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Ch35Word - Ch 35 p 1 CHAPTER 35 The Wave Nature of Light...

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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  θ 1  being the  angle  between  the wave fronts and  the surface.   The reflecting wave fronts are parallel,  with the angle of reflection  θ 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   θ 1  =  AD /sin   θ 2   . Thus we have sin  θ 1  = sin  θ 2   ,   or    θ 1  =  θ 2   . 2. For constructive interference, the path  difference is a multiple of the wavelength: d  sin  θ  =  m λ ,    m  = 0, 1, 2, 3, … . For the fifth order, we have (1.6 × 10 –5  m) sin 9.8 °  = (5) λ , which gives  λ  = 5.4 × 10 –7  m =       540 nm . 3. For constructive interference, the path  difference is a multiple of the wavelength: d  sin  θ  =  m λ ,    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  μ m . 4. For constructive interference, the path  difference is a multiple of the wavelength: d  sin  θ  =  m λ ,    m  = 0, 1, 2, 3, … . We find the location on the screen from y  =  L  tan  θ . For small angles, we have sin  θ  ˜ tan  θ , which gives y  =  L ( m λ / d ) =  mL λ / d . For adjacent fringes, ? m  = 1, so we have ? y  =  L λ ? m / d ; 0.065 m = (5.00 m) λ (1)/(0.048 × 10 –3  m), which gives  λ  = 6.24 × 10 –7  m =       0.62  μ m . The frequency is f  =  c / λ  = (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  θ  =  m λ ,    m  = 0, 1, 2, 3, … . We find the location on the screen from y  =  L  tan  θ . For small angles, we have Use Word 6.0c or later to view Macintosh picture.
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Ch. 35   p. 2 sin  θ  ˜ tan  θ , which gives y  =  L ( m λ / d ) =  mL λ / d . For adjacent fringes, ? m  = 1, so we have ? y   =  L λ ? m / d ; = (3.6 m)(656 × 10 –9  m)(1)/(0.060 × 10 –3  m) = 3.9 × 10 –2  m =       3.9 cm . 6. For constructive interference, the path  difference is a multiple of the wavelength: d  sin  θ  =  m λ ,    m  = 0, 1, 2, 3, … . We find the location on the screen from y  =  L  tan  θ .
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