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hw 40 - nguyen(jmn727 homework 40 Turner(59070 This...

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nguyen (jmn727) – homework 40 – Turner – (59070) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the setup of a single slit experiment. The wavelength of the incident light is λ = 480 nm . The slit width and the distance between the slit and the screen is specified in the figure. y 1 6 . 1 m 550 μ m S 1 S 2 θ viewing screen Find the position y = y 1 of the first inten- sity minimum. Use a small angle approxima- tion sin θ = tan θ . Correct answer: 5 . 32364 mm. Explanation: Let : λ = 480 nm , L = 6 . 1 m , and a = 550 μ m . y 1 L a S 1 S 2 θ viewing screen δ a sin θ a bracketleftBig y L bracketrightBig For single slit diffraction, 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 first minimum, the phase angle difference is β 1 = 2 π and the path length difference 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 5 . 32364 mm . 002 10.0 points Consider the setup of a single slit experiment. y 6 L a S 1 S 2 θ viewing screen × 15 Find the height y 6 where the sixth mini- mum occurs. Use a small angle approxima- tion sin θ = tan θ . 1. y 6 = 15 2 λ L a 2. y 6 = 6 λ L a correct 3. y 6 = 11 2 λ L a 4. y 6 = 7 λ L a 5. y 6 = 5 λ L a 6. y 6 = 4 λ L a 7. y 6 = 17 2 λ L a 8. y 6 = 8 λ L a 9. y 6 = 9 2 λ L a 10. y 6 = 13 2 λ L a Explanation: Let : k 2 π λ .

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nguyen (jmn727) – homework 40 – Turner – (59070) 2 The first minimum is at β = 2 π , where β = 2 φ = 2 π , and φ = π is the phase difference of the two rays for destructive interference.
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hw 40 - nguyen(jmn727 homework 40 Turner(59070 This...

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