# The difference in path lengths for waves in the double slit experiment

Fig.1

I don't quite understand the diagram, because it shows $$L_1$$ and $$L_2$$ as parallel, even though they are supposed to meet at the same point. I believe the idea is that $$\Delta L$$ approaches $$d \sin \theta$$ as the distance to the screen increases. However, that is not exactly the case, and here is my proof:

Fig.2

I want to get a formula for $$AF - BF$$ in terms of $$AF$$, $$\theta_1$$ and $$\theta_2$$, where these angles measure the angle between the horizontal gridlines and $$AF$$ and $$BF$$, respectively.

As we can see, we have dotted line that is parallel to $$AF$$, and a dotted line between them that is perpendicular to both of them. We have an even more finely dotted line $$CD$$ that is perpendicular to $$BF$$.

Now, first off, we can establish the following:

\begin{align} AF - BF &= AF - BC - CF \\ &= AF - BC - \sqrt{AF^2 + AC^2} \\ &= AF - BC - \sqrt{AF^2 + \cos(\theta_1)^2d^2} \end{align}

So, now we must get another expression for $$BC$$. As we know from fig. 1 (using some trigonometry), $$BE = d\sin(\theta_1)$$. In fig. 2, we see that $$BC = BE - DE$$.

The angle between $$BF$$ and the dotted line is $$\theta_2 - \theta_1$$. Therefore, we have the following two identities:

$$CE= \sin(\theta_2 - \theta_1)\cdot BE = \sin(\theta_2 - \theta_1) d\sin(\theta_1)$$

$$CD = \sin(\theta_2 - \theta_1)\cdot BC$$

The line $$DE$$ can be given a new expression by the Pythagorean theorem:

$$DE = \sqrt{CE^2 -DC^2} = \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2 }$$

As already established, $$BC = BE - DC$$. So, we have the following expression for $$BC$$:

$$BC = d\sin(\theta_1) - \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2}$$

Thus, we have our final expression for the difference:

$$AF - BF = AF - d\sin(\theta_1) + \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2} - \sqrt{AF^2 + \cos(\theta_1)^2d^2}$$

Now, as the distance to the screen, denoted as $$d_s$$, increases relative to the distance between the slits, denoted as $$d$$, the difference in $$\theta_1$$ and $$\theta_2$$ decreases, which is fairly intuitive and simple to prove. Thusly, asking what $$AF - BF$$ approaches as $$d/d_s$$ approaches zero is the sake as asking what $$AF - BF$$ approaches as $$\theta_2 - \theta_1$$ approaches zero. Thusly:

\begin{align} \lim_{\theta_2 - \theta_1 = 0} \Delta L &= \lim_{\theta_2 - \theta_1 = 0}AF - BF \\[2ex] &= \lim_{\theta_2 - \theta_1 = 0}AF - d\sin(\theta_1) + \sqrt{(\sin(\theta_2 - \theta_1) d\sin(\theta_1))^2 - (\sin(\theta_2 - \theta_1)\cdot BC)^2} - \sqrt{AF^2 + \cos(\theta_1)^2d^2} \\[2ex] &= AF - d\sin(\theta_1) - \sqrt{AF^2 + \cos(\theta_1)^2d^2} \end{align}

So, what gives? Is it just my trig that is off, or have I misunderstood what the diagram is trying to say to begin with?

• For $d$ of 100 microns and $L$ of 5 meters, what is the difference between the two? Sep 29 at 18:07
• Yeah, maybe like complete the square under the root, then take the square root, and try to expand in some small parameter....the approximation makes sense in the large $L/d$ limit, and I don't immediately see something wrong in your approach, so I think it's just massaging the last line. Sep 29 at 18:30
• Apply a binomial formula, binomial expansion, and you get $$AF-\sqrt{AF^2-\cos^2(\theta_1)d^2}=\frac{\cos^2(\theta_1)d^2}{AF+\sqrt{AF^2-\cos^2(\theta_1)d^2}},$$ so the whole limit expression is dominated by $-d\sin\theta_1$. Sep 30 at 5:46

Let $$R_b=\bar {BF}$$ and $$R_a=\bar {AF}$$, and also the angle $$\alpha = \measuredangle [ \bar {AB}, \bar {BF}]$$. Then using the Taylor exapansion of $$\sqrt{1+x}=1+\frac{1}{2}x+\mathcal O(x^2)$$ we get $$R_a = \sqrt{R_b^2+d^2-2dR_b \cos(\alpha)} = R_b\sqrt{1+\frac{d^2}{R_b^2}-2\frac{d}{R_b} \cos(\alpha)}\\ =R_b\left(1-\frac{d}{R_b} \cos(\alpha) + \mathcal O \left(\frac{d^2}{R_b^2}\right)\right)$$ in other words, asymptotically the path difference $$\delta L$$ is $$\delta L = R_b - R_a = d \cos(\alpha) + \mathcal O \left(\frac{d^2}{R_b^2}\right)$$ showing, to a 1st order error, that it is independent of the distance between the point of interference and the slit.