Version 001 Interference jean (AP Phy MHS 2012) 1

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1 Version 001 Interference jean AP Phy MHS 01) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. sound m Concept points Will the light from two very close stars produce an interference pattern? wavelength 0.16 m 7 m 1. Yes; strong light from the two stars creates very clear interference patterns.. No; light from two very close stars is not likely to have the same frequency. correct 3. No; the light is too strong to produce a discernable interference pattern. 4. Yes; if they are very close to each other, the light has large overlapping area. What is the distance between the two slits? Correct answer: m. Basic Concept: The rules for determining interference maximum or minimum are the same for sound waves and light waves. Thus, the path length difference is δ d d 1 nλ, 1) where n 1 for the first maximum. Solution: Double Slit interference. d d 1 d y d y y + d Light from a pair of stars will not produce an interference pattern because the waves of light from the two separate sources have different frequencies and thus are incoherent; when combined they smudge. Interference occurs when light from a single source divides and recombines. AP B 1998 MC 51 A points Plane sound waves of wavelength 0.16 m are incident on two narrow slits in a box with nonreflecting walls, as shown in the figure below. Ataperpendiculardistanceof7mfromthe center of slits, a first order maximum occurs at a point which is m from the central maximum. L Let : λ 0.16 m, L 7 m, and y m, The approximation sinθ δ d requires L d, which does NOT apply here; the signals are NOT traveling nearly parallel to each other. Wemust goback tothedefinitionsand basic concepts of constructive and destructive interference. From the picture and using the Pythagorean Theorem, the wave from the upper slit travels a distance d 1 L + y d )

2 Version 001 Interference jean AP Phy MHS 01) and the wave from the lower slit travels a distance d L + y + d ) [d +d 1 ][d d 1 ] d d 1, so [d +d 1 ]nλ L + y + d L ) y d ) y +yd+ d 4 y +yd d 4 yd. Since d 1 7 m) + m 0.56 m 7.08 m and d 7 m) + m m m, ) ) d nλ[d +d 1 ] y 1)0.16 m) m) [ m)+7.08 m)] m. Alternative Approximate Solution: Since the receiver is at the first maximum, n 1. From trigonometry, tanθ y L. For constructive interference using the approximation), nλ δ d sinθ. This approximation assumes that L d, which is only good toafew percent in thiscase. Solvingfor d, we have d nλ sinθ nλ [ y sin arctan L)] sin 1)0.16 m) [ )] m arctan 7 m m. Such an estimate, if L > d, is usually fairly good,butmaynotbecloseenoughtogiveone percent accuracy. Small Angle Approximation: The small angle approximation assumes that sinθ tanθ θ, where θ is in radians. Thus sinθ δ d y, which gives L d δ L y 0.56 m with percent error m 0.56 m % Error 100% m %. Bright Band Separation 003 part 1 of ) 10.0 points Two slitsareilluminatedby a606nm light. The angle between the zeroth-order bright band at the center of the screen and the fourth-order bright band is If the screen is 179 cm from the double-slit, how far apart is this bright band from the central peak? Correct answer: cm. Basic Concept: Solution: y L tanθ. y Ltanθ 179 cm)tan15.5 ) cm. 004 part of ) 10.0 points What is the distance between the two slits?

3 Version 001 Interference jean AP Phy MHS 01) 3 Correct answer: mm. Basic Concept: dsinθ mλ Solution: The fourth-order bright band occurs when m 4; therefore, d mλ sinθ 4606 nm) sin15.5 ) mm. Double Slit Delta Wave short points A screen is illuminated by monochromatic light whose wave length is λ, as shown. The distance from the slits to the screen is L. 8. δ 7λ and φ 14π 9. δ 9 λ and φ 9π 10. δ 5 λ and φ 5π For bright fringes δ dsinθ m λ, wherem 0, ±1, ±, ±3,soforthesecond bright fringe m ), δ mλ λ and δ λ φ π φ 4π. d S 1 S θ L y viewing screen At the second bright fringe on the screen position y on the screen), δ the corresponding path length difference) and φ the phase angle difference) is given by 1. δ 13. δ 11 λ and φ 13π λ and φ 11π 3. δ 7 λ and φ 7π 4. δ 6λ and φ 1π 5. δ 4λ and φ 8π 6. δ 3λ and φ 6π 7. δ λ and φ 4π correct Hewitt CP9 8 P points Consider two ways that light might hypothetically get from its starting point S to its final point F by being reflected by a mirror at either point A or point B. Since light travels at a fixed speed in air, the path of the least time will also be the path of the least distance. 33 cm S F 46 cm 46 cm A B C 33 cm Findthedifference inthepathlengthssaf and SBF. Correct answer: cm. Given : SA 33 cm and, AB 46 cm.

4 Version 001 Interference jean AP Phy MHS 01) 4 The path SBF is SBF SA +AB + SA +AB BC +CF 33 cm) +46 cm) cm. The path SAF is SAF SA+ AC +CF 33 cm)+ 9 cm) +33 cm) cm. The path difference is SF SAF SBF cm cm cm. An Optical Coating 007 part 1 of ) 10.0 points A light ray is traveling in a medium with an index of refraction n 1 and it isreflected at the boundary of a second medium with an index of refraction n. Considering the change of the relative phases φ due to their reflections, which of the following conditions is correct? 1. If n 1 > n, then φ π and if n 1 < n, then φ π.. If n 1 > n, then φ 0 and if n 1 < n, then φ π. correct 3. If n 1 > n, then φ 0 and if n 1 < n, then φ π. 4. If n 1 > n, then φ 0 and if n 1 < n, then φ If n 1 > n, then φ π and if n 1 < n, then φ π. 6. If n 1 > n, then φ π and if n 1 < n, then φ π. If a light ray is traveling in a medium with an index of refraction n 1 and it is reflected at the boundary of a second medium with an index of refraction n, the phase change is φ 0 when n 1 > n and isπ when n 1 < n. 008 part of ) 10.0 points Consider the optical coating on a glass lens where the index of refraction of the coating is n, where n is greater than the index of refraction of the air. θ 0 θ 1 t air n lens Assume: The index of refraction of the lens is greater than that of the coating. To minimize the reflection of a ray with a wavelength λ 500nm and incident angle θ 0,whatistheminimumnonzerothickness t of the coating? 1. t nλ. t 3nλ 4 3. t nλ 4. t λ n 5. t λ n 6. t 3λ 4n 7. t nλ 4 8. t nλ 8 9. t λ 8n 10. t λ 4n correct

5 Version 001 Interference jean AP Phy MHS 01) 5 Destructive interference occurs when the difference between the phase angle of the incidentrayreflectedfromtheoutersurfaceofthe coating ray 1) and the phase angle of the ray reflected from the inner surface of the coating ray)areatπ, 3π,5π etc. Thephase differences aredueto thepathdifference ofthe two rays and the change of their relative phases due to reflections. Putting them together, it gives π, 3π, 5π... t)k n + π π tk n. With k n π λ n, it implies that the minimum thickness is given by t λ 4n. Film Between Air and Water 009 part 1 of 3) 10.0 points Consider a horizontal plane of thin film with a thickness t. This film is located in between air and water see sketch). Light is directed from the air downward through the film and into the water, perpendicular to the surfaces. The index of refraction of the film n The index of refraction of water is n Thewavelengthoftheincidentwaveintheair is λ, and its frequency in air is f c λ. Water Air t n 1 The wavelength λ water of the light in the water is 1. λ water 1 n λ correct. λ water n 1 n λ 3. λ water n λ Basic Concepts: Interference phenomenon due to a thin film. Solution: The frequency of light is constant from medium to medium. In the water, the n wavelength of the light is λ water λ n. The frequency will remain the same. 010 part of 3) 10.0 points The frequency of the light in the water will 1. decrease. stay the same correct 3. increase See Part part 3 of 3) 10.0 points The smallest non-zero thickness t min leading to no reflection is 1. t min λ n 1. t min λ 3. t min λ 4. t min λ n 5. t min λ 4n 6. t min λ n 1 correct 7. t min λ n 8. t min λ 4n 1 9. t min λ 4 1 Air n n 1.33 The phase angle difference contributed by

6 reflection is given by Version 001 Interference jean AP Phy MHS 01) 6 φ refl φ refl 1 φ refl π 0 π. The minima occur when the phase angle differences of the two light rays are φ t λ 1 π +π π, 3π,. The smallest non-zero thickness occurs at t λ 1 π +π 3π t min λ 1 λ n 1.