refracted and the wavelength of the light shortens. According to the equation! = w"y L

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1 Sectio 10.2: Sigle-Slit Diffractio Tutorial 1 Practice, page Whe light travels from air to a medium that is deser tha air, such as ater, the light is refracted ad the avelegth of the light shortes. Accordig to the equatio! = "y, is proportioal to Δy, so he is reduced, the cetral maximum ill also be reduced. So the cetral maximum ould be arroer if the equipmet ere submerged i ater. 2. Give:! = " 10 #7 m; = 43 µm = m; = 3.0 m Required: Δy Aalysis: Rearrage the equatio! = "y!y = " Solutio:!y = " = (3.0 m(7.328 # 10$7 # 10 $7 m 4.3 # 10 $5 m = 5.1# 10 $2 m to solve for the distace betee adjacet miima;!y = 5.1 cm Statemet: The separatio of adjacet miima is 5.1 cm. 3. Give: = 3.00! 10 "6 m; # 1 = 25.0 Required: Aalysis: The agle betee the first dark friges is equal to the agle for the idth of the cetral maximum, hich is tice the agle for the first dark frige, give by siθ =. I this case, = 1 ad 2! = 25.0 ;! = si". Solutio: 2θ = 25.0 θ = 12.5! = si" = (3.0 # 10 $6 msi12.5! = 6.49 # 10 $7 m Statemet: The avelegth of the light is m. 4. Give: θa = 56 ; θb = 34 a Required: b Aalysis: The formula that relates the agles, the commo avelegth, ad the slit sizes is a siθ = ; a si! a = " ad b si! b = ". Divide the equatios to determie the ratio. b Copyright 2012 Nelso Educatio td. Chapter 10: Applicatios of the Wave Nature of ight

2 Solutio: a si! a b si! b = " " a b = si! b si! a = si34 si56 a b = 0.67 Statemet: The ratio of the slit idths, a b, is Sectio 10.2 Questios, page Give: sigle-slit diffractio;! = 794 m = 7.94 " 10#7 m; = 1.0 m = 9, y 9 = 6.48 cm = m Required: Aalysis:!y = y 9 Solutio:!y = y 9 =! "y ;!y = " = "!y = m 9!y = 7.2 "10 3 m = (7.94 #10$7 m(1.0 m 7.2 #10 3 m = 1.1 #10 3 m Statemet: The idth of the slit is 1.1!10 "4 m. 2. Give: sigle-slit diffractio;! = 600 m = 6.00 " 10 #7 m; $ 1 = 6.9 Required: Aalysis: The first dark frige is located here siθ =. Rearrage the equatio siθ = to solve for slit idth; =! si". Copyright 2012 Nelso Educatio td. Chapter 10: Applicatios of the Wave Nature of ight

3 Solutio: =! si" 1 = 6.00 #10$7 m si6.9 = 5.0 #10 $6 m Statemet: The idth of the slit is m. 3. Give: sigle-slit diffractio; = 450 m = m; = 10.0 m; = 0.15 mm = 1.5! 10 "4 m Required: y, the distace betee the first ad third dark friges Aalysis: Δ y = ; y = 2!y Solutio:!y = " = (4.50 #10$7 m(10.0 m 1.50 #10 $4 m = 3.0 #10 $2 m y = 2!y = 6.0 #10 $2 m y = 6.0 cm Statemet: The distace betee the first ad third dark friges is 6.0 cm. 4. Give: sigle-slit diffractio; = 550 m = m; = 2.0 m; y 1 = 5.5 mm = 5.5! 10 "3 m Required: Aalysis: Rearrage the equatio y1 =! to solve for slit idth; = y 1 Solutio: =! y 1 = (5.50 "10#7 m(2.0 m 5.50 "10 #3 m = 2.0 "10 #4 m = 0.20 mm Statemet: The idth of the slit is 0.20 mm. 5. Give: sigle-slit diffractio; = 630 m = m; = 3.0 m; = 0.25 mm = 2.5! 10 "4 m Required: 2Δy, the idth of the cetral maximum Aalysis: Multiply the equatio!y = " by 2 to obtai 2Δy. Copyright 2012 Nelso Educatio td. Chapter 10: Applicatios of the Wave Nature of ight

4 Solutio: 2!y = 2" ( 3.0 m = #10$7 m 2.50 #10 $4 m = 1.5#10 $2 m ( 2!y= 1.5 cm Statemet: The idth of the cetral maximum is 1.5 cm Give: Δ y= cm = m; = cm = m; = 60.0 cm= 0.60 m Required: Aalysis: Rearrage the equatio!y = " "y to solve for avelegth;! = Solutio:! = "y = (2.95#10$4 m(1.20 #10 $3 m 6.0 #10 $1 m! = 5.90 #10 $7 m 7 Statemet: The avelegth of the yello light is m. 7. (a The distace betee successive maxima i sigle-slit diffractio is give by Δ y =. If I double the avelegth,, the the distace Δy ill also double. The agles of the maxima ad the miima ould be approximately doubled. (b If I multiplied both the avelegth,, ad the slit idth,, i the equatio Δ y = by 2, the 2s ill cacel each other out. Therefore, there ill be o effect o Δy. The iterferece patter ill be the same. 8. Blue light has a average avelegth of 475 m, ad gree light has a average avelegth of 510 m. If I replaced the blue light ith the gree light, the I ould be icreasig the avelegth. Therefore, spacig of the itesity maxima ould be greater. 9. Give: sigle-slit diffractio Required: θ 10 Aalysis: Assume the idth of a typical dooray is = 0.92 m. Assume the visible light has a avelegth of 500 m. Rearrage the equatio si! = " to solve for the agle; $ # '! = si "1 % & ( Copyright 2012 Nelso Educatio td. Chapter 10: Applicatios of the Wave Nature of ight

5 $ # ' Solutio:! = si "1 % & ( $ (10( 5.00 *10 "7 m ' = si "1 & % & 0.92 m ( = si "1 5.43*10 "6! 10 = 3.1*10 "4 ( Statemet: The agle of the teth miimum for a dooray that is 0.92 m ide is To improve the resolutio of a digital image, I could use more pixels per square cetimetre. Or, I could use a ider aperture (size of slit to icrease the resolutio. Hoever, the ider aperture ould reduce the depth of field (rage of the focus. 11. I ould be able to resolve the double stars i Mizar ith the telescope because, i additio to elargig the image, the telescope s aperture is ider tha the aperture i my eye. The ider aperture icreases the resolutio, alloig me to see the to stars. 12. I a double-slit iterferece patter, there are more itesity maxima tha i a sigle-slit iterferece patter. I the double-slit iterferece patter, there is less space betee friges because the secod slit causes additioal destructive iterferece. Copyright 2012 Nelso Educatio td. Chapter 10: Applicatios of the Wave Nature of ight

λ = ) = d sin! n Analysis: Rearrange the equation dsinθn to solve for wavelength; # d sin! n = n " 1 & $ ( ) n " 1 2 Solution: ( m) sin 3.

λ = ) = d sin! n Analysis: Rearrange the equation dsinθn to solve for wavelength; # d sin! n = n 1 & $ ( ) n 1 2 Solution: ( m) sin 3. Sectio 9.5: Iterferece of ight Waves: Youg s Double-Slit Eperimet Tutorial Practice, page 82. Give: ouble-slit iterferece; 5; θ 5 3.8 ; 0.02 mm.2 0 5 m Require: λ siθ λ to solve for wavelegth; si! & (