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final_exam_review [2014/06/09 07:39]
wikimanager [Review problem 2]
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) None of the above   * [....] E) None of the above
-<color green></​color> ​+<color green>de Broglie wavelength equation: (λ)=h/p. If p=m*v then the electron'​s de Broglie wavelength is as follows: (λ)=h/​(me*v) while the proton'​s de Broglie wavelength is (λ)=h/​(mp*v). Since the proton and the electron share the same final speed (v) then the the electron will have a larger wavelength than the proton due to its smaller mass in the denominator.</​color>​
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   * [....] E) a 90$^\circ$ phase change in the reflected beam.    * [....] E) a 90$^\circ$ phase change in the reflected beam. 
-<color green></​color>​+<color green>If n1>n2 then there is no phase change; however, if n1<n2 then the phase changes by 1/2 lambda. In this case the beam of light is traveling in glass which has a greater index of refraction than air does. </​color>​
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   * <color green>​$v_{23}=0.34\,​c$</​color>​   * <color green>​$v_{23}=0.34\,​c$</​color>​
   * <color green>​$v_{21}=?​$</​color>​   * <color green>​$v_{21}=?​$</​color>​
-    * <color green>​$v_{21}=\frac{v_{23} - v_{13}}{1-\frac{v_{22}\,​\cdot\,​v_{13}}{c^2}}$ $=\frac{0.34\,​c -0.21\,​c}{1-\frac{(0.34\,​c)(0.21\,​c)}{c^2}}$ $=\frac{0.13\,​c}{0.9286} =0.14\,​c$</​color>​+    * <color green>​$v_{21}=\frac{v_{23} - v_{13}}{1-\frac{v_{23}\,​\cdot\,​v_{13}}{c^2}}$ $=\frac{0.34\,​c -0.21\,​c}{1-\frac{(0.34\,​c)(0.21\,​c)}{c^2}}$ $=\frac{0.13\,​c}{0.9286} =0.14\,​c$</​color>​
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   * [....] E) 1.5 m   * [....] E) 1.5 m
-<color green></​color>​+  - <color green>Use the single-slit diffraction equation for dark fringes: </​color>​ 
 +    * <color green>​$W\sin(\theta)=m\lambda$ to 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.txt · Last modified: 2014/06/10 19:28 by wikimanager