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 Author Topic: Pattern for Newton's Ring  (Read 16660 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: February 14, 2010, 04:27:56 pm » posted from:Taipei,T\'ai-pei,Taiwan

This simulation is created due to the following request from Re: Effect of a Medium on a Double Slit Interference Pattern
Quote
dear sir
the interference simulations on introdusing a medium inthe path really superb
why dont you try measurement in newtons rings?
best wishes
raman

The following message is from http://en.wikipedia.org/w/index.php?title=Newton's_rings

The phenomenon of Newton's rings, named after Isaac Newton, is an interference pattern caused by the reflection of light between two surfaces - a spherical surface and an adjacent flat surface. When viewed with monochromatic light it appears as a series of concentric, alternating light and dark rings centered at the point of contact between the two surfaces. When viewed with white light, it forms a concentric ring pattern of rainbow colors because the different wavelengths of light interfere at different thicknesses of the air layer between the surfaces. The light rings are caused by constructive interference between the light rays reflected from both surfaces, while the dark rings are caused by destructive interference. Also, the outer rings are spaced more closely than the inner ones. Moving outwards from one dark ring to the next, for example, increases the path difference by the same amount λ, corresponding to the same increase of thickness of the air layer λ/2. Since the slope of the lens surface increases outwards, separation of the rings gets smaller for the outer rings.

The radius of the Nth Newton's bright ring is given by $r_N= \left[\left(N - {1 \over 2}\right)\lambda R\right]^{1/2}$

(image from http://scienceworld.wolfram.com/physics/NewtonsRings.html)
$r^2+(d-R)^2=R^2$, so $d=\sqrt{R^2-r^2}$
Reflection from hemisphere undergoes a phase change of $\Delta \phi=0$. Reflection from the plane gives a phase change of $\Delta \phi=\pi$.
The phase difference between rays reflected off the plane and hemisphere is therefore
$\delta=k(2d)-\pi=\frac{4\pi n d}{\lambda}-\pi$for $d<, and the interference pattern goes as  $\cos^2\delta$
where $n$ is the index of refraction.

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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: February 14, 2010, 04:32:46 pm » posted from:Taipei,T\'ai-pei,Taiwan

Here is another one:

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