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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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final_exam_review.txt · Last modified: 2014/06/10 19:28 by wikimanager

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