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

Question

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?

Answer 1

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.

Answer 2

I don't quite understand the diagram, because it shows L1 and L2 as parallel, even though they are supposed to meet at the same point.

the interferences are located infinitely far from the slits, you need a lens so that the two rays can interfere