# Double Slits

1. Apr 20, 2010

### 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])

2. Apr 20, 2010

### 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.

3. Apr 21, 2010

### 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?

4. Apr 21, 2010

### 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.

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

5. Apr 21, 2010

### Oerg

but isnt 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 cant possibly move more than a fringe. Is there anything wrong with my reasoning?

6. Apr 21, 2010

### kuruman

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.