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phy142kk:lectures:fr1 [2015/05/11 10:03]
kkumar
phy142kk:lectures:fr1 [2015/05/11 10:14] (current)
kkumar [Diffraction grating]
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 {{reflectionphasechange.png}} {{reflectionphasechange.png}}
 +===== Air wedge and Other Effects =====
 +
 +
 +{{airwedge.png}}
 +
 +For a single wavelength dark stripes will occur whenever ​
 +
 +$2t=m\lambda$<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;</​html>​ (m=0,​1,​2,​..)
 +
 +and bright stripes will occur whenever ​
 +
 +$2t=(m+\frac{1}{2})\lambda$<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;</​html>​ (m=0,​1,​2,​..)
 +
 +
 ===== Anti reflective coating ===== ===== Anti reflective coating =====
 +
 +Ideally we would use a coating that produced an equal amount of reflection at both interfaces, but there is no suitable material with the required refractive index, ​ $n=1.26$, ​ so we use magnesium flouride MgF<​sub>​2</​sub>​.
  
 {{antreflectivecoating.png}} {{antreflectivecoating.png}}
  
-As the two reflections both occur from more optically dense media they both experience a  ​$\frac{\lambda}{2}$ ​phase change ​on reflection. To have the light be out of phase we need to light that goes through the coating to have advanced by $\frac{\lambda}{2}for destructive interference ​to occur. Critically, when destructive interference occurs ​the light is not lost, but is instead transmitted. As the wavelength of light in a medium is given by $\lambda=\frac{\lambda_{0}}{n}$ where $n$ is the refractive index of the medium and $\lambda_{0}$ is the wavelength of the light in free space, the thickness of the coating should be $\frac{\lambda}{4n_{2}}$.+As the two reflections both occur from more optically dense media they both experience a phase change of $\pion reflection which corresponds ​to advancing ​the wave by $\frac{\lambda}{2}$ . 
  
-In practice ​the light incident will not all be the same wavelength, so the thickness ​of the coating ​is typically chosen ​to work optimally in the center of the visible band (~550nm).+To have the light be out of phase we need to light that goes through ​the coating to have advanced by $\frac{\lambda}{2}$ for destructive interference to occur
  
-===== Air wedge =====+Critically, when destructive interference occurs the light is not lost, but is instead transmitted. ​
  
-{{airwedge.png}}+As the wavelength of light in a medium is given by $\lambda=\frac{\lambda_{0}}{n}$ where $n$ is the refractive index of the medium and $\lambda_{0}$ is the wavelength of the light in free space, the thickness of the coating should be $\frac{\lambda}{4n_{2}}$.
  
-For a single wavelength dark rings will occur whenever ​ 
  
-$2t=m\lambda$<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;</​html>​ (m=0,1,2,..)+===== Intensity for double slit interference =====
  
-and bright rings will occur whenever ​+As we now have the intensity in terms of the phase difference $\delta$
  
-$2t=(m+\frac{1}{2})\lambda$<​html>&​nbsp;&​nbsp;&​nbsp;&​nbsp;&​nbsp;</​html>​ (m=0,​1,​2,​..)+$I_{\theta}=I_{0}\cos^{2}\frac{\delta}{2}$
  
-For white light different colors will experience constructive interference at different thicknesses,​ leading ​to the colorful lines we see when an air gap is under normal light.+we should return ​to our diagram to find the intensity in terms of $d$ and $\theta$
  
 +{{twoslitphase.png}}
 +
 +We can see that 
 +
 +$\frac{\delta\lambda}{2\pi}=d \sin \theta$
 +
 +giving
 +
 +$\delta=\frac{2\pi}{\lambda}d\sin\theta$
 +
 +and the intensity is
 +
 +$I_{\theta}=I_{0}\cos^{2}(\frac{\pi d \sin \theta}{\lambda})$
 +
 +{{figure_34_14.jpg?​800}}
  
 ===== Intensity pattern for a single slit ===== ===== Intensity pattern for a single slit =====
Line 105: Line 135:
  
 but that the larger the number of slits from which diffraction occurs the sharper the maxima will be. but that the larger the number of slits from which diffraction occurs the sharper the maxima will be.
 +
 +===== n slits =====
 +
 +{{nslitgrating.png}}
 +
 +n=2 $\frac{I_{\theta}}{I_{0}}=\frac{(2+2\cos\delta)}{4}$
 +
 +n=3 $\frac{I_{\theta}}{I_{0}}=\frac{3+4\cos\delta+2\cos 2\delta}{9}$
 +
 +n=4 $\frac{I_{\theta}}{I_{0}}=\frac{4+6\cos\delta+4\cos 2\delta+2\cos 3\delta}{16}$
 +
 +For any $n$
 +
 +$\frac{I_{\theta}}{I_{0}}=\frac{n+\sum\limits_{k=1}^{n-1}2(n-k)\cos(k\delta)}{n^{2}}$
  
 ===== Polarizers ===== ===== Polarizers =====
phy142kk/lectures/fr1.1431352999.txt ยท Last modified: 2015/05/11 10:03 by kkumar
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