Question

Public Radio station KXPR-FM in Sacramento broadcasts at 88.9 MHz. The radio waves pass between two tall skyscrapers that are 15.0 m apart along their closest walls.

(a) At what horizontal angles, relative to the original direction of the waves, will a distant antenna not receive any signal from this station?

(b) If the maximum intensity is 3.50 W>m2 at the antenna, what is the intensity at _5.00_ from the center of the central maximum at the distant antenna?

Short answer

a) $\pm 13.0^{\circ },\pm 27.7^{\circ },\pm 42.4^{\circ },\pm 64.1^{\circ }$

b)$2.08W/m^2$$2.08W/m^2$

Step-by-step solution

 Given.

$f=88.9MHz=88.9\times {10}^{6}Hz,a=15.0m,c=3.0\times {10}^{8}m/s,I_0=3.50W/m^2,\theta =\pm 5.{00}^{\circ }$

Ray diagram.

This is a view from above which means that the wave lines are horizontal, as the author mentioned.

Concept.

This problem is similar to the one-slit kind of problem whereas the distance between the two skyscrapers is the width of the whole a.

First of all, we need to find the wavelength which is given by

$v=\lambda f$

In this case, v = c so

$c=\lambda f$

Hence,

$\lambda =\frac{c}{f}$ (1)

We know that the angle of the minimum fringe in the one-slit experiments is given by

$sin{\theta }_m=\frac{m\lambda }{\alpha }$

Whereas$m=\pm 1,\pm 2,\pm 3,...$

Substitute from (1)

$sin{\theta }_m=\frac{m\lambda }{af}$

Solving for${\theta }_m$

${\theta }_m=si{n}^{-1}\frac{m\lambda }{af}$

Now we can easily find the horizontal angles in which horizontal angles will a distant antenna will receive no signal from this station by substituting the values of m into the last equation. Thus, we will substitute the given and m's values.

Calculation.

When m=1,

$$\begin{gathered} {\theta }_1=si{n}^{-1}\left(\frac{1\times 3.0\times {10}^8}{15.0\times 88.9\times {10}^6}\right) \\ {\theta }_1=\pm 13.0^{\circ } \end{gathered}$$

When m=2,

$$\begin{gathered} {\theta }_2=si{n}^{-1}\left(\frac{2\times 3.0\times {10}^8}{15.0\times 88.9\times {10}^6}\right) \\ {\theta }_2=\pm 27.7^{\circ } \end{gathered}$$

When m=3,

$$\begin{gathered} {\theta }_3=si{n}^{-1}\left(\frac{3\times 3.0\times {10}^8}{15.0\times 88.9\times {10}^6}\right) \\ {\theta }_3=\pm 42.4^{\circ } \end{gathered}$$

When m=4,

$$\begin{gathered} {\theta }_4=si{n}^{-1}\left(\frac{4\times 3.0\times {10}^8}{15.0\times 88.9\times {10}^6}\right) \\ {\theta }_4=\pm 64.1^{\circ } \end{gathered}$$

When m=5,

${\theta }_5=si{n}^{-1}\left(\frac{5\times 3.0\times {10}^8}{15.0\times 88.9\times {10}^6}\right)$

Since the term of $\left(\frac{2\times 3.0\times {10}^8}{15.0\times 88.9\times {10}^6}\right)$is greater than one, so no angle available 2x3.0 x 105 here and this means that the last angle is for m=4.

 Step 5: Calculating intensity.

We know that the intensity is given by

$I=I_0{\left(\frac{\sin {\displaystyle \frac{\beta }{2}}}{{\displaystyle \frac{\beta }{2}}}\right)}^2$

We also know that

$\beta =\frac{2\pi a\sin \theta }{\lambda }$

Substitute from (1)

$\beta =\frac{2\pi af\sin \theta }{c}$

Substitute into the intensity formula above, noting that 2 cancels

$I=I_0{\left(\frac{\sin \left({\displaystyle \frac{\mathrm{\pi af}\sin \mathrm{\theta }}{c}}\right)}{\frac{2\mathrm{\pi af}\sin \mathrm{\theta }}{c}}\right)}^2$

Substitute the given

$I=3.50\times {\left(\frac{\sin {\displaystyle }\left(\frac{\pi \times 15.0\times 88.9\times {10}^6\times \sin {5.00}^{\circ }}{3.0\times {10}^8}\right)}{\frac{\pi \times 15.0\times 88.9\times {10}^6\times \sin {5.00}^{\circ }}{3.0\times {10}^8}}\right)}^2$