21 Optics 6 - Can we see any interference without a laser Some math the slits are two coherent sources The distances to the observation points are

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Unformatted text preview: Can we see any interference without a laser? Some math: the slits are two coherent sources . The distances to the observation points are r 1 and r 2 . Their difference θ sin 1 2 d r r = − L y L y / ) / ( tan 1 ≈ = − θ L dy r r / 1 2 ≈ − for small angles θ , small y/L Constructive (a bright strip) θ sin 1 2 d r r = − L dy r r / 1 2 ≈ − λ θ m d = sin Destructive (a dark strip) λ θ ) 2 / 1 ( sin + = m d θ Approximation used: L d << for small L y / In the case when 1 / << L y L y / tan sin ≈ ≈ θ θ L y 1 r 2 r y ) ( y S 4 S 2 S Positions of the bright and dark fringes (maxima and minima of interference) d L y λ = Δ d L m y bright λ = d L m y dark λ ) 2 / 1 ( + = The distance between the fringes: d L y / λ = Δ θ ... 2 , 1 , ± ± = m Angular positions of the bright and dark fringes: d m / sin λ θ = d m / ) 2 / 1 ( sin λ θ + = ) ( cos 4 ) sin ( cos 4 2 2 y L d S d S S λ π λ θ π = = Does this look any familiar?...
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This note was uploaded on 12/13/2011 for the course PHYS 2C PHYS 2C taught by Professor Groisman during the Spring '11 term at UCSD.

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21 Optics 6 - Can we see any interference without a laser Some math the slits are two coherent sources The distances to the observation points are

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