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Unformatted text preview: Ch. 35 p. 1 CHAPTER 35 The Wave Nature of Light; Interference 1. We draw the wavelets and see that the incident wave fronts are parallel, with the angle of incidence 1 being the angle between the wave fronts and the surface. The reflecting wave fronts are parallel, with the angle of reflection 2 being the angle between the wave fronts and the surface. Both sets of wave fronts are in the same medium, so they travel at the same speed. The perpendicular distance between wave fronts is BC = AD = c ? t . From the triangles, we see that AB = BC /sin 1 = AD /sin 2 . Thus we have sin 1 = sin 2 , or 1 = 2 . 2. For constructive interference, the path difference is a multiple of the wavelength: d sin = m , m = 0, 1, 2, 3, . For the fifth order, we have (1.6 10 5 m) sin 9.8 = (5) , which gives = 5.4 10 7 m = 540 nm . 3. For constructive interference, the path difference is a multiple of the wavelength: d sin = m , m = 0, 1, 2, 3, . For the third order, we have d sin 18 = (3)(610 10 9 m), which gives d = 5.9 10 6 m = 5.9 m . 4. For constructive interference, the path difference is a multiple of the wavelength: d sin = m , m = 0, 1, 2, 3, . We find the location on the screen from y = L tan . For small angles, we have sin tan , which gives y = L ( m / d ) = mL / d . For adjacent fringes, ? m = 1, so we have ? y = L ? m / d ; 0.065 m = (5.00 m) (1)/(0.048 10 3 m), which gives = 6.24 10 7 m = 0.62 m . The frequency is f = c / = (3.00 10 8 m/s)/(6.24 10 7 m) = 4.8 10 14 Hz . 5. For constructive interference, the path difference is a multiple of the wavelength: d sin = m , m = 0, 1, 2, 3, . We find the location on the screen from y = L tan . For small angles, we have U s e W o r d 6 . 0 c o r l a t e r t o v i e w M a c i n t o s h p i c t u r e . Ch. 35 p. 2 sin tan , which gives y = L ( m / d ) = mL / d . For adjacent fringes, ? m = 1, so we have ? y = L ? m / d ; = (3.6 m)(656 10 9 m)(1)/(0.060 10 3 m) = 3.9 10 2 m = 3.9 cm ....
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 Spring '08
 ROSS
 Light

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