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To calculate the diffraction pattern for two (or any number of) slits, we need to generalize the method we just used for a single slit. That is, across each slit, we place a uniform distribution of point sources that radiate Huygens wavelets , and then we sum the wavelets from all the slits.
Example 14.1: Double-Slit Experiment Suppose in the double-slit arrangement, d =0.150mm, L =120cm, λ=833nm, and y =2.00cm . (a) What is the path difference δ for the rays from the two slits arriving at point P? (b) Express this path difference in terms of λ. (c) Does point P correspond to a maximum, a minimum, or an intermediate condition?
To calculate the diffraction pattern for two (or any number of) slits, we need to generalize the method we just used for a single slit. That is, across each slit, we place a uniform distribution of point sources that radiate Huygens wavelets, and then we sum the wavelets from all the slits.
Summary. Young’s double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sinθ = mλ(form = 0, 1, −2, 2, −2,...) d sin. .
Find the angular positions of the interference maxima within the central peak of a double-slit diffraction for a monochromatic light of wavelength 628 nm on slits of width of \(1.5\ \mu\text{m}\) separated by \(4.0\ \mu\text{m}\text{.}\)
Theory. When plane monochromatic light wave front of wavelength, λ, falls normally on a system of two slits of equal widths, b, separated by a distance, g, the luminous intensity I of the beam diffracted in the direction φ is given by: 2 sin ( . b . sin / ) I ( ) I cos. 2 ( . g . sin / ). .....................( 1) 0 ( . b .
• Light with λ=440 nm passes through a double slit and produces the diffraction pattern above. What is (a) the slit width ‘a’ (b) the slit separation ‘d’? •I=I0 [sin(φ/2)/(φ/2)]2 (cos(δ/2))2 • first minimum of diffraction envelope is at θ~50=5π/180 rads φ/2= πasinθ/λ= π • a= λ/sinθ= 5.048 µm