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In the following setup, we have two point sources of light producing monochromatic, spherical light waves in-phase of wavelength $\lambda$, and a screen positioned in a plane prependicular to the line going through the sources. We assume a Fraunhofer situation where the screen is very far away from the sources compared to the distance between the sources $d$ and the wavelength $\lambda$, and so the distance from the sources to the screen is simply called $R$.

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Concentric interference circles will be seen on the screen, and we wish to find their radii. It is easily seen through the usual Fraunhofer-diffraction arguments that the path difference between the two sources is $d\cos \theta$, and so for maxima we have $d\cos \theta=n\lambda$.

Given that the angle $\theta$ is small, we can claim as usual that $\cos \theta \approx 1-\frac12 \theta ^2$ and also $R\theta \approx r$ where $r$ is the radius of the circle, what we're looking for. Substituting and isolating, we get $$r \approx \sqrt2R \sqrt{1-\frac{\lambda n}{d}}$$

However, the same problem can be solved without the small-angle approximation, and with only the Fraunhofer approximation (yes, I know the Fraunhofer approximation sort of has implicit in it that the angle is small, but nevertheless I show that this can be solved without approximating the trigonometric functions, bear with me). Backtracking a bit to where we only had $d\cos \theta=n\lambda$, we now note that $\cos \theta = \frac{R}{\sqrt{r^2+R^2}}$. Substituting and isolating, we now get

$$r =R \sqrt{\frac{d^2}{\lambda^2 n^2}-1}$$

However, I don't see how these two solutions agree. They seem completely different, not only in the values they predict - they're not even defined on the same region (of $n,d,\lambda$-space)! Nor can I find any series that reduces one of them to the other. What is going on here? Why don't the two solutions agree?

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Your first derivation looks fine but I am not so sure about the second one particularly as the distance $r$ is related to the "inverse" of the wavelength.

Let me start by looking at what would happen on a circular screen of radius $R$ as shown in the diagram below.

enter image description here

The path difference to position $P$ is $\Delta x = AP - BP$.

Using the cosine rule for triangle $APC$ and $CPB$ gives $AP^2-BP^2 = 2dR \cos \theta$.

If $\theta$ is small then $AP+BP \approx 2R$ and so the path difference is $\Delta x = AP - BP = d \cos \theta$ which in agreement with what you stated.

Note that my circular screen is very close to your flat screen in this region and your expression for $r \,(\approx \sqrt2R \sqrt{1-\frac{\lambda n}{d}})$ is valid.

Now what happens if the angles are not small?

Well if the angles are close to $\frac \pi 2$ then you have the familiar 2 slit interference arrangement with an angle $\phi = \frac \pi 2 - \theta$ generally used and the path difference at a screen a distance $R$ away approximately equal to $d \sin \phi$.

However your screen is at right angles to my screen at $D$ and moves away from my screen as the angle $\theta$ increases.

Consider that at position $P$ on my screen there will be a maximum so that $AP-BP$ must be an integer number of wavelengths.
In that direction $\theta$ at position $Q$, $AQ - BQ$ will not be an integer number of wavelengths.

Also looking at your formula for $\cos \theta$ it is no longer equal to $\frac{R}{\sqrt{r^2+R^2}}$ rather it is $\dfrac {DQ(=r)}{CQ}$ and $CQ$ is not equal to $R$

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