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Extra Credit

# Extra Credit - 37 1 7 14 25 33 34 38 3 7 11 21 DC B A...

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θ θ B A C θ D 37 - 1, 7, 14, 25, 33, 34; 38 - 3, 7, 11, 21 Chapter 37 1) Wave length = 632.8nm λ Distance d = .2mm Distance L = 5m Width of bright = ∆y λLd = . × - ×( )(. × - ) ∆y 632 8 10 9m 5m 2 10 3m = . 1 58cm 7) Distance d = .25mm Wave length = 546.1nm λ Distance L = 1.2m a) m = 1 Distance of 1 st bright = ∆y mλLd = × . × - ×( . )(. × - ) ∆y 1 546 1 10 9m 1 2m 25 10 3m = . 2 62mm b) Distance between 1 st and 2 nd = ∆y λLd = . × - ×( . )(. × - ) ∆y 546 1 10 9m 1 2m 25 10 3m = . 2 62mm 14)

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Path difference = - δ DC 4B = - dsinθ2 dsinθ1 = ( - ) d sinθ2 sinθ1 The path difference should be an integral multiple of the wavelength so that λ ( - )- = d sinθ2 sinθ1 0 where = ,± ,± m 0 1 2 for interference maximum 25) Refractive index n = 1.25 Red wavelength λR = 640nm Blue wavelength λB = 512nm = mλR 2nt Thickness of oil = t mλR2n = ×( × - ) × . 2 640 10 9m 2 1 25 = . × - 5 12 10 7m = 512nm 33) 34) Displacement d = . × - 3 82 10 4m Number of fringes crossed n = 1700 Wavelength = λ 2dn = ×( . × - ) 2 3 82 10 4m 1700 = . × - 4 49 10 7m = 449nm Chapter 38 3) Distance L = 50cm Wavelength = 690nm λ Distance between 1 st & 3 rd ∆y = 3mm = = sinθ1 m1λa where m1 1 3 rd minima distance = y2 m2λLa = - ∆y λLam2 m1 = - ∆y λLa3 1
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