hw07_sol - 7. (a) We use Eq. 36-3 to calculate the...

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7. (a) We use Eq. 36-3 to calculate the separation between the first ( m 1 = 1) and fifth 2 (5 ) m = minima: ∆∆ yD D m a D a m D a mm == F H G I K J ==− sin . θ λλ λ 21 bg Solving for the slit width, we obtain a Dm m y = = ×− = λ 6 400 550 10 5 1 035 25 ch . .. (b) For m = 1, . × m a λ 1 550 10 22 10 6 4 The angle is = sin –1 (2.2 × 10 –4 ) = 2.2 × 10 –4 rad.
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10. Let the first minimum be a distance y from the central axis which is perpendicular to the speaker. Then sin θ =+ = = yD y ma a 22 12 ch λλ (for m = 1). Therefore, () 2 100m 41.2m . 11 0.300m 3000Hz 343m s 1 s DD y aa f v == = = λ− ⎡⎤ ⎣⎦
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12. We will make use of arctangents and sines in our solution, even though they can be “shortcut” somewhat since the angles are small enough to justify the use of the small angle approximation. (a) Given y / D = 15/300 (both expressed here in centimeters), then θ = tan 1 ( y / D ) = 2.86°. Use of Eq. 36-6 (with a = 6000 nm and λ = 500 nm) leads to ( ) 6000nm sin 2.86 sin 1.883rad . 500nm a α π° π == = λ Thus, 2 sin 0.256 .
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This note was uploaded on 04/05/2011 for the course PHYSICS 133 taught by Professor Steigman during the Fall '09 term at Ohio State.

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hw07_sol - 7. (a) We use Eq. 36-3 to calculate the...

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