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final_exam_review [2014/06/10 06:07]
hy2 [Review question 6]
final_exam_review [2014/06/10 19:28] (current)
wikimanager [Review question 6]
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 +====Live clarifications from the June 10 exam====
 +  * **Problem 2C**: "in the opposite direction"​ means "​opposite to the motion of the ship"
 +  * **Problem 2D**: "how long" means "find the duration (time) of"
 +  * **Problem 3**: in **parts D, E, F** use  "mass of the electron",​ not the mass of <​sup>​90</​sup>​Y found in **part B**
 ======Final exam review====== ======Final exam review======
 The following questions and problems are courtesy of Justin Dunlap The following questions and problems are courtesy of Justin Dunlap
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   * [....] E) 1.5 m   * [....] E) 1.5 m
-<color green>​Use the Single-Slit Dark Fringe Diffraction ​equation: ​Wsin(theta)=m(lambdato find angle at which light bends. ​Θ=arcsin(mλ/W). Then plug in Θ into the linear distance equation ​(L=tanΘ*Y) to solve for the screen to slit distance, L. </​color>​+  - <color green>​Use the single-slit diffraction ​equation ​for dark fringes</​color>​ 
 +    * <color green>​$W\sin(\theta)=m\lambdato find the angle at which the light bends. ​</​color>​ 
 +    * <color green>​Second dark fringe means $m=2$. </​color>​ 
 +    * <color green>​Therefore $\theta=\arcsin\big(\!\frac{2\lambda}{W}\!\big)$</​color>​ 
 +  - <color green>Then plug in $\theta$ ​into the linear distance equation ​</​color>​ 
 +    * <color green>$y=L\tan\theta\;​$, ​to solve for the screen to slit distance, ​//L//</​color>​ 
 +  - <color green>​The final formula is </​color>​ 
 +    * <color green>​$L=\frac{y}{\tan\left[\arcsin\big(\!\frac{2\lambda}{W}\!\big)\right]}$</​color>​
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   * [....] E) It depends on whether you are moving towards or away from Randy.   * [....] E) It depends on whether you are moving towards or away from Randy.
-<​color ​green></​color>​+<​color ​blue>The speed of light in a vacuum $\left(c=3.00\times 10^8\frac{\text m}{\text s}\right)$ is the same in all inertial frames of reference.</​color>​
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   * [....] E) 311 nm   * [....] E) 311 nm
-<​color ​green></​color> ​+<​color ​brown>Find the peak frequency:​ 
 +f=(5.88*10^8)(6000)=3.53*10^14 HZ 
 +solve for the wavelength using the calculated frequency and the speed of light constant, c. λ= (3*10^8)/​(3.53*10^14)=8.5*10^-7m=850nm. 
 +</​color> ​
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final_exam_review.1402380423.txt.gz · Last modified: 2014/06/10 06:07 by hy2