Light of wavelength ?= 475.0 nm strikes the first screen with slit A. The second

Question

Light of wavelength ?= 475.0 nm strikes the first screen with slit A. The second screen, x1 = 0.7 meters behind the first screen, has two slits, B and C . The third screen is x2 = 1.5 meters behind the second screen. It has slit O, which is level with slit A. A lightmeter measures the light intensity at the slit O. When light is sent through slit A and measured at the slit O with either slit B or slit C open one slit at a time, the intensity at the point O is the same: I0 = 0.5 W/m2. (The slit widths can always be adjusted so that this is true, but for this problem you can/should ignore the width of all slits.) Slit B is at height y1 = 2.0 mm above slit A. Slit C is at height y2 = 1.0 mm above slit A. Note that the drawing is not drawn to scale. 1) What is the light intensity measure at the point O when both slits B and C are open? I = 2) Now we move slits B and C in such a way that both paths ABO and ACO are made longer by 0.2 ?m. The new intensity measured at point O is I = 3) What is the smallest wavelength greater than 475.0 nm for which the intensity at point O will be zero? ? =

Explanation / Answer

a) .2563 b) .2563 c) 479 For a, you have to find the path difference between ABO and ACO. This is 3.142851 E-6. Plug this into phi = 2pi (delta) / lambda. This gets you 41.572874. Now you should have everything you need to solve I = 4(I1) cos^2 (phi/2). B should be the same, because you're increasing both ABO and ACO by the same amount, therefore not changing the path difference.

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