Provided everything remains constant, does the fringe separation, that is the distance between adjacent fringes become further apart for higher order maxima?

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Consider the above diagram. From the diagram it appears that the distance between adjacent maxima is getting larger with distance from the central maxima. My intuition agrees with this. However, I have heard that the distance between adjacent fringes is constant. Assume this is true. How so?

Double slit interference pattern: Fringes are equally spaced and of equal widths.

The above statement seems highly unintuitive. Shouldn't the spacing between adjacent fringes get larger, as the wave spreads out. How do I satisfy myself with the above statement? Proof?


1 Answer 1


The equation you need to consider is that which gives the angle $\theta_{\rm n}$ for the $n^{\rm th}$ bright fringe, $n\lambda = d \sin \theta_{\rm n}$, where $\lambda$ is the wavelength and $d$ the separation of the slits.

The angular separation of adjacent bright fringes is $\theta_{\rm n+1} -\theta_{\rm n}$.

If the angle $\theta_{\rm n}$ is small then an approximation can be made $\sin \theta_{\rm n}\approx \theta_{\rm n}$.

So the angular separation of the fringes is approximately $\frac{(n+1)\lambda}{d} - \frac{n\lambda}{d} = \frac \lambda d$ which is constant.
The separation of the fringes on a screen a distance $D$ from the double slit is $x_{\rm n+1}-x_{\rm n}$ where $x$ is the distance from the central maximum and $\theta \approx \frac x L$.

Thus $x_{\rm n+1} - x_n \approx \frac {\lambda L}{d}$ - constant fringe separation.

However the sine function is not linear with respect to $\theta$ and decreases less per increase in $\theta$ as $\theta$ increases.
This means that the distance between adjacent maxima will get bigger as $\theta$ increases and effect you may have well observed when looking at a spectrum using a diffraction grating which in really much more than a double slit but with many more slits.


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