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I was wondering what would happened if we titled the screen in Young's double slit experiment (YDSE) by a certain angle say $\theta$. What effect would it have on fringe patterns and their intensities on an axis parallel to the plane of the sheet?

I would speculate that there shouldn't be any signification change if $\theta$ is small as $D$(distance between slit and screen) $>>$ $d$(distance between the slots). So the path difference of light from the two slots stays the same as before, i.e. $∆x=d\sin(\alpha)$, where $\alpha$ is the angle subtended by any point on screen at the center of the two slits. So remaining quantities can be derived similarly.

I also know that if $\theta =90°$, there would be no interference pattern and if the screen is overhead the two slits we would get circular fringes.

What would happen in this case?

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  • $\begingroup$ Your intuition is right. However, the fringe pattern would change a bit away from its center. If you make a drawing and do the trig, you'll see this. $\endgroup$ – S. McGrew Oct 2 at 13:43
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On the assumption that the screen is far away from the two slits and the angles involved are small the fringe separation is $\Delta x = \frac {\lambda D}{d}$ where $\lambda$ is the wavelength, $d$is the slit separation and $D$ the two slit to screen distance.

So $\Delta x \propto D$.

A diagram with the angles exaggerated is shown below with the radial lines emanating from the twin slits being the positions of fringe maxima.

enter image description here

The fringes on the slanted screen 3 are not equally spaced.
The intensity of the slits above the central maximum fringe will be greater than the corresponding fringes below because the light is spread out more as the fringe separation is larger below.

Given that you are dealing with slits, ie have cylindrical symmetry, looking from above at a screen which is slightly inclined to the plane of my diagram you would observe fringes as shown in the diagram.

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