Quiz06b-s - = sin 1 p 2(2 a a cm ) 5 a a cm P = sin 1 p 4 5...

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Quiz 6b Name: Date: Phys224 Spring 2010 Dr. P. Hanlet Concept (3 points each) 1. The intensity I of an electromagnetic wave (having electric Feld v E ) is proportional to: (a) E (b) E (c) E 2 (d) E 4 (e) 4 E 1
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Problem (5 pts) Show your work!!! If I can read it, I will give you partial credit!!! Correct answers without work will NOT get full credit. 2. Microwaves of wavelength λ =2 . 0 cm from a radio transmitter are aimed at two narrow parallel slits in an aluminum plate. The slits are separated by a distance of d =5 . 0 cm as shown in the Fgure. λ v n θ d r 1 r 2 D y The variables are: λ =2 . 0 cm d =5 . 0 cm D =? y =? θ =? m =2 (a) At what angles, θ , at some distance D from the slits, will the second maximum be found? (Note that for this part of the problem, n =1 .) For constructive interference, i.e. a maximum, the equation governing the condition is: d sin θ = Such that: θ = sin 1 p d P Substituting values:
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Unformatted text preview: = sin 1 p 2(2 a a cm ) 5 a a cm P = sin 1 p 4 5 P Such that: = 0 . 927 rad = 53 . 1 2 (b) If a diamond block of index of refraction n =2 . 4 is placed behind one of the slits, what thickness of diamond would cause the second maximum, to become a minimum? So that the maximum becomes a minimum, a wavelength shift of n 2 is required, where: n = n This can be accomplished by adding the diamond block at the upper slit. The number, N , of wavelengths, n , crossing a length L is: N = L n = L = N n = N n = 1 2 (2 cm ) 2 . 4 Finally, in order to cause a microwave maximum at the screen to become a minimum, the length of the diamond block must be: L = 0 . 42 cm . Pretty BIG diamond!!! 3...
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Quiz06b-s - = sin 1 p 2(2 a a cm ) 5 a a cm P = sin 1 p 4 5...

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