Ch. 28 HW Solutions
2.) (a) The speakers are two coherent sources at the same location. They emit in phase so
the amplitudes are added: 2A.
(b) The speakers are 90 cm apart, or half a wavelength. Along the x-axis their waves are
180◦ out of phase so the interference is destructive: 0A.
(c) Same as part (b): 0A.
(d) The speakers are now 180 cm apart along the x-axis, or one full wavelength. The interference is then fully constructive so the combined amplitude is 2A.
7.) The wavelength of light is λ = 546 nm. Bright fringes occur for constructive interference, ∆s = nλ.
(a) The zeroth order fringe on the screen is equidistant from the slits, so ∆s = 0 m.
(b) The first order bright fringe occurs for ∆s = λ = 546 nm.
(c) The second order bright fringe occurs for ∆s = 2λ = 1092 nm.
(d) The third order bright fringe occurs for ∆s = 3λ = 1638 nm.
9.) The wavelength of light is λ = 589 nm, the slit width is d = 0.100 mm, and the
slit-to-screen distance is h = 1.50 m. Recall that for h d, ∆s/d = x/h.
(a) For the third bright fringe, interference is constructive and n = 3, so ∆s = 3λ. So
x = 3hλ/d = 2.65 cm. (b) For destructive interference, ∆s = (n + 1/2)λ. The first dark
fringe occurs at n = 0, second at n = 1, and third at n = 2. So x = 2.5hλ/d = 2.21 cm.
11.) 8000 lines/cm means the width between slits is d = 1/8000 cm/line = 0.000125 cm. (a)
For n = 2, θ = 72◦ . So λ = d sin θ/2 = 594 nm.
(b) Third order maximum would be at θ = sin−1 (3λ/d) = sin−1 (1.42) which is undefined.
There is no third order maximum for this situation.
17.) For single slit diffraction, θc = sin−1 (λ/w). For w = 0.020 mm and λ = 579 nm,
θc = 1.66◦ .
18.) For the previous problem and a screen h = 50 cm away, we have a triangle with
angle θc , adjacent side h, and opposite side x. tan θc = x/h or x = h tan θc = 1.45 cm. This
is the distance from the bright spot to the first dark spot, so the width of the bright spot
will be two times this size. w = 2.90 cm.
20.) For the two objects to be resolved, their angular separation must be greater than
θc . For λ = 550 nm and aperture diameter D = 102 cm, sin θc = 1.22λ/D. θc = 3.77 × 10−5
degrees. Set up a triangle with this angle, the earth-moon distance h = 3.84 × 108 m, and
the separation of the two objects, x. Then tan θc = x/h, so the objects must be at least
x = h tan θc = 252.6 m apart to be resolved.
28.) For the variable air (µair = 1.00) gap in question, one of the two reflections will
have an inverted phase. For near zero reflection we have destructive interference, 2d = nλ
where λ = 546 nm. To find the first four widths to get zero reflection we use n = 0, 1, 2, 3
(n = 0 causes destructive interference because one of the rays has an inverted phase). d = 0
nm, 273 nm, 546 nm, 819 nm.
29.) Same setup but for strong reflection we have constructive interference, 2d = (n + 1/2)λ,
where n = 0, 1, 2, 3. d = 136.5 nm, 409.5 nm, 682.5 nm, 955.5 nm
32.) The observed color of the reflected light is given by constructive interference of the
light. There is only one phase reversal in this setup (µair < µoil but µoil > µwater here).
Constructive interference is given for 2d = (n + 1/2)λ/µoil where µoil = 1.48. The second
fringe is in a less thick position (smaller n value). The thickness between the red fringes is
∆d = λ/2µoil = 202.7 nm.
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