ch36-p049 - 49. (a) Since d = (1.00 mm)/180 = 0.0056 mm, we...

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mm 11 2 2 λ λ = for which the smallest possible choices are m 1 = 5 and m 2 = 4. Returning to Eq. 36-25, then, we find () 4 1 10 mm) sin sin sin 0.36 21 . 0.0056 mm m d θ −− ⎛⎞ λ 5(4.0× == = = ° ⎜⎟ ⎝⎠ (c) There are no refraction angles greater than 90°, so we can solve for “ m max ” (realizing it might not be an integer): max 4 22 sin90 0.0056 mm 11 10 mm dd m ° = λλ 5 . 0 × where we have rounded down. There are no values of m (for light of wavelength
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