Question

The two point sources in Fig 35-61 emit coherent waves. Show that all curves (such as the one shown), over which the phase difference for rays ${\mathit{r}}_{\mathbf{1}}$and ${\mathit{r}}_{\mathbf{2}}$in a constant, are hyperbolas. (Hint: A constant phase difference implies a constant difference in length between ${\mathit{r}}_1$and ${\mathit{r}}_{\mathbf{2}}$).

Short answer

All curves are hyperbola.

Step-by-step solution

The given data:

Two rays are $r_1$and $r_2$.

Formula of distance between two points:

The distance between two points is the length of a line that connects two points in a plane.

The formula for finding the distance between two points is usually given by,

$\mathbf{d}\mathbf{=}\sqrt{{\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}{\mathbf{y}}_{\mathbf{2}}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}}$

This formula is used to find the distance between any two points in the coordinate plane or x-y plane.

According to the question:

Let $S_1\left(a,0\right)$and $S_2\left(-a,0\right)$.

Let the point with constant path difference is $\left(x,y\right)\left(P\right)$.

localid="1663399680029" $$\begin{gathered} S_{2}P-S_{1}P=c \\ \sqrt{{\left(x+a\right)}^2+y^2}-\sqrt{{\left(x-a\right)}^2+y^2}=c \end{gathered}$$

All curves over the phase difference: 

Squaring both of the sides of equation (1) , and you have

$$\begin{gathered} 2{(x+a)}^2+2y^2-c^2=2\sqrt{{(x+a)}^2+y^2}\sqrt{{(x-a)}^2+y^2} \\ {\left({\left(x+a\right)}^2+y^2-\frac{{\displaystyle c^2}}{{\displaystyle 2}}\right)}^2=\left({\left(x+a\right)}^2+y^2\right)\left({\left(x-a\right)}^2+y^2\right) \\ {(x+a)}^4+y^4+\frac{c^4}{4}+2y^2{(x+a)}^2-y^2{c}^2-c^2{(x+a)}^2={(x+a)}^2{(x-a)}^2+{(x+a)}^2{y}^2+{(x-a)}^2{y}^2+y^4 \end{gathered}$$

$$\begin{gathered} y^4+2y^2{(x+a)}^2-y^2{c}^2={(x+a)}^2{(x-a)}^2+{(x+a)}^2{y}^2+{(x-a)}^2{y}^2+y^4+{(x+a)}^4-\frac{c^4}{4}+c^2{(x+a)}^2 \\ 2y^2{(x+a)}^2-y^2{c}^2-{(x+a)}^2{y}^2-{(x-a)}^2{y}^2={(x+a)}^2{(x-a)}^2+{(x+a)}^4-\frac{c^4}{4}+c^2{(x+a)}^2 \end{gathered}$$

After solving the above equation gives:

$$\begin{gathered} y^2({(x+a)}^2-c^2-{(x-a)}^2)={(x+a)}^2({(x-a)}^2+{(x+a)}^2+c^2)-\frac{c^4}{4} \\ {y}^2(4xa-c^2)={(x+a)}^2(2x^2+2a^2+c^2)-\frac{c^4}{4} \\ {(x+a)}^2(2x^2+2a^2+c^2)-y^2(4xa-c^2)=\frac{c^4}{4} \end{gathered}$$

$\begin{array}{rcl}{(x+a)}^2\times \frac{4(2x^2+2a^2+c^2)}{c^4}-y^2\times \frac{4(4xa-c^2)}{c^4}& =& 1\\ \frac{x^2}{a^2}-\frac{y^2}{{\beta }^2}& =& 1\end{array}$

Thus, the equation obtain is:

$\frac{x^2}{a^2}-\frac{y^2}{{\beta }^2}=1$

Here, $\beta$is a constant,

Above equation is an equation of hyperbola.

Hence, all curves are hyperbola.