next up previous
Next: About this document ... Up: No Title Previous: No Title

Useful Relations for the Final Exam in 123, WI 1999.


    $\displaystyle v~=~\nu\lambda
\qquad
\qquad ,
\qquad
\qquad
\nu~=~{\omega\over 2...
...\qquad
k~=~{2\pi\over\lambda}
\qquad
\qquad ,
\qquad
\qquad
v~=~{\omega\over k}$  
       
    $\displaystyle \nu_{\rm obs.}~=~\nu_0 \ \left( { c_s \pm v_{\rm obs.} \over c_s ...
...ight)
\qquad
\qquad
\qquad ,
\qquad
\qquad
\qquad
\nu_b~=~\vert\nu_1-\nu_2\vert$  
       
    $\displaystyle I~=~{P\over A}
\qquad
\qquad ,
\qquad
\qquad
c_{\rm air}~=~v_{\rm...
...40\ {\rm m/s}
\qquad
\qquad ,
\qquad
\qquad
c~=~2.99792458\times 10^8~{\rm m/s}$  
       
    $\displaystyle {\rm SL}~=~10\ \log_{10}\left({I\over I_0}\right)
\qquad
\qquad ,...
...quad
I_0~=~10^{-12}\ {\rm W/m^2}
\qquad
\qquad ,
\qquad
\qquad
P~=~{dE\over dt}$  
       
    $\displaystyle \nu_n = n\ {v\over 2 L}
\qquad
\qquad ,
\qquad
\qquad
\nu_m = (2 m+1) \ {v\over 4 L}
\qquad
\qquad ,
\qquad
\qquad
A~=~4\pi r^2$  
       
    $\displaystyle v = \sqrt{T\over\mu} \ =\ \sqrt{B\over\rho}
\qquad
\qquad ,
\qquad
\qquad
y(x\ ,\ t) = A\ \sin\left( k x - \omega t + \phi\right)$  
       
    $\displaystyle {\bf S}~=~{1\over\mu_0}{\bf E}\times {\bf B}
\qquad ,
\qquad
I~=~...
...{\bf E}_{\rm max}\vert^2
\qquad ,
\qquad
\mu_0~=~4\pi\times 10^{-7}~{\rm T m/A}$  
       
    $\displaystyle n_1\sin\theta_1~=~n_2\sin\theta_2
\qquad
\qquad ,
\qquad
\qquad
\...
...p}\vert~=~{E\over c}
\qquad
\qquad ,
\qquad
\qquad
{\bf F}~=~{d{\bf p}\over dt}$  
       
    $\displaystyle {1\over O}+{1\over I}~=~{1\over f}
\qquad
\qquad ,
\qquad
\qquad
...
...quad
\qquad
{1\over O}+{1\over I}~=~(n-1)\left({1\over R_1}-{1\over R_2}\right)$  
       
    $\displaystyle I(\theta)~=~I_0 \cos^2\beta \left({\sin\alpha\over\alpha}\right)^...
...r\lambda}
\qquad
\qquad ,
\qquad
\qquad
\alpha~=~{\pi a \sin\theta\over\lambda}$  
       
    $\displaystyle a\sin\theta~=~m\lambda
\qquad
\qquad ,
\qquad
\qquad
d\sin\theta~=~m\lambda
\qquad
\qquad ,
\qquad
\qquad
d\sin\theta~=~(m+{1\over 2})\lambda$  
       
    $\displaystyle r~=~\sqrt{m\lambda R}
\qquad
\qquad ,
\qquad
2 n d~=~m\lambda
\qquad
\qquad ,
\qquad
2 n d~=~(m+{1\over 2})\lambda$  
       
    $\displaystyle P~=~{{\bf F}\cdot {\bf n}\over A}
\qquad
\qquad ,
\qquad
\qquad
a...
...quad
\qquad ,
\qquad
\qquad
\phi~=~{\bf k}\cdot {\bf x}~=~2\pi {x\over \lambda}$  
       
    $\displaystyle I~=~I_0\cos^2\theta
\qquad
\qquad ,
\qquad
\qquad
I~=~{1\over 2} I_0\cos^2\theta
\qquad
\qquad ,
\qquad
\qquad
\tan\theta_p~=~{n_2\over n_1}$  
       
    $\displaystyle \sin\theta_c~=~{n_2\over n_1}
\qquad
\qquad ,
\qquad
\qquad
m~=~-...
...u_{\rm obs.}~=~\nu_0 \ \
{\sqrt{1-{v^2\over c^2}}\over 1-{v\over c}\cos\theta }$  


next up previous
Next: About this document ... Up: No Title Previous: No Title
Martin Savage
1999-03-12