This shows you the differences between two versions of the page.

Both sides previous revision Previous revision Next revision | Previous revision | ||

phy142kk:lectures:fr1 [2015/05/11 10:08] kkumar [Anti reflective coating] |
phy142kk:lectures:fr1 [2015/05/11 10:14] (current) kkumar [Diffraction grating] |
||
---|---|---|---|

Line 73: | Line 73: | ||

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 $\frac{\lambda}{4n_{2}}$. | ||

+ | |||

+ | ===== 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 ===== | ||

Line 111: | 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 ===== |