Physics Solution Manual for 1100 and 2101

# 1444 24444 4 3 half wavelength net phase change due

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Unformatted text preview: ND THE WAVE NATURE OF LIGHT ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (e) The first intensity minimum occurs when the difference in path lengths is second when the difference is 2. 3 2 λ , and the third when the difference is 5 2 (a) The angle θ that specifies the m bright fringe is given by sin θ = m th If sin θ ≈ θ , then θ = m λ d λ d 1λ 2 , the λ. (Equation 27.1). . Thus, if λ and d are both doubled, the angle does not change. 3. (b) According to the discussion in Section 27.2, the difference in path lengths of the light waves increases by one wavelength as one moves from one bright fringe to the next one farther out. 4. d = 1.1 × 10−5 m 5. (d) The wavelength λwater of the light in water is related to the wavelength λvacuum in a vacuum by λwater = λvacuum / nwater (Equation 27.3). Since the index of refraction of water nwater is greater than one, the wavelength decreases when the apparatus is placed in water. The angle θ that specifies the mth bright fringe is given by sin θ = m λ d (Equation 27.1). When the wavelength decreases, the angle also decreases. 6. (d) The down-and-back distance traveled by the light wave in the film is 2400 nm. The wavelength of the light within the film is (see Equation 27.3) λfilm = λvacuum/nfilm = ( 600 nm ) /1.5 = 400 nm . Thus, the down-and-back distance is equivalent to λ (2400 nm) film = 6λfilm . 400 nm 7. (c) In drawings 1 and 2, light travels through a material with a smaller refractive index toward a material with a larger refractive index, so the reflection at the boundary occurs with a phase change that is equivalent to one-half a wavelength in the material. In drawings 3 and 4, light travels through a material with a larger refractive index toward a material with a smaller refractive index, so there is no phase change upon reflection at the boundary. Chapter 27 Answers to Focus on Concepts Questions 1429 8. (e) In film 4, the reflection at the top surface occurs with a phase change that is equivalent to one-half a wavelength in the film, since light travels from a material with a smaller refractive index (nair = 1.0) toward a material with a larger refractive index (nfilm = 1.5). The reflection at the bottom surface occurs without a phase change, since light travels from a material with a larger refractive index (nfilm = 1.5) toward a material with a smaller refractive index (nair = 1.0). The down-and-back distance traveled by the light is one-half of a wavelength. Therefore, the net phase change is one wavelength, which leads to constructive interference. 9. (b) A bright fringe occurs at a point where the thickness of the air wedge has a certain value. As the thickness of the air wedge becomes smaller, the fringe moves to the right where the air wedge is thicker. 10. t = 2.6 × 10−6 m 11. (c) The amount of diffraction depends on the angles that locate the dark fringes on either side of the central bright fringe (see Section 27.5). These angles (one for each value of the integer m) are related to the wavelength λ of the light and the width W of the slit by sin θ = m ( λ / W ) , (Equation 27.4). When λ is much less than W, θ ≈ 0°, and there is very little diffraction. 12. (c) The width of the central bright fringe is determined by the angle θ that locates the first (m = 1) dark fringe on either side of the central bright fringe (see Section 27.5). This angle is given by sin θ = (1) λ / W (Equation 27.4), where λ is the wavelength and W is the width of the slit. If the width of the slit decreases, the angle increases. 13. (c) The angles that locates the dark fringes on either side of the central bright fringe are given by sin θ = mλ / W , where λ is the wavelength, W is the width of the slit, and m = 1, 2, 3, … (Equation 27.4). If the wavelength is doubled and the width of the slit is doubled, the angles remain the same. Hence, the diffraction patterns remain the same. 14. θ = 5.44º 15. (d) The best possible resolving power occurs when the angle subtended by two objects is a minimum (see Section 27.6) The minimum angle θmin between two objects that are just resolved is θmin ≈ 1.22λ / D (Equation 27.6), where λ is the wavelength and D is the diameter of the lens through which the light passes. If the diameter of the lens is fixed, the minimum angle is proportional to the wavelength. Thus, blue light, having the smallest wavelength, gives the smallest angle and, hence, the best resolving power. 16. (c) The minimum angle θmin between two objects that are just resolved is θmin ≈ 1.22λ / D (Equation 27.6), where λ is the wavelength and D is the diameter of the lens through which the light passes. When the light intensity decreases, the diameter of your pupil increases so 1430 INTERFERENCE AND THE WAVE NATURE OF LIGHT as to admit more light. Therefore, the minimum angle decreases, and your ability to resolve the two dots, increases. 17. smin = 24 m 18. (d) Because there is always a central bright maximum for each wavelength, there is an orange (red plus yellow) fringe at θ = 0°. According to Equation 2...
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