Solutions10

# Solutions10 - Physics 208 Assignment 10 Solutions RT...

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Unformatted text preview: Physics 208, Assignment 10, Solutions, 4/6/09, RT Problem 1. (a) Deflection x 2 = 0 . 018 m at a screen L = 2 . 0 m distant corresponds to an angular deflection θ 2 = tan − 1 . 018 2 = 0 . 0089976 also ≈ . 018 2 = 0 . 0090000 r , where, just this once, both the exact formula and the small angle formula have been displayed. For visible light the small angle version is almost always valid to practical accuracy. For microwaves the exact formula should normally be used (because the wavelength is not negligible compared to the apparatus dimensions). (b) For a slit of width a , the condition for second minimum is a sin θ 2 = 2 λ . So a ≈ 2 λ θ 2 = 2 × 440 × 10 − 9 . 009 = 0 . 98 × 10 − 4 m . Problem 2. (a) Increasing λ by the ratio 660/440=1.5 would increase transverse an- gles and lengths proportionally. So x 2 → 1 . 5 × . 018 = 0 . 027 m and θ 2 → 1 . 5 × . 009 = 0 . 0135 r (b) Then doubling a cuts x 2 in half, x 2 → . 0135 m. Problem 3. The path of a ray entering the eye is somewhat complicated because it passes from vacuum to cornea to lens medium to vitreous hu- mor. The equation i − 1 + p − 1 = f − 1 needs, therefore, to be modified somewhat. But we need not be concerned with this since we can assume that the person looking at the two oncoming headlights of a car (at a dis- tance L and separated, let us guess, by D = 1 . 5 m) focuses the two images onto the retina, which we take to be L ret ....
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Solutions10 - Physics 208 Assignment 10 Solutions RT...

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