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draft_page_2 [2014/04/17 22:54]
veckert [Problem 54 part B]
draft_page_2 [2014/05/18 22:33] (current)
tom_grass [Practice/work here if other pages are temporarily locked by other editors]
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 +====Equation Sheet Ch.28====
 +
 +28-1:  Conditions for Bright Fringes (Constructive Interference) In a Two Slit Experiment:  ​
 +    *$d sin θ = m\lambda$. ​ $d$ is the slit separation
 +    *$m=0,​±1,​±2,​±3.,​..$
 +    *$m=0$ occurs at $θ=0$, this is the central bright fringe. ​
 +    *Positive values of $m$ are above the central bright fringe, negative values are below. ​
 +    *Solving for $θ:  θ = sin^{-1} (m\frac{\lambda}{d})$
 +
 +28-2:  Conditions for Dark Fringes (Destructive Interference) in a Two Slit Experiment:  ​
 +      * $d sin θ = (m-\frac{1}{2})\lambda.$ ​ $m = 1,2,3...$ (above central bright fringe)
 +      * $d sin θ = (m+\frac{1}{2})\lambda.$ ​ $m = -1,​-2,​-3...$ (below central bright fringe)
 +      * Solving for $θ:  θ = sin^{-1} [(m ± \frac{1}{2})\frac{\lambda}{d}]$. ​ $+ or -$ depending on location. ​
 +
 +28-3  Linear Distance from Central Fringe:  ​
 +        * $y = L tan θ$.  L is the distance to the screen. ​
 +        * Solving for $θ$ of a bright fringe: ​ $θ = tan^{-1} (m\frac{y}{L})$
 +        * Solving for $\lambda: ​ \lambda = \frac{d}{m}sinθ$
 +        * Solving for $θ$ of a dark fringe: ​ $θ = sin^{-1}[(m±\frac{1}{2})\lambda/​d]$. ​ $+ or -$ depending on location.
 +
 +28-12  Conditions for Dark Fringes in Single-Slit Interference:  ​
 +       * $Wsin θ = m\lambda$. ​  $m = ±1,​±2,​±3...$
 +       * Solving for $\lambda: \lambda = \frac{Wsinθ}{m}$ \
 +       * Solving for $θ:  θ = sin^{-1}(\frac{m\lambda}{W})$
 +
 +28-14  First Dark Fringe for the Diffraction Pattern of a Circular Opening:  ​
 +      * $sinθ = 1.22\frac{\lambda}{D}$
 +
 +28-15  Rayleigh'​s Criterion:
 +     * $θ_{min} = 1.22\frac{\lambda}{D}$
 +     * Note: $\lambda$ is dependent on the diffraction of the material that the light is traveling through. ​ If the diffraction is $n, \lambda$ becomes $\frac{\lambda}{n}$
 +
 +28-16: ​ Constructive Interference in a Diffraction Grating:  ​
 +     * $d sin θ = m\lambda$. ​   $m = ±1,​±2,​±3...$
 +     * Solving for $d:  d = \frac{m\lambda}{sinθ}$
 +
 ===Problem 1.3.13.13=== ===Problem 1.3.13.13===
 Two pendulums (or, //​pendula//​) are made of identical 1 kg masses suspended on two weightless strings, 40.0 and 40.5 cm in length. If these pendulums are deflected from vertical by 5 cm and released at the same time, how long will it take for them to get completely "out of step" with each other? Two pendulums (or, //​pendula//​) are made of identical 1 kg masses suspended on two weightless strings, 40.0 and 40.5 cm in length. If these pendulums are deflected from vertical by 5 cm and released at the same time, how long will it take for them to get completely "out of step" with each other?
draft_page_2.1397775280.txt.gz · Last modified: 2014/04/17 22:54 by veckert