91Ó°ÊÓ

Constructive interference is at$\mathit{n}\mathit{λ}$

and destructive interference is at$\frac{\left(2n+1\right)\mathbf{λ}}{\mathbf{2}}$.

Where n is an integer.

We know that the interference at point P is a constructive interference.

This means that the path difference between the two waves of the two sources must be an integer number of X. We need to find the wavelength , which is the wavelength of the two sources. Noting that we are given three values of x in which source B where located to make the two waves interfere constructively at P. The difference between ant two locations of them is about 6.0 m.

${\mathrm{x}}_2-{\mathrm{x}}_1=216\mathrm{m}-210\mathrm{m}=6\mathrm{m}$

And

${\mathrm{x}}_3-{\mathrm{x}}_2=222\mathrm{m}-216\mathrm{m}=6\mathrm{m}$

Hence, the wavelength is given by

$\mathrm{λ}=6.0m$

Now we find the frequency using$v=f\mathrm{λ}$

Since the source is light we get speed of light and then

$f=\frac{c}{\mathrm{λ}}$

$$\begin{gathered} f=\frac{3.0×{10}^8}{6.0} \\ f=50MHz \end{gathered}$$

We know that the path difference should be

$∆x=(m+\frac{1}{2})\mathrm{λ}$

In this case we are looking for

$240-x=(m+\frac{1}{2})\mathrm{λ}$

Solve for x

$\mathrm{x}=240-(\mathrm{m}+\frac{1}{2})\mathrm{λ}$

From this equation above, the smallest vale of m will give the largest value of x

Hence we put m=0;

role="math" localid="1664094326632" $$\begin{gathered} \mathrm{x}=240-(0+\frac{1}{2})\mathrm{λ} \\ \mathrm{x}=240-\frac{1}{2}\mathrm{λ} \\ \mathrm{x}=240-\frac{1}{2}6.0 \\ \mathrm{x}=237\mathrm{m} \end{gathered}$$

Hence, the maximum distance at which interference is destructive is 237m