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Unformatted text preview: 10. (a) We note that ray 1 travels an extra distance 4 L more than ray 2. To get the least possible L which will result in destructive interference, we set this extra distance equal to half of a wavelength: 420.0 nm 4 52.50 nm 2 8 8 L L λ λ = ⇒ = = = . (b) The next case occurs when that extra distance is set equal to 3 2 λ . The result is 3 3(420.0 nm) 157.5 nm 8 8 L λ = = = . 11. (a) We wish to set Eq. 3511 equal to 1/ 2, since a halfwavelength phase difference is equivalent to a π radians difference. Thus, L n n min . . = − = − = = λ 2 2 1 620 145 1550 155 b g b g nm 2 1.65 nm m. µ (b) Since a phase difference of 3 2 (wavelengths) is effectively the same as what we required in part (a), then L n n L = − = = = 3 3 3 1 5 5 4 6 5 2 1 λ 2 b g b g min . . µ µ m m . 14. (a) We use Eq. 3514 with m = 3: θ = F H G I K J = × × L N M M O Q P P = − − − − sin sin . . 1 1 9 6 2 550 10 7 70 10 0 216 m d λ m m rad. c h (b) θ = (0.216) (180°/ π ) = 12.4°. 16. (a) For the maximum adjacent to the central one, we set m = 1 in Eq. 3514 and obtain ( )( ) 1 1 1 1 1 sin sin 0.010rad. 100 m m d θ − − = ⎡ ⎤ ⎛ ⎞ = = = ⎢ ⎥ ⎜ ⎟ ⎝ ⎠ ⎣ ⎦ λ λ λ (b) Since y 1 = D tan θ 1 (see Fig. 3510(a)), we obtain y 1 = (500 mm) tan (0.010 rad) = 5.0 mm....
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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.
 Fall '09
 STEIGMAN
 Physics

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