# Physics 203 at Portland State 2014

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final_exam_review
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final_exam_review [2014/06/06 21:46]
ps4 [Review problem 1]
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) 18.2$^\circ$ ​     * [....] E) 18.2$^\circ$ ​

-<color green>​each slit gains an extra lamda --> λ=600nm = 0.6μm --> sinΘ=(λ/D(0.6/1.7) --> 20.7 degrees</​color>​+Steps:
+  - <color green>For the 1<​sup>​st</​sup>​ maximum, ​each slit gains one extra wavelength of path compared to the adjacent slit</​color>​
+  ​<color green>$\lambda=600\,​{\text{nm}}$ $=0.6\,​\mu{\text ​m}$</​color>​
+  ​<color green>$\sin\theta=\frac{1\,​\cdot\,​\lambda}{D=\frac{0.6\,\mu{\text m}}{1.7\,\mu{\text m}}$ $=0.353$</​color>​
+  ​<color green>$\theta=\arcsin 0.353$ $=20.7^\circ$</​color>​

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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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* [....] E) $\;4.05 \times 10^{-7}\,​$rad   * [....] E) $\;4.05 \times 10^{-7}\,​$rad

-<color green>​1.22λ/--> (1.22•620x10^9)/5.6 --> ​1.35x10^-7</​color>​+<color green>Use Rayleigh criterion and small-angle approximation ($\sin\theta_\text{min}=$ $\tan\theta_\text{min}$ $=\theta_\text{min}$,​ in radians):</​color>​
+  * <color green>​$\theta_\text{min}=$ $\frac{1.22\lambda}{D}=$ $\frac{1.22\,​\cdot\,​620\times 10^{-9}\,{\text m}}{5.6\,{\text m}}=$ $1.35\times 10^{-7}\,$rad</​color>​

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* [ <color green>​X</​color>​ ] E) $\;0.14\,c$   * [ <color green>​X</​color>​ ] E) $\;0.14\,c$

-<color green></​color>​+  * <color green>$v_{13}=0.21\,​c$</​color>​
+  * <color green>​$v_{23}=0.34\,​c$</​color>​
+  * <color green>​$v_{21}=?​$</​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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-   ​*$v_{13}=.21c$
-   ​*$v_{23}=.34c$
-   ​*$v_{21}=?​$
-  -$v_{21}=\frac{v_{23} - v_{13}}{1-\frac{(v_{22})(v_{13})}{c^2}}$
-  -$=\frac{.34c -.21c}{1-\frac{(.34c)(.21c)}{c^2}}$
-  -$=\frac{.13c}{.9286} =.14c$
====Review question 6==== ====Review question 6====
Light of wavelength 687 nm is incident on a single slit 0.75 mm wide.  At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the screen?  ​ Light of wavelength 687 nm is incident on a single slit 0.75 mm wide.  At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the screen?  ​
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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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* [ <color green>​X</​color>​ ] E) It is doubled.   * [ <color green>​X</​color>​ ] E) It is doubled.

-<​color ​green></​color>​+<​color ​blue>Establish the relationship between the frequency and the momentum of a photon:</​color> ​
+  * <color blue>​$\lambda = \frac{c}{f}$ </​color>​
+  * <color blue>$p = \frac{h}{\lambda}$ $= \frac{hf}{c}$ $=\left(\frac{h}{c}\right)\cdot f$</​color>​
+    * <color blue>the ratio in brackets is the Planck constant divided by the speed of light constant, it is also a constant</​color>​

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* [....] E) 1.90    * [....] E) 1.90

-<color green></​color>​ +<color green>Use the fact that the wavelength of light in a material is inversely proportional to its refraction index</​color>​
-<color green> ​λvac/λfilm = 469/360 = 1.3</​color>​+  ​* ​<color green>$\frac{\lambda_\text{vac}}{\lambda_\text{film}} ​=$ $\frac{n_{\,​\text{film}}}{n_\text{vac}}$</color>
+  * <color green>​$n_\text{vac}=1$</​color>​
+  * <color green>​$n_{\,​\text{film}} \frac{\lambda_\text{vac}}{\lambda_\text{film}}$ $= \frac{469\,​{\text{nm}}}{360\,​{\text{nm}}} ​= 1.3$</​color>​
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====Review question 11==== ====Review question 11====
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* [....] E) $\;​4.1\times 10^{-8}\,​$kg ​     * [....] E) $\;​4.1\times 10^{-8}\,​$kg ​

-<color green>(c^2/2G)*R = (3 10^8)^2 ​* 10^-15/(2(6.7 10^-11)) = 6.7 10^-11 kg</​color>​ +<color green>Use the Schwarzschild radius formula:</​color>​
+  * <color green>​$R ​\frac{2GM}{c^2}$</color>
+  ​<color green>$M = \frac{c^2}{2G}R$ $\frac{(3 \times ​10^8\,​\frac{\text m}{\text s})^2}{2\,\cdot\,6.7 \times ​10^{-11}\,\frac{ {\text m}^3}{ {\text{kg}}\cdot{\text s}^2}}\cdot 10^{-15}\,​{\text m}$ $= 6.7 \times ​10^{-11}\,$kg</​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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* a) If the beam of light is incident normal to the film, what is the minimum thickness of the film that will make it look bright? (4 pts)   * a) If the beam of light is incident normal to the film, what is the minimum thickness of the film that will make it look bright? (4 pts)
* <color green> 94 nm</​color>​     * <color green> 94 nm</​color>​
-    ​* <​color ​black> $\lambda_film$ ​[$\lambda_ vac$]/(n= 500 nm 1.33 3(500) / 4 = 376 </​color>​ +      ​* <​color ​blue> $\lambda_\text{film} ​=$\frac{\lambda_\text{vac}}{n}=\frac{500\,{\text{nm}}}{1.33} \approx\frac{3}{4}\!\cdot\!500\,​{\text{nm}}= 376\,​{\text{nm}}$​</​color>​ - * <​color ​black2(thickness) = [$\lambda_film$]/</​color>​ + * <​color ​blue>To get a bright spot, the round-trip within the film should give a half-wavelength delay, since the dissimilar reflection is already introducing a half-a-period delay:</​color>​ - thickness =$\lambda_film$/n = 376/n 94nm + * <color blue>$2t = \frac{\lambda_\text{film}}{2}\;​$, where //t// is the thickness of the film </​color>​ - ​ + <color blue>$t = \frac{\lambda_\text{film}}{4}$​$= \frac{376\,​{\text{nm}}}{4} ​94\,​$nm</​color>​ - * b) At this thickness, what is one other wavelength of light that would work to get a bright appearance? (4 pts) + * b) At this thickness, what is one other wavelength of light <color darkorange>​(in vacuum)</​color> ​that would work to get a bright appearance? (4 pts) - * <color green> ​282 nm</​color>​ + * <color green> ​166 nm</​color>​ - * <​color ​black> 2(thickness) = 3/2$\lambda$</​color>​ + * <​color ​blueThe next bright fringe happens when the round-trip delay in the film is$\frac{3}{2}$wavelengths ​(in addition to the the half-cycle delay due to dissimilar reflections)</​color>​ - * <​color ​blackthickness = 3/4$\lambda$​3$\lambda$/4 </​color>​ + * <color blue>$2t \frac{3}{2}\lambda_\text{film}$</​color>​ - 2n(thickness)/​$\lambda$- (0.5) + * <​color ​blue>$\lambda_\text{film}=$\frac{4t}{3}$</​color>​
-    * (2n/$\lambda$)thickness ​m + 0.+        <color blue> ​$\lambda_\text{vac}=$ $n\lambda_\text{film}=$\frac{4nt}{3}=\frac{4\,​\cdot\,​1.33\,​\cdot\,​94\,​{\text{nm}}}{3}=166.7\,$nm</color> - * thickness ​[$\lambda$/2n] [m+0.5] + - *$\lambda= 500/3  +
-    * 166 nm +
-       +
-      ​+
* c) What is the next thickness that would make it look bright (using the original wavelength of light)? (4 pts)   * c) What is the next thickness that would make it look bright (using the original wavelength of light)? (4 pts)
* <color green> 282 nm</​color>​     * <color green> 282 nm</​color>​
-    ​* <​color ​blackthickness ​</​color>​ +      ​* <​color ​blue>Assuming $\lambda_\text{vac}=500\,$nm, the next bright fringe will again require $\frac{3}{2}$ wavelengths worth of round-trip delay in the film, to add up to two full wavelengths after adding the dissimilar reflection effect</​color>​
-      ​+        * <color blue> $2t = \frac{3}{2}\lambda_\text{film}$</​color>​
+        * <color blue> $t = \frac{3}{4}\lambda_\text{film}$</​color>​
+        * <color blue> $t = \frac{3}{4}\!\cdot\!376\,​{\text{nm}}=$ $282\,​{\text{nm}}$</​color>​
+      ​* <color blue>​Alternatively,​ use the thin-film formula with $m=1$</​color>​
+        * <color blue>​$\frac{2nt}{\;​\;​\lambda_\text{vac}} - \frac{1}{2} = m$</​color>​
+        * <color blue>​$\frac{2nt}{\;​\;​\lambda_\text{vac}} = m + \frac{1}{2}$</​color>​
+        * <color blue>$t = \frac{\lambda_\text{vac}}{2n}\left(m+\frac{1}{2}\right)$ $=\frac{\lambda_\text{vac}}{2n}\left(\frac{3}{2}\right)=$ $\frac{500\,​{\text{nm}}\,​\cdot\,​3}{2\,​\cdot\,​1.33\,​\cdot\,​2}=$ $282\,​{\text{nm}}$</​color>​
* d) What would you observe at the top of the film where the soap is very thin? Why? (3 pts)   * d) What would you observe at the top of the film where the soap is very thin? Why? (3 pts)
+    * <color blue> As $t\rightarrow 0$, the delay due to extra round-trip path becomes small (close to zero).</​color> ​
+    * <color blue>The extra $\pi$ phase delay due to dissimilar reflection will lead to destructive interference,​ i.e. a dark spot</​color>​
* <color green>We can use both the dark and the bright fringe equations to see what happens as the thickness //t// becomes very small.</​color>​     * <color green>We can use both the dark and the bright fringe equations to see what happens as the thickness //t// becomes very small.</​color>​
* <color green>​The equation that allows for an integer //m// will make sense in the case when //t// is small.</​color>​     * <color green>​The equation that allows for an integer //m// will make sense in the case when //t// is small.</​color>​
-      * <color green>​Bright ​Fringe: $\,​\frac{2nt}{\lambda_\text{vac}}-\frac{1}{2} =m$</​color>​+      * <color green>​Bright ​fringe: $\,​\frac{2nt}{\lambda_\text{vac}}-\frac{1}{2} =m$</​color>​
* <color green>As //t// becomes smaller (//t// -> 0), we have $\;​m\rightarrow -\frac{1}{2}$. </​color>​         * <color green>As //t// becomes smaller (//t// -> 0), we have $\;​m\rightarrow -\frac{1}{2}$. </​color>​
* <color green>​This is a non-integer number, so it is impossible for bright fringes to occur: the top of the film cannot appear bright.</​color>​         * <color green>​This is a non-integer number, so it is impossible for bright fringes to occur: the top of the film cannot appear bright.</​color>​
-      * <color green>​Dark ​Fringe: $\,​\frac{2nt}{\lambda_\text{vac}} =m$</​color>​+      * <color green>​Dark ​fringe: $\,​\frac{2nt}{\lambda_\text{vac}} =m$</​color>​
* <color green>As //t// becomes smaller (//t// -> 0), we have $\;​m\rightarrow 0$</​color>​         * <color green>As //t// becomes smaller (//t// -> 0), we have $\;​m\rightarrow 0$</​color>​
* <color green> This gives an integer //m// for dark fringes so the top of the film can appear dark. Therefore, if the film is very thin, it will look dark.</​color>​         * <color green> This gives an integer //m// for dark fringes so the top of the film can appear dark. Therefore, if the film is very thin, it will look dark.</​color>​
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====Review problem 2==== ====Review problem 2====
-The [[wp>​Large Hadron Collider]] (LHC) at [[wp>​CERN]] in Europe can accelerate protons (rest mass, $m_p = 1.673\times 10^{27}\,​$kg,​ $c = 2.998\times 10^8\frac{\text m}{\text s}$) to a speed of $v = 0.999999991\,​c$.+The [[wp>​Large Hadron Collider]] (LHC) at [[wp>​CERN]] in Europe can accelerate protons (rest mass, $m_p = 1.673\times 10^{-27}\,$kg, $c = 2.998\times 10^8\frac{\text m}{\text s}$) to a speed of $v = 0.999999991\,​c$.

-  * a) What is the rest energy (in MeV) of a proton? (1 eV $= 1.602\times 10^{19}\,​$J) (5pts) +  * a) What is the rest energy (in MeV) of a proton? (1 eV $= 1.602\times 10^{-19}\,$J) (5pts)
-    * <color green>​...</​color>​ +    ​* <color blue>Use the Einstein formula and unit conversion:</​color>​
-      * <color green>​...</​color>​+      ​* <color green>$E_0=m_pc^2$ $=1.673\times 10^{-27}\,​{\text{kg}}\cdot(2.998\times 10^8\frac{\text m}{\text s})^2=$ $1.504\times 10^{-10}\,​{\text J}$</​color>​
+      * <color green>$\frac{1.504\times 10^{-10}\,​{\text J}}{1.602\times 10^{-19}\,​\frac{\text J}{\text{eV}}}=$ $9.39\times 10^8\,​{\text{eV}}$ $=938.6\,​$MeV</​color>​
* b) Find the magnitude of the relativistic momentum of the protons in the LHC. Comparing it with the rest nonrelativistic value, is it bigger or smaller? By what factor? (5 pts)   * b) Find the magnitude of the relativistic momentum of the protons in the LHC. Comparing it with the rest nonrelativistic value, is it bigger or smaller? By what factor? (5 pts)
-    * <​color ​green>​...</​color>​ +    * <​color ​blue>$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-0.999999991^2}}\approx$ $\frac{1}{\sqrt{2\,​\cdot\,​9\times 10^{-9}}}$ $=7454$</​color>​
-      * <​color ​green>​...</​color>​+    * <color blue>​$p_\text{rel}=\frac{m_pv}{\sqrt{1-0.999999991^2}}$ $=7454\cdot 1.673\times 10^{-27}\,​{\text{kg}}\cdot 0.999999991\,​c$ $=3.74\times 10^{-15}\frac{\text{kg}\cdot{\text m}}{\text s}$</​color>​
+    * <​color ​blue>$p_\text{non-rel}=m_pv$ $=1.673\times 10^{-27}\,​{\text{kg}}\cdot 0.999999991\,​c$ $=5.01\times 10^{-19}\frac{\text{kg}\cdot{\text m}}{\text s}$ </​color>​
+    * <color blue>The relativistic value is larger by a factor of 7454</​color>​
* c) Compute a protons total energy in the LHC (in GeV). How much more energy is this compared to the rest energy? (5 pts)   * c) Compute a protons total energy in the LHC (in GeV). How much more energy is this compared to the rest energy? (5 pts)
-    * <​color ​green>...</​color>​ +    * <​color ​blue>$E=\frac{m_pc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ $=7454\cdot 1.673\times 10^{-27}\,​{\text{kg}}\cdot c^2$ $=1.121\times 10^{-6}\,​{\text J}$ $=6994\,​$GeV</​color>​
-      * <color green>...</​color>​+    * <color green>$\frac{E}{E_0}=$ $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=$ $\frac{6994\,​{\text{GeV}}}{0.9386\,​{\text{GeV}}}$ $=7454$</​color>​

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