91Ó°ÊÓ

$f=88.9MHz=88.9×{10}^{6}Hz,a=15.0m,c=3.0×{10}^{8}m/s,I_0=3.50W/m^2,\mathrm{θ}=Â\pm 5.{00}^{∘}$

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

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

$v=\mathrm{λ}f$

In this case, v = c so

$c=\mathrm{λ}f$

Hence,

$\mathrm{λ}=\frac{c}{f}$ (1)

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

$sin{\mathrm{θ}}_m=\frac{m\mathrm{λ}}{\mathrm{Î\pm }}$

Whereas$m=Â\pm 1,Â\pm 2,Â\pm 3,...$

Substitute from (1)

$sin{\mathrm{θ}}_m=\frac{m\mathrm{λ}}{af}$

Solving for${\mathrm{θ}}_m$

${\mathrm{θ}}_m=si{n}^{-1}\frac{m\mathrm{λ}}{af}$

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.

When m=1,

When m=2,

$$\begin{gathered} {\mathrm{θ}}_2=si{n}^{-1}\left(\frac{2×3.0×{10}^8}{15.0×88.9×{10}^6}\right) \\ {\mathrm{θ}}_2=Â\pm 27.7^{∘} \end{gathered}$$

When m=3,

$$\begin{gathered} {\mathrm{θ}}_3=si{n}^{-1}\left(\frac{3×3.0×{10}^8}{15.0×88.9×{10}^6}\right) \\ {\mathrm{θ}}_3=Â\pm 42.4^{∘} \end{gathered}$$

When m=4,

$$\begin{gathered} {\mathrm{θ}}_4=si{n}^{-1}\left(\frac{4×3.0×{10}^8}{15.0×88.9×{10}^6}\right) \\ {\mathrm{θ}}_4=Â\pm 64.1^{∘} \end{gathered}$$

When m=5,

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

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

We know that the intensity is given by

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

We also know that

$\mathrm{β}=\frac{2\mathrm{π}a\sin \mathrm{θ}}{\mathrm{λ}}$

Substitute from (1)

$\mathrm{β}=\frac{2\mathrm{π}af\sin \mathrm{θ}}{c}$

Substitute into the intensity formula above, noting that 2 cancels

$I=I_0{\left(\frac{\sin \left({\displaystyle \frac{\mathrm{Ï€²¹´Ú}\sin \mathrm{θ}}{c}}\right)}{\frac{2\mathrm{Ï€²¹´Ú}\sin \mathrm{θ}}{c}}\right)}^2$

Substitute the given

$I=3.50×{\left(\frac{\sin {\displaystyle }\left(\frac{\mathrm{π}×15.0×88.9×{10}^6×\sin {5.00}^{∘}}{3.0×{10}^8}\right)}{\frac{\mathrm{π}×15.0×88.9×{10}^6×\sin {5.00}^{∘}}{3.0×{10}^8}}\right)}^2$