Question

In Fig 35-53, a microwave transmitter at height a above the water level of a wide lake transmits microwaves of wavelength $\mathit{\lambda }$toward a receiver on the opposite shore, a distance x above the water level. The microwaves reflecting from the water interfere with the microwaves arriving directly from the transmitter. Assuming that the lake width D is much greater than a and x, and that $\mathit{\lambda }\mathbf{\ge }\mathit{a}$, find an expression that gives the values of x for which the signal at the receiver is maximum. (Hint: Does the reflection cause a phase change?).

Short answer

Thus, the value of x is $x=\left(m+\frac{1}{2}\right)\left(\frac{\lambda D}{2a}\right)$.

Step-by-step solution

The difference of the distances.

Assume that the reflected wave travels a distance $L_2$and the wave that gets directly to the receiver travels a distance $L_1$. Since water has a larger refraction index than air, the final wave experiences a half-wavelength phase change upon reflection. The difference needs to be an odd multiple of a half wavelength in order to produce constructive interference at the receiver.

Consider the diagram below:

The ray incident on the water, the water line, and the right triangle on the left that is created by the vertical line from the water to the transmitter T, the water line, and the water, yield $D_a=\frac{a}{\tan \theta }$. The vertical line from the water to the receiver, R, the reflected ray, and the water line form the right triangle, which is on the right, and it leads to $D_b=\frac{x}{\tan \theta }$. Due to $D_a+D_b=D$,

$\tan \theta =\frac{a+x}{D}$

Use the identity ${\sin }^2\theta =\frac{{\tan }^2\theta }{\left(1+{\tan }^2\theta \right)}$to show that,

$\sin \theta =\frac{\left(a+x\right)}{\sqrt{D^2+{\left(a+x\right)}^2}}$

This means,

$L_{2a}=\frac{a}{\sin \theta }=\frac{a\sqrt{D^2+{\left(a+x\right)}^2}}{\left(a+x\right)}$

And

$L_{2b}=\frac{a}{\sin \theta }=\frac{x\sqrt{D^2+{\left(a+x\right)}^2}}{\left(a+x\right)}$

Therefore, write as follows:

$\begin{array}{rcl}{L}_2& =& L_{2a}+L_{2b}\\ & =& \frac{\left(a+x\right)\sqrt{D^2+{\left(a+x\right)}^2}}{\left(a+x\right)}\\ & =& \sqrt{D^2+{\left(a+x\right)}^2}\end{array}$

Evaluate the value of x.

Using the binomial theorem, with $D^2$ large and$a^2+x^2$ small, we approximate this expression:

$L_2\approx D+\frac{{\left(a+x\right)}^2}{2D}$

The distance travelled by the direct wave is $L_1=\sqrt{D^2+{\left(a-x\right)}^2}$. Using the binomial theorem, the expression is written as:

$L_1\approx D+\frac{{\left(a-x\right)}^2}{2D}$

Thus, solve the expression as follows:

$\begin{array}{rcl}{L}_2-L_1& \approx & D+\frac{a^2+2ax+x^2}{2D}-D-\frac{a^2-2ax+x^2}{2D}\\ & =& \frac{2ax}{D}\end{array}$

Setting this equal to $\left(m+\frac{1}{2}\right)\lambda$, where m is a zero or a positive integer, we get

$x=\left(m+\frac{1}{2}\right)\left(\frac{\lambda D}{2a}\right)$

Similarly, the condition for destructive interference is:

$\begin{array}{rcl}{L}_2-L_1& \approx & \frac{2ax}{D}\\ & =& m\lambda \end{array}$

Hence, the value x of is $x=\left(m+\frac{1}{2}\right)\left(\frac{\lambda D}{2a}\right)$.