Chapter 45: Problem 48

Question: A ship approaching harbour is transmitting at a wavelength $\lambda~=~3.43~{\rm m}$ from its antenna located $h~=~23~{\rm m}$ above sea-level. The receiving station antenna is located $H~=~160~{\rm m}$ above sea-level. What is the horizontal distance between the ship and the receiving tower when radio contact is momentarily lost for the first time? Assume that the calm ocean reflects radio waves perfectly according to the law of reflection. See Figure 27, Capter 45.

Solution: Radio contact will be lost when the receiving antenna moves through the first interference minimum from the forward direction, created by the signal from the ship and its reflection. The distance between the effective two sources is $d~=~2 h~=~46~{\rm m}$, and there is a $\pi$ phase change at the reflection from the water surface.

In the very forward direction $\theta~=~0$, there is a minimum due to the $\pi$ phase change, and then there is a maximum located at angle satisfying $d\sin\theta~=~{1\over 2}\lambda$. As the ship moves toward the receiver, the first minimum, corresponding to loss of radio contact occurs when $d\sin\theta~=~\lambda$. For small angles. we use $\sin\theta\sim\theta$, and hence $2 h \theta~=~\lambda$. The distance to the ship, the height of the receiving antenna and the angle of the signal are related by $H~=~D\theta$, in the limit of small angles. and so adio contact is lost when $2 h H~=~ D\lambda$, or when $D~=~2.15~{\rm km}$.