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phy142kk:lectures:fr1 [2015/05/11 10:07]
kkumar [Air wedge and Other Effects]
phy142kk:lectures:fr1 [2015/05/11 10:14] (current)
kkumar [Diffraction grating]
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 ===== Anti reflective coating ===== ===== Anti reflective coating =====
- 
-For most lenses we want as much of the incident light to be transmitted as possible. ​ 
- 
-Suppose we take a glass lens with refractive index $n=1.52$. We can see from the reflectance equation ​ 
- 
- 
-$R=(\frac{n_{0}-n_{1}}{n_{0}+n_{1}})^2=(\frac{1-1.52}{1+1.52})^2=0.043$ 
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-that about 4% of the incident light is reflected. ​ 
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-This percentage can be reduced by the use of an [[wp>​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>​. 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>​.
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 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 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 $\frac{\lambda}{4n_{2}}$.
  
-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).+ 
 +===== Intensity for double slit interference ===== 
 + 
 +As we now have the intensity in terms of the phase difference $\delta$ 
 + 
 +$I_{\theta}=I_{0}\cos^{2}\frac{\delta}{2}$ 
 + 
 +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 =====
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 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/fr1.1431353249.txt · Last modified: 2015/05/11 10:07 by kkumar
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