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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 ===== | ||

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- | For most lenses we want as much of the incident light to be transmitted as possible. | ||

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- | Suppose we take a glass lens with refractive index $n=1.52$. We can see from the reflectance equation | ||

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- | |||

- | $R=(\frac{n_{0}-n_{1}}{n_{0}+n_{1}})^2=(\frac{1-1.52}{1+1.52})^2=0.043$ | ||

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- | that about 4% of the incident light is reflected. | ||

- | |||

- | 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 ===== |