CourseNotes.67 - R lying on a flat glass plate and...

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Figure 49: Figure for Problem 35.23 We now need to use some physical reasoning to figure out when the two will interfere destruc- tively at the largest x P . Notice that if x P goes off to infinity, then the difference between the two goes to zero. Hence, the maximum x P which leads to destructive interference is when the difference is 1 2 . p x 2 P + d 2 - x P λ = 1 2 This is a quadratic equation in x P that we can easily solve. x 2 P + d 2 = ± x P + λ 2 ² 2 x 2 P + d 2 = x 2 P + λx P + λ 2 4 λx P = d 2 - λ 2 4 x P = d 2 λ - λ 4 Hence, our solution is x P = 7 . 87 μm . Problem 35.75: Figure 50a shows a lens with radius of curvature
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Unformatted text preview: R lying on a flat glass plate and illuminated from above by light with wavelength λ . Figure 50b a photograph taken from above the lens) shows that circular interference fringes (called Newton’s rings) appear, associated with the variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assuming r/R ± 1. Figure 50: Figure for Problem 35.75 67...
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