Physics

Physics · optics

Problem

d. Now, suppose the hemispherical shell is removed, and we instead observe the interference pattern on the top wall. To the nearest integer, what is the total number of fringes that appear on the top wall? You may assume that d≪Ld \ll L. Given constants: ## Mirror Mirror on the Wall Consider a square room with side length LL. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength λ\lambda is placed a distance dd above the center of the mirror, where λ≪d≪L\lambda \ll d \ll L. <image_1> *Remember that the phase of light reflected by a mirror changes by 180∘180^{\circ}. Context question: a. On the right wall, an interference pattern emerges. What is the distance yy between the bottom corner and the closest bright fringe above it? Hint: you may assume λ≪y≪L\lambda \ll y \ll L as well. Context answer: y=λL8dy=\frac{\lambda L}{8 d} Context question: c. Now suppose we place a transparent hemispherical shell of thickness ss and index of refraction nn over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell. hemispherical shell <image_2> At what yy is the fringe closest to the bottom-most corner now? (You may find it convenient to use ⌊x⌋\lfloor x\rfloor, the largest integer below xx.) Context answer: y=L4d((n−1)s−λ⌊(n−1)sλ−12⌋−λ2)y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)

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