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Unformatted text preview: terry (ect328) – homework 40 – Turner – (59130) 1 This printout should have 10 questions. Multiplechoice 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 = 550 nm . The slit width and the distance between the slit and the screen is specified in the figure. y 1 7 . 5 m 860 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: 4 . 79651 mm. Explanation: Let : λ = 550 nm , L = 7 . 5 m , and a = 860 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 4 . 79651 mm . 002 10.0 points Consider the setup of a single slit experiment. y 3 L a S 1 S 2 θ viewing screen × 15 Find the height y 3 where the third mini mum occurs. Use a small angle approxima tion sin θ = tan θ . 1. y 3 = 7 2 λ L a 2. y 3 = 3 λ L a correct 3. y 3 = 1 λ L a 4. y 3 = 5 2 λ L a 5. y 3 = 11 2 λ L a 6. y 3 = 5 λ L a 7. y 3 = 4 λ L a 8. y 3 = 9 2 λ L a 9. y 3 = 3 2 λ L a 10. y 3 = 2 λ L a Explanation: Let : k ≡ 2 π λ . terry (ect328) – homework 40 – Turner – (59130) 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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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.
 Fall '09
 Turner

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