Inc. 22  27.3 Thin Film Interference (1 of 7) Because of reflection and refraction, two light waves enter the eye when light shines on a thin film Because of the extra distance traveled, there can be interference between the two waves. ( 1 ) 𝐿2 − 𝐿1 = m+ 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0 , 1 , 2 , 3 , … 2 Refraction: changes wavelength of light. Wavelength inside thin film 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 𝜆 𝑓𝑖𝑙𝑚= 𝑛 Always use of film Copyright ©2022 John Wiley & Sons, Inc. 23  27.3 Thin Film Interference (1 of 7) Because of reflection and refraction, two light waves enter the eye when light shines on a thin film Because of the extra distance traveled, there can be interference between the two waves. ( 1 ) 𝐿2 − 𝐿1 = m+ 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0 , 1 , 2 , 3 , … 2 The path length difference that light travels inside a film of thickness is: 𝐿2 − 𝐿1 =2 𝑡 Copyright ©2022 John Wiley & Sons, Inc. 24 27.3 Thin Film Interference (2 of 7) When light travels through a material with a smaller refractive index towards a material with a larger refractive index, reflection at the boundary occurs along with a phase change that is equivalent to one-half of a wavelength in the film. (1 ) 𝐿2 − 𝐿1 = m+ 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0 , 1 , 2 , 3 , … 2 m+ 1 2 𝐿2 − 𝐿1 =( m ) 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0 , 1 , 2 ,3 ,… Copyright ©2022 John Wiley & Sons, Inc. 25  27.3 Thin Film Interference (2 of 7) Destructive interference in thin film (Summary): 𝐿2 − 𝐿1 =2 𝑡 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 m+ 1 𝜆 𝑓𝑖𝑙𝑚= 𝑛 ( ) 1 2 𝐿2 − 𝐿1 = m+ 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0 , 1 , 2 , 3 , … 2 NOT IN FORMULA SHEET! 𝐿2 − 𝐿1 =( m ) 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0, 1, 2 ,3 ,… Remember for exam! 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 2 𝑡 =( m ) ; 𝑚=0 , 1 , 2 , 3 ,… 2 𝑡=( m ) 𝜆 𝑓𝑖𝑙𝑚 ;𝑚=0 , 1 , 2 ,3 ,… 𝑛 Copyright ©2022 John Wiley & Sons, Inc. 26 Example 3 A Colored Thin Film of Gasoline A thin film of gasoline floats on a puddle of water. Sunlight falls perpendicularly on the film and reflects into your eyes. The film has a yellow hue because destructive interference eliminates the color of blue (469 nm) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.33. (a) What is the wavelength and frequency of the light inside the film? (b) Determine the minimum non-zero thickness of the film. −9 𝜆 𝑣𝑎𝑐𝑢𝑢𝑚=469 × 10 𝑚 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 (a)𝜆 𝑓𝑖𝑙𝑚= 𝑛 𝜆 𝑓𝑖𝑙𝑚=? −9 𝑛=1.40 469 × 10 𝑚 𝜆 𝑓𝑖𝑙𝑚= 1.40 −9 𝜆 𝑓𝑖𝑙𝑚=335×10 𝑚 Always use of film Copyright ©2022 John Wiley & Sons, Inc. 27 Frequency inside film is same as frequency in vacuum! () 26.2 The Refraction of Light (9 of 9) When light passes from a medium of larger refractive index into one of smaller refractive index: Speed decreases Frequency remains constant Wavelength and speed decreases, frequency (colour) remains constant! (𝑣 2 < 𝑣 1) ( 𝜆 ¿ ¿2< 𝜆1 )¿ ( 𝑓 ¿ ¿2= 𝑓 1 )¿ Copyright ©2022 John Wiley & Sons, Inc. 28 Example 3 A Colored Thin Film of Gasoline A thin film of gasoline floats on a puddle of water. Sunlight falls perpendicularly on the film and reflects into your eyes. The film has a yellow hue because destructive interference eliminates the color of blue (469 nm) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.33. (a) What is the wavelength and frequency of the light inside the film? (b) Determine the minimum non-zero thickness of the film. 𝜆 𝑣𝑎𝑐𝑢𝑢𝑚=469 × 10 𝑚 Frequency inside medium does not change! −9 𝜆 =335×10 −9 𝑚(a) 𝑓 𝑓𝑖𝑙𝑚= 𝑓 𝑣𝑎𝑐𝑢𝑢𝑚 𝑓𝑖𝑙𝑚 𝑛=1.40 𝑐= 𝑓 𝑣𝑎𝑐𝑢𝑢𝑚 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 𝑓 𝑓𝑖𝑙𝑚=? 𝑐 𝑓 𝑓𝑖𝑙𝑚=¿ 𝑣𝑎𝑐𝑢𝑢𝑚 𝑓 = 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 3 × 10 8 𝑚. 𝑠 − 1 𝑓 𝑓𝑖𝑙𝑚= (469× 10 − 9 𝑚) 14 𝑓 𝑓𝑖𝑙𝑚=6.4 × 10 𝐻𝑧 At any point be able to switch betweenCopyright frequency ©2022 Johnand Wileywavelength! & Sons, Inc. 29 MORE COMPLICATED WAY TO THE SAME THING A thin film of gasoline floats on a puddle of water. Sunlight falls perpendicularly on the film and reflects into your eyes. The film has a yellow hue because destructive interference eliminates the color of blue (469 nm) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.33. (a) What is the wavelength and frequency of the light inside the film? (b) Determine the minimum non-zero thickness of the film. 𝜆 𝑣𝑎𝑐𝑢𝑢𝑚=469 × 10 𝑚 Speed of light slows down in film −9 𝜆 =335×10 −9 𝑚(a)𝑣 = 𝑓 𝑓𝑖𝑙𝑚 𝜆 𝑓𝑖𝑙𝑚 𝑛= c → 𝑣 = 𝑐 𝑓𝑖𝑙𝑚 𝑐 𝑛=1.40 =𝑓 𝑓𝑖𝑙𝑚 𝜆 𝑓𝑖𝑙𝑚𝑣 𝑛 𝑛 𝑓 𝑓𝑖𝑙𝑚=? 𝑐 𝑓 𝑓𝑖𝑙𝑚 = 𝑛 𝜆 𝑓𝑖𝑙𝑚 3 ×108 𝑚. 𝑠 − 1 𝑓 𝑓𝑖𝑙𝑚= (1.40)(3 35 ×10− 9 𝑚) 14 𝑓 𝑓𝑖𝑙𝑚=6.4 × 10 𝐻𝑧 At any point be able to switch betweenCopyright frequency ©2022 Johnand Wileywavelength! & Sons, Inc. 30 Example 3 A Colored Thin Film of Gasoline A thin film of gasoline floats on a puddle of water. Sunlight falls perpendicularly on the film and reflects into your eyes. The film has a yellow hue because destructive interference eliminates the color of blue (469 nm) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.33. (a) What is the wavelength and frequency of the light inside the film? (b) Determine the minimum non-zero thickness of the film. 𝜆 𝑣𝑎𝑐𝑢𝑢𝑚=469 × 10 𝑚(b) The condition for destructive interference is −9 𝜆 𝑓𝑖𝑙𝑚=335×10 −9 𝑚 𝑛=1.40 2 𝑡=( m ) 𝜆 𝑓𝑖𝑙𝑚=;𝑚=0 , 1, 2 ,3 ,… −15 𝑓 𝑓𝑖𝑙𝑚=1.12 ×10 𝐻𝑧 ( m ) 𝜆 𝑓𝑖𝑙𝑚 𝑡= 𝑡 =? 2 ( 1 ) 335× 10 − 9 𝑚 𝑚=1 𝑡= 2 −7 𝑡=1.68×10 𝑚 𝑡=168𝑛𝑚 Copyright ©2022 John Wiley & Sons, Inc. 31 Conceptual Example 4 Multicolored Thin Films Under natural conditions, thin films, like gasoline on water or like the soap bubble in the figure, have a multicolored appearance that often changes while you are watching them. Why are such films multicolored and why do they change with time? Different colors arise because the thickness is different in different places on the film. Moreover, the fact that the colors change as you watch them indicates that the thickness is changing with time Copyright ©2022 John Wiley & Sons, Inc. 32 27.3 Thin Film Interference (6 of 7) The wedge of air formed between two glass plates SKIP causes an interference pattern of alternating dark and bright fringes. Copyright ©2022 John Wiley & Sons, Inc. 33 27.3 Thin Film Interference (7 of 7) Copyright ©2022 John Wiley & Sons, Inc. 34 27.4 The Michelson Interferometer A schematic drawing of a Michelson interferometer. SKIP Copyright ©2022 John Wiley & Sons, Inc. 35 27.5 Fresnel’s Biprism Fresnel’s biprism experiment SKIP is used to measure the wavelength of unknown sources. Copyright ©2022 John Wiley & Sons, Inc. 36  Review of Chapter 17 17.3 Diffraction (1 of 3) The bending of a wave around an obstacle or the edges of an opening is called diffraction. Copyright ©2022 John Wiley & Sons, Inc. 37  Review of Chapter 17 17.3 Diffraction (2 of 3) longer wavelength more spreadout wider slit (normal door) less spread out  single slit – first minimum sin   D Copyright ©2022 John Wiley & Sons, Inc. 38  Review of Chapter 17 17.3 Diffraction (2 of 3) 𝝀  single slit – first minimum sin   D Copyright ©2022 John Wiley & Sons, Inc. 39  Review of Chapter 17 17.3 Diffraction (2 of 3) longer wavelength more spreadout short wavelength long wavelength (red) (blue) less spreadout more sinspreadout 𝜃= 𝜆 single slit – first minimum 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 40  Review of Chapter 17 17.3 Diffraction (2 of 3) wider slit less spread out large slit (normal door) small slit (slit in lab) less spreadout more spreadout 𝜆 sin 𝜃 = single slit – first minimum 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 41 27.6 Diffraction (1 slit) (1 of 8) Diffraction is the bending of waves around obstacles or the edges of an opening. Huygens’ principle Every point on a wave front acts as a source of tiny wavelets that move forward with the same speed as the wave; the wave front at a later instant is the surface that is tangent to the wavelets. Copyright ©2022 John Wiley & Sons, Inc. 42 27.6 Diffraction (3 of 8) Copyright ©2022 John Wiley & Sons, Inc. 43 27.6 Diffraction (4 of 8) This top view shows five sources of Huygens’ wavelets. Wavelets acts as a if there are a bunch of tiny slits. Copyright ©2022 John Wiley & Sons, Inc. 44 27.6 Diffraction (5 of 8) These drawings show how destructive interference leads to the first dark fringe on either side of the central bright fringe. Dark fringes for single slit diffraction 𝐿2 − 𝐿1 =m 𝜆 ;𝑚=0 , 1 , 2 , 3 ,… 𝑊𝑠𝑖𝑛𝜃=m 𝜆 ;𝑚=0,1 ,2 ,3 ,… Copyright ©2022 John Wiley & Sons, Inc. 45 27.6 Diffraction (7 of 8) Copyright ©2022 John Wiley & Sons, Inc. 46  Review of Chapter 17 17.3 Diffraction (3 of 3) longer wavelength more spreadout short wavelength long wavelength (red) (blue) less spreadout Circular opening – first minimum more spreadout sin 𝜃 =1.22 𝜆 𝐷 Copyright ©2022 John Wiley & Sons, Inc. 47  Review of Chapter 17 17.3 Diffraction (3 of 3) longer wavelength more spreadout short wavelength long wavelength (red) (blue) less spreadout Circular opening – first minimum more spreadout sin 𝜃 =1.22 𝜆 𝐷 distance from center to first dark line Copyright ©2022 John Wiley & Sons, Inc. 48  Review of Chapter 17 17.3 Diffraction (3 of 3) wider slit less spread out large slit (normal door) small slit (slit in lab) less spreadout more spreadout 𝜆 Circular opening – first minimum sin 𝜃 =1.22 𝐷 distance from center to first dark line Copyright ©2022 John Wiley & Sons, Inc. 49 27.5 Circular diffraction When light moves through a circular opening, the first minimum is the same as that of sound (with small ) First minimum of a circular diffraction pattern 𝜆 𝜃 ≈ 1.22 𝐷 𝐷=𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 h𝑜𝑙𝑒 in radians! () Copyright ©2022 John Wiley & Sons, Inc. 50  Example: Calculating the diameter of a circular slit Light of frequency Hz is incident on a small circular hole. The light travels to a wall located 3.0 m from the hole and makes a first diffraction minimum at. What is the diameter of the opening? 14 𝑓 =5× 10 Hz 𝜆 (𝜃𝑚𝑖𝑛 ∈radians ) 𝜃 ≈ 1.22 ∘ 𝜋 𝐷 𝜃 𝑚𝑖𝑛 =45 ¿ −7 4 𝜆 6 × 10 𝑚 𝐷=? 𝐷=1.22 ¿ 1.22 𝜃 𝜋 /4 𝜆=? ¿ 9.32 × 10 −7 𝑚 in radians! () ¿ 932 𝑛𝑚 𝑐= 𝜆 𝑓 𝑐 𝜆= 𝑓 3 × 108 𝑚. 𝑠− 1 𝜆= 14 − 1 ¿ 6 × 10 −7 𝑚 5× 10 𝑠 Copyright ©2022 John Wiley & Sons, Inc. 51 27.6 Diffraction (8 of 8) When the source and the screen on which diffraction pattern is observed are at infinite distances from the SKIP obstacle or aperture causing diffraction and the rays from the source are parallel, it is called Fraunhofer diffraction When the source, or screen, or both are at a finite distance from the obstacle or aperture and the rays from the source are not parallel, it is called Fresnel diffraction Copyright ©2022 John Wiley & Sons, Inc. 52 27.7 Resolving Power (1 of 5) SKIP Three photographs of an automobile’s headlights, taken at progressively greater distances. Copyright ©2022 John Wiley & Sons, Inc. 53 27.7 Resolving Power (3 of 5) SKIP Copyright ©2022 John Wiley & Sons, Inc. 54 27.7 Resolving Power (4 of 5) Rayleigh criterion Two point objects are SKIP just resolved when the first dark fringe in the diffraction pattern of one falls directly on the central bright fringe in the diffraction patter of the other.   min 1.22 D Copyright ©2022 John Wiley & Sons, Inc. 55 27.7 Resolving Power (5 of 5) Conceptual Example 8 What You See is Not What You Get The French postimpressionist artist Georges Seurat developed a technique SKIP of painting in which dots of color are placed close together on the canvas. From sufficiently far away the individual dots are not distinguishable, and the images in the picture take on a more normal appearance. Why does the camera resolve the dots, while his eyes do not? Copyright ©2022 John Wiley & Sons, Inc. 56 27.8 The Diffraction Grating (1 of 5) Diffraction, but consisting of a large number of closely spaced, parallel slits is called a diffraction grating. Copyright ©2022 John Wiley & Sons, Inc. 57 27.8 The Diffraction Grating (2 of 5) Diffraction gratings have a bunch of mini triangles. The conditions shown here lead to the first- and second- order intensity maxima in the diffraction pattern. Copyright ©2022 John Wiley & Sons, Inc. 58  27.8 The Diffraction Grating (3 of 5) The bright fringes produced by a diffraction grating are much narrower than those prodd by a double slit. Principal Double slit maxima of a diffractionDiffraction grating grating dsin 𝜃=m 𝜆 ;𝑚=0,1,2 ,3 ,… distance between slits (1/d =slits per cm) Works the same as double slit! Copyright ©2022 John Wiley & Sons, Inc. 59 Example: Diffraction grating Red light (664 nm) is used in a diffraction grating with slits separation of 4400 slits/cm. The screen is located a distance 2.75 m from the slits. (a) Find the angle of the second and the third-order bright fringe. (b) Find the distance between the two bright fringes −9 𝜆=664×10 𝑚 (a) 𝑑= 1 4400𝑐 𝑚− 1 −6 𝑑=2.27×10 𝑚 cm 1𝑚 m 100𝑐𝑚 𝑚=3 𝑚=2 Diffraction grating questions have a bonus step! Finding d in m! 𝑚=3 Copyright ©2022 John Wiley & Sons, Inc. 𝑚=2 60 Example: Diffraction grating Red light (664 nm) is used in a diffraction grating with slits separation of 4400 slits/cm. The screen is located a distance 2.75 m from the slits. (a) Find the angle of the second and the third-order bright fringe. (b) Find the distance between the two bright fringes 𝜆=664×10 𝑚 (a)sin 𝜃= m 𝜆 ;𝑚=0 ,1 ,2 , 3 , … −9 −6 d 𝑑=2.27×10 𝑚 𝑚2 𝑛𝑑 =2 𝑚3 𝑟𝑑 =3 𝜃=sin −1 m 𝜆 d ( ) ( ) ( ) −9 (2)(664 × 10 −9 𝑚 ) (3)(664 × 10 𝑚 ) 𝜃 2=? 𝜃2=sin− 1 𝜃 3=sin − 1 2.27 × 10−6 𝑚 2.27× 10 − 6 𝑚 𝜃 3=? ∘ ∘ 𝜃 2= 35.8 𝜃 3= 61.34 𝑚=3 𝑚=2 𝑚=3 Copyright ©2022 John Wiley & Sons, Inc. 𝑚=2 61 Example: Diffraction grating Red light (664 nm) is used in a diffraction grating with slits separation of 4400 slits/cm. The screen is located a distance 2.75 m from the slits. (a) Find the angle of the second and the third-order bright fringe. (b) Find the distance between the two bright fringes 𝜆=664×10 𝑚 (a)sin 𝜃= m 𝜆 ;𝑚=0 ,1 ,2 , 3 , … −9 −6 d 𝑑=2.27×10 𝑚 𝑚2 𝑛𝑑 =2 𝑚3 𝑟𝑑 =3 𝜃=sin −1 m 𝜆 d ( ) ( ) ( ) −9 (2)(664 × 10 −9 𝑚 ) (3)(664 × 10 𝑚 ) 𝜃 2=? 𝜃2=sin− 1 𝜃 3=sin − 1 2.27 × 10−6 𝑚 2.27× 10 − 6 𝑚 𝜃 3=? ∘ ∘ 𝜃 2= 35.8 𝜃 3= 61.34 (b) 𝑦 3 − 𝑦 2¿5.03 =? −1.98 m¿ 3.05m 𝑦 3𝑚=3 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒¿ 𝑦 𝑚=2 𝑡𝑎𝑛 𝜃= 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐿 𝜃 𝜃𝑦 3 2 2 𝑦 2=𝐿𝑡𝑎𝑛 𝜃¿2(2.75𝑚)𝑡𝑎𝑛(35.8 ¿) 1.98 m ∘ 𝐿 𝑦 3=𝐿𝑡𝑎𝑛 𝜃¿3(2.75𝑚)𝑡𝑎𝑛(61.34∘) m 𝑚=3 Copyright ©2022 John Wiley & Sons, Inc. 𝑚=2 62 27.7 The Diffraction Grating (4 of 5) When light falls on a cd, it creates the colour of the rainbow Why? CD’s act as diffraction gratings! Copyright ©2022 John Wiley & Sons, Inc. 63 Example 9 Separating Colors With a Diffraction Grating A mixture of violet (410 nm) light and red (660 nm) light falls onto a grating 2.75 m away that contains 1.0 104 lines cm. (a) For both colours, find the angle that locates the 1st-order maximum. (b) Find the distance between the red and blue maxima 1 (a) 𝑑= 1 × 10 4 𝑐 𝑚− 1 Diffraction grating questions −6 𝑑=1 ×10 m have a bonus step! Finding d in m! cm 1𝑚 m 100𝑐𝑚 𝑟𝑒𝑑 𝑏𝑙𝑢𝑒 𝑏𝑙𝑢𝑒 𝑟𝑒𝑑 Copyright ©2022 John Wiley & Sons, Inc. 64  Example 9 Separating Colors With a Diffraction Grating A mixture of violet (410 nm) light and red (660 nm) light falls onto a grating 2.75 m away that contains 1.0 104 lines cm. (a) For both colours, find the angle that locates the 1st-order maximum. (b) Find the distance between the red and blue maxima 𝑑=1 ×10 −6 m m𝜆 sin 𝜃= (a) ;𝑚=0 ,1 ,2 , 3 , … 𝜆 𝑣𝑖𝑜𝑙𝑒𝑡 =410× 10 − 9 𝑚 d 𝜆 𝑟𝑒𝑑 =660× 10 𝑚 𝑚=1 −9 𝜃=sin −1 m 𝜆 d ( ) ( ) −9 ( ) −9 (1)(660 × 10 𝑚 ) 𝜃𝑣𝑖𝑜𝑙𝑒𝑡 =?𝜃𝑣𝑖𝑜𝑙𝑒𝑡 =sin− 1 (1)(410 ×−10 𝑚 )𝜃 =sin −1 𝑟𝑒𝑑 1× 10 6 m 1 × 10 − 6 m 𝜃𝑟𝑒𝑑 =? ∘ 𝜃 =41.3 ∘ 𝜃 𝑣𝑖𝑜𝑙𝑒𝑡 =24.2 𝑟𝑒𝑑 𝑟𝑒𝑑 𝑏𝑙𝑢𝑒 𝑏𝑙𝑢𝑒 𝑟𝑒𝑑 Copyright ©2022 John Wiley & Sons, Inc. 65  Example 9 Separating Colors With a Diffraction Grating A mixture of violet (410 nm) light and red (660 nm) light falls onto a grating 2.75 m away that contains 1.0 104 lines cm. (a) For both colours, find the angle that locates the 1st-order maximum. (b) Find the distance between the red and blue maxima 𝑑=1 ×10 −6 m m𝜆 sin 𝜃= (a) ;𝑚=0 ,1 ,2 , 3 , … 𝜆 𝑣𝑖𝑜𝑙𝑒𝑡 =410× 10 − 9 𝑚 d 𝜆 𝑟𝑒𝑑 =660× 10 𝑚 𝑚=1 −9 𝜃=sin −1 m 𝜆 d ( ) ( ) −9 ( ) −9 (1)(660 × 10 𝑚 ) 𝜃𝑣𝑖𝑜𝑙𝑒𝑡 =?𝜃𝑣𝑖𝑜𝑙𝑒𝑡 =sin− 1 (1)(410 ×−10 𝑚 )𝜃 =sin −1 𝑟𝑒𝑑 1× 10 6 m 1 × 10 − 6 m 𝜃𝑟𝑒𝑑 =? ∘ 𝜃 =41.3 ∘ 𝜃 𝑣𝑖𝑜𝑙𝑒𝑡 =24.2 𝑟𝑒𝑑 (b) 𝑦 𝑟 − 𝑦 𝑏=? 𝑦 𝑟𝑟𝑒𝑑 m ¿ 1.18 m 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒¿ 𝑦 𝑏𝑙𝑢𝑒 𝑡𝑎𝑛 𝜃= 𝑦𝑏 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐿 𝜃 𝜃𝑣 𝑟 𝑦 2=𝐿𝑡𝑎𝑛 𝜃¿𝑣(2.75𝑚)𝑡𝑎𝑛(24.2∘¿) 1.24 m 𝐿 𝑦 3=𝐿𝑡𝑎𝑛 𝜃¿𝑟 (2.75𝑚)𝑡𝑎𝑛(41.3∘) m 𝑏𝑙𝑢𝑒 Copyright ©2022 John Wiley & Sons, Inc. 𝑟𝑒𝑑 66 27.9 X-Ray Diffraction (1 of 2) Diffraction can be used to tell us about the separation of slits. Many things in nature have gaps through which light diffracts Ex. High frequency x-rays diffract from crystal structure of atoms! Reverse engineer Diffraction pattern is used to Diffraction pattern determine the structure of molecules! Copyright ©2022 John Wiley & Sons, Inc. 67 27.9 X-Ray Diffraction (2 of 2) Diffraction was used to discover the structure of life itself… DNA! Copyright ©2022 John Wiley & Sons, Inc. 68 Copyright Copyright © 2022 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein. Copyright ©2022 John Wiley & Sons, Inc. 69 Problems Copyright ©2022 John Wiley & Sons, Inc. 70 Problems Copyright ©2022 John Wiley & Sons, Inc. 71 Problems Copyright ©2022 John Wiley & Sons, Inc. 72 Problems Copyright ©2022 John Wiley & Sons, Inc. 73 Problems a) b) c) d) Copyright ©2022 John Wiley & Sons, Inc. 74 Problems Copyright ©2022 John Wiley & Sons, Inc. 75 Problems Copyright ©2022 John Wiley & Sons, Inc. 76 Problems Copyright ©2022 John Wiley & Sons, Inc. 77