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## GR0877 Problem 89

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) The reflection coefficient $R$ is defined in terms of the incident and reflected probability current density $j$: $\displaystyle R = \frac{|j_{refl}|}{|j_{inc}|}$, where $\displaystyle j_{refl} = - \frac{\hbar k}{m} |B|^2$ and $\displaystyle j_{inc} = \frac{\hbar k}{m} |A|^2$. Thus, $\displaystyle R = \frac{|B|^2}{|A|^2}$. According to the standard boundary conditions, $\psi$ is always continues and $d \psi/dx$ is continues (except at points where the potential becomes infinite). Therefore, for the point $x = 0$ one has $A + B = C$ for continuity of $\psi$ and $A k_1 - Bk_2 = Ck_2$ for continuity of $d \psi/dx$. From these two equations $\displaystyle \frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2}$ and $\displaystyle R = \frac{|B|^2}{|A|^2} = \left(\frac{k_1 - k_2}{k_1 + k_2}\right)^2$.

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1. 15.10.2011 at 09:30

Fastest way to solve this one is by observing the limit as k2->k1 (such that the added potential step gets smaller). You can see that only (D) has the limit of R=0.

2. 10.11.2011 at 07:42

You may also notice that from Fresnel’s equations (d) is by definition the reflection coeffiecent.

3. 11.10.2012 at 03:51

I believe there is a typo in that the first derivative continuity should give k_1A-k_1B=k_2C since the reflected and incident waves have the same wavenumber in this problem.

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