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Amplitude modification in double slit interference

Physics Asked by Tony Marshle on January 17, 2021

Over the surface of interference, illumination varies. We have calculated so perfectly about it so that we know exactly how the variation chart looks like. Even our vision renders it as in that chart.

But should it render so nicely? In the laboratory, we are free to alter our position and angle of vision; then what about the change of phase in the ‘reflected back to eye’ waves? I do understand how the ‘on the surface’ interfered waves, though got separated after reflection from the surface, are, on their way, by refraction on eye lens, met again on the eye nerve; so that our ‘interference combination’ stays an ‘interference combination’. But the wave phase change may be brutal, in that, the interference pattern might be distorted on the eye, and ‘really bright point’ may appear horribly dark, and conversely.

What am I missing?

One Answer

To simplify the explanation I will assume that the eye is just a glass convex lens with air either side and the image is formed on the retina.
So it is like a camera.

First of all remember what a lens does to produce a sharp image of an object.
Each point on the object has waves emanating from it in phase and travelling in a variety of directions.
Those waves travel through the air and some of them go through the glass of the lens and then travel through the air converging at a point still in phase.
Where those waves converge is a point on the image which exactly corresponds to a point on the object.
Some of the waves have travelled faster through more air but then travel slower through less glass near the edge of the lens but taken exactly the same time as those waves which have travelled faster through less air but then travel slower through more glass near the centre of the lens.
The key thing is that the waves start off in phase from the object and finish up in phase at the corresponding point on the image because it takes all the waves exactly the same time to travel from a point on the object to a corresponding point on the image irrespective of the path that they have taken. So for every point on the object there is a a corresponding point on the image.

Now consider a situation where you are looking at the region which is remote from the source of light where the interference actually occurs.
You do not need a screen to view the interference fringes, all you need to to is to focus the eye onto an area in the region where the interference pattern occurs.
In practice the focussing on "mid-air" can be helped by holding a translucent screen in that region so that the interference pattern can be seen on the screen.
Then by slowly removing the translucent screen the eye can be made to focus on the interference fringes.
You see the interference fringes because an image of the interference (the object) pattern is formed on the retina.
Note that all waves from the two slits which then overlap in the region between the eye and the two slits and produce a maximum and so are in phase will continue on and those which go through the lens (eye) will arrive at the retina still in phase and so produce a maximum.

Now have the screen present and look at the screen from the side closest to the two slits.
At a maximum of intensity the waves from the two slits arrive in phase at a point on the screen and are scattered from that point in phase.
Some of those scattered waves then travel through the lens and arrive in phase at the retina so you get a maximum.
And the same thing happens for each point on the screen producing an image of the point on the refine. You eye is focussing on the screen on which the interference pattern can be seen and your eye collects the light which is coming off the screen and focusses it on the retina.

If the lens has defects then the one-to-one correspondence between is not perfect as the waves tale different times to travel between a point on the object to a point on the image and you get a distorted image.

Answered by Farcher on January 17, 2021

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