solution_pdf40 - tovar (jdt436) homework 40 Turner (59070)...

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tovar (jdt436) – homework 40 – Turner – (59070) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider the setup oF a single slit experiment. The wavelength oF the incident light is λ = 530 nm . The slit width and the distance between the slit and the screen is specifed in the fgure. y 1 3 . 8 m 410 μ m S 1 S 2 θ viewing screen ±ind the position y = y 1 oF the frst inten- sity minimum. Use a small angle approxima- tion sin θ = tan θ . Correct answer: 4 . 9122 mm. Explanation: Let : λ = 530 nm , L = 3 . 8 m , and a = 410 μ m . y 1 L a S 1 S 2 δ a sin θ a b y L B ±or single slit di²raction, destructive in- terFerence occurs when, a 2 sin θ = λ 2 , or sim- ply when, δ a sin θ = λ . Thus, between the two end rays which correspond to the frst minimum, the phase angle di²erence is β 1 = 2 π and the path length di²erence is δ 1 = λ . The small angle approximation gives us y 1 L = tan θ 1 θ 1 sin θ 1 = δ 1 a , or y 1 = δ 1 a L = λ L a = sturt 4 . 9122 mm . 002 10.0 points Consider the setup oF a single slit experiment. y 7 L a S 1 S 2 × 15 ±ind the height y 7 where the seventh mini- mum occurs. Use a small angle approximation sin θ = tan θ . 1. y 7 = 6 λ L a 2. y 7 = 15 2 λ L a 3. y 7 = 7 λ L a correct 4. y 7 = 9 λ L a 5. y 7 = 13 2 λ L a 6. y 7 = 8 λ L a 7. y 7 = 11 2 λ L a 8. y 7 = 5 λ L a 9. y 7 = 17 2 λ L a 10. y 7 = 19 2 λ L a Explanation: Let : k 2 π λ .
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tovar (jdt436) – homework 40 – Turner – (59070) 2 The frst minimum is at β = 2 π , where β = 2 φ = 2 π , and φ = π is the phase diFerence o± the two rays ±or destructive inter±erence. The seventh minimum occurs at β = 14 π , which corresponds to a path diFerence be- tween two end rays o± δ = β k = 14 π p 2 π λ P = 7 λ .
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solution_pdf40 - tovar (jdt436) homework 40 Turner (59070)...

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