GiancoliCh35Sol

# GiancoliCh35Sol - Solutions to WebAssign HW#5 Giancoli Ch...

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Solutions to WebAssign HW #5 - Giancoli Ch. 35-2, 3, 8, 25, 31, 36, 53, 54, and 80; 2. The angle from the central maximum to the first dark fringe is equal to half the width of the central maximum. Using this angle and Eq. 35-1, we calculate the wavelength used. ( ) () 11 1 22 34 32 16 sin sin 2.60 10 mm sin 16 7.17 10 mm 717 nm D D θθ λ θλ θ −− =∆= °= ° =→= = × ° = × = 3. The angle to the first maximum is about halfway between the angles to the first and second minima. We use Eq. 35-2 to calculate the angular distance to the first and second minima. Then we average these to values to determine the approximate location of the first maximum. Finally, using trigonometry, we set the linear distance equal to the distance to the screen multiplied by the tangent of the angle. ( ) 1 99 12 66 1 sin sin 1 580 10 m 2 580 10 m sin 8.678 sin 17.774 3.8 10 m 8.678 = 13.23 tan 10.0 m tan 13.23 2.35 m mm m Dm D y ⎛⎞ =→ = ⎜⎟ ⎝⎠ ×× == ° = = ° + °+ ° ° ° = l 8. ( a ) There will be no diffraction minima if the angle for the first minimum is greater than 90 . ° We set the angle in Eq. 35-1 equal to 90

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GiancoliCh35Sol - Solutions to WebAssign HW#5 Giancoli Ch...

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