Determining Photon Paths in a Double Slit System

[kof9595995]

Homework Statement

Show that in order to be able to determine through which slit a double slit system each photon passes without destroying the double silt diffraction pattern, the condition
$$\delta y\delta {p_y} \ll \frac{h}{{4\pi }}$$
must be satisfied. Since this condition violated the uncertainty principle, it can not be met

Homework Equations

$$d\sin \theta = \lambda$$
$$\sin \theta = \frac{{\delta {p_y}}}{p}$$
$$\lambda = \frac{h}{p}$$

The Attempt at a Solution

In order not to destroy the pattern, the angle should not be large enough to shift one maximum to its adjacent maximum. So $$\sin \theta \ll \frac{\lambda }{d}$$, and then we have
$$\frac{{\delta {p_y}}}{p} \ll \frac{\lambda }{d}$$
To figure out which slit the photon passes through, we must have
$$\delta y \ll \frac{d}{2}$$
Combine these two and use de broglie's relation $$\lambda = \frac{h}{p}$$
We can get
$$\delta y\delta {p_y} \ll \frac{h}{2}$$

But it seems to me the extra $$2\pi$$ just comes out from nowhere. I'm really pulling my hair off on this quetsion

Homework Statement

Homework Equations

The Attempt at a Solution

 

[V Sri]

I was stuck with this problem as well, and your approach provided the spark. Thanks so much .
However following up on your argument, let's look at it this way:
"In order not to destroy the pattern, the angle should not be large enough to shift one maximum to its adjacent MINIMUM". My point here is that if the bright band/s merge with the dark band/s, THAT would destroy the interference pattern . So Sin Θ < λ/2d (would this then be right, btw ?). Hence ΔPy/P < λ/2d.
Next, to figure out which slit the photon passes through, and quite arbitrarily frankly, I assume Δy < d/8 (as opposed to d/2)
Pulling all this together we then get: ΔPy/Δy < h/16, and since h/16 < h/4π.
Hence ΔPyΔy < h/4π !
Can anybody do better ?!