# Physics 203 at Portland State 2014

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final_exam_review
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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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