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6_Ch 24 College Physics ProblemCH24 Wave Optics

# 6_Ch 24 College Physics ProblemCH24 Wave Optics - 308...

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Unformatted text preview: 308 CHAPTER 24 Problem Solutions 24.1 ∆ybright = ym+1 − ym = ( 632.8 × 10 = λL −9 d ( m + 1) − d m ) ( 5.00 m ) 0.200 × 10 −3 m 24.2 λL m= λL d = 1.58 × 10 −2 m = 1.58 cm (a) For a bright fringe of order m, the path difference is δ = mλ , where m = 0 , 1, 2,… At the location of the third order bright fringe, m = 3 and δ = 3 λ = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 µ m 1 (b) For a dark fringe, the path difference is δ = m + λ , where m = 0 , 1, 2, … 2 At the third dark fringe, m = 2 and 1 5 δ = 2 + λ = ( 589 nm ) = 1.47 × 10 3 nm = 1.47 µ m 24.3 2 2 (a) The distance between the central maximum and the first order bright fringe is λL ∆y = ybright − ybright = , or m =1 m=0 d ∆y = λL d = ( 546.1 × 10 −9 m ) ( 1.20 m ) 0.250 × 10 −3 m = 2 .62 × 10 −3 m = 2 .62 mm (b) The distance between the first and second dark bands is ∆y = ydark 24.4 m =1 − ydark m=0 = λL d = 2.62 mm as in (a) above. λL 1 m + , the spacing between the first and second dark fringes is 2 d λL 3 1 λL ∆y = . Thus, the required distance to the screen is − = d 2 2 d From ydark = L= ( ∆y ) d = ( 4.00 × 10−3 m ) ( 0.300 × 10-3 m ) = λ 460 × 10 −9 m 2.61 m ...
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