Understanding Young's Double Slit Experiment: Why 4.5 Fringes Shift?

[Oerg]

I have a problem on Young's double slits which says that the central bright fringe shifts 4.5? fringes after a piece of plastic is placed in front one of the slits.

My question is: Why is the central bright fringe shifted for 4.5 fringes? Shouldn't it be 0.5 fringe instead since maximum interference is already achieved then? (Because 4 wavelengths of optical path difference is the same as none at all [no considering the coherence length])

 

[kuruman]

You are quite right is saying that 4 wavelengths are the same as none at all, but we are not talking about the phase change here. A thicker piece of plastic will shift the central maximum more than a thinner piece, i.e. the shift depends on how many wavelengths you can fit within a particular thickness of plastic.

 

[Oerg]

well, since there is an optical path difference then there is a phase change and the condition becomes a\sin{\theta}=OPD-\frac{OPD}{\lambda}+m\lambda. Shouldn't this be the case?

 

[kuruman]

Consider what happens to the point where the central maximum was initially. With insertion of the plastic, this point is occupied by what used to be the minimum between fringes 4 and 5. Therefore, displacing air of thickness d with plastic of thickness d and index of refraction n adds 4.5 wavelengths to the central path, i.e.

\frac{d}{\lambda/n}=\frac{d}{\lambda}+4.5

 

[Oerg]

but isn't the central fringe supposed to be the fringe where \theta gives a maximum interference for the lowest \theta? Because the sin function has a decreasing gradient, the size of the the central fringe is the largest so the central fringe can't possibly move more than a fringe. Is there anything wrong with my reasoning?

 

[kuruman]

but isn't the central fringe supposed to be the fringe where \theta gives a maximum interference for the lowest \theta? Because the sin function has a decreasing gradient, the size of the the central fringe is the largest so the central fringe can't possibly move more than a fringe. Is there anything wrong with my reasoning?

The central fringe appears where the phase difference is zero (or path length difference is zero wavelengths). Fringe 4 appears where the phase difference is 4*2π (or path length difference is 4 wavelengths).

When you add the plastic to one slit, you are essentially adding 4.5λ to any path from that slit to the screen. The central maximum correspondingly moves to a point on the screen that is 4.5λ farther away from the other slit in order to preserve the criterion that the difference between the two paths is zero wavelengths.