91Ó°ÊÓ

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 $D_a=\frac{a}{\tan \mathrm{θ}}$. 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 $D_b=\frac{x}{\tan \mathrm{θ}}$. Due to $D_a+D_b=D$,

$\tan \mathrm{θ}=\frac{a+x}{D}$

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

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

This means,

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

And

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

Therefore, write as follows:

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

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

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

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:

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

Thus, solve the expression as follows:

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

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

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

Similarly, the condition for destructive interference is:

$\begin{array}{rcl}{L}_2-L_1& ≈& \frac{2ax}{D}\\ & =& m\mathrm{λ}\end{array}$

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