Two-Ray Ground-Reflection Model: Understanding the Difference in Phase Offsets at the Receiver

Currently, I'm trying to understand the Two-Ray Ground-Reflection Model. During my research, I came across the following simplification expressing the received power as follows (taken from Wikipedia):

The part that is not clear to me is $$e^{-j\Delta\phi}$$: How is that derived from $$e^{-j2\pi l/\lambda}$$ and $$e^{-j2\pi (x + x\prime)/\lambda}$$?

I know that $$\phi$$ can be calculated with $$\phi = 2\pi d/\lambda$$, where $$d$$ is the distance between the transmitter and the receiver.

In the above equation, the ground-reflected path is expressed as $$(x + x\prime)$$ and $$l$$ is the direct line-of-sight path.

Moreover, the approximation only holds if the signal is "narrow band relative to the inverse delay spread $$1/\tau$$, so that $$s ( t ) \approx s ( t − τ )$$".

Why is that important?

Attempting to take $$e^{-j2\pi l/\lambda}$$ as a common factor we get $$\left|e^{-j2\pi l/\lambda} \right|^2 \left| \frac{\sqrt{G}}{l} +\Gamma(\theta)\frac{e^{-j2\pi (x+x\prime)/\lambda}}{(x+x\prime) (e^{-j2\pi l/\lambda})} \right|^2$$
On the above, you might have to use the simplifications $$\frac{e^a}{e^b} = e^{a-b}$$ and $$|e^{j\ c}|=1$$.
They might be interested in finding out the locations or angles for which the signal and its reflected version destructively interfere and result in low signal strength. If a signal $$s(t)$$ and its time delayed version $$s(t-\xi)$$ needs to cancel out, for all values of $$t$$, then the assumption above ($$s(t)\approx \color{red}{-}^{\dagger}s(t-\xi)$$) is required. The word narrow band is simply used to indicate that the signal needs to looks like a sinusoidal wave (which has the required property of periodicity mentioned above).
$$\dagger$$ The minus sign may be provided by the act of reflection. I am not sure.