In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle’s “shadow region.” Previously, television signals had a wavelength of about $\mathbf{50}\mathbf{}\mathbf{cm}$, but digital television signals that are transmitted from towers have a wavelength of about $\mathbf{10}\mathbf{}\mathbf{mm}$. (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of $\mathbf{5}\mathbf{}\mathbf{m}$width between two adjacent buildings. What is the angular spread of the central diffraction maximum (out to the first minima) for wavelengths of (b)localid="1664270683913" $\mathbf{50}\mathbf{}\mathbf{cm}$and (c) localid="1664270678997" $\mathbf{10}\mathbf{}\mathbf{mm}$?

Previously signal wavelength is $50\mathrm{cm}$and for digital television, the signal wavelength is $10\text{mm}$.

The slit width,$\text{a}=5\text{m}$.

The expression for the minima in the diffraction pattern is given as follows.

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ $\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\mathbf{a}$is the slit width, $\mathit{\lambda }$is the wavelength,$\mathbf{m}$is the number of fringes in the envelope, andlocalid="1663139465827" $\mathit{\theta }$is the angle of diffraction

From equation (1), it can be concluded that the angle of the diffraction is directly proportional to the wavelength.

Since the wavelength decreases from $50\text{cm}$to $10\text{mm}$. Therefore, the angle of diffraction is also decreases. Thus, decreasing the diffraction of the signal into a shadow region is the obstacle.

Hence decreasing the diffraction of the signal into a shadow region is the obstacle.

Recall the equation (1),

$$\begin{gathered} a\sin \theta =m\lambda \\ \sin \theta =\frac{m\lambda }{a} \\ \theta ={\sin }^{-1}\left(\frac{m\lambda }{a}\right) \end{gathered}$$ (2)

The angular spread is double of the angle of diffraction.

Multiply the equation (2) by 2.

$2\theta ={\sin }^{-1}\left(\frac{m\lambda }{a}\right)$

Substitute $1$for $m$, $50\text{cm}$for $\lambda$, and $5\text{m}$for$a$into the above equation.

localid="1664271350398" $$\begin{gathered} 2\theta =2{\sin }^{-1}\left(\frac{1\times 50\times {10}^{-2}\text{m}}{5\text{m}}\right) \\ 2\theta =2{\sin }^{-1}\left(0.1\right) \\ 2\theta =2\times 5.73° \\ 2\theta =11.4° \end{gathered}$$

Hence the angular spread is$11.4°$.

Recall the equation (1),

$$\begin{gathered} a\sin \theta =m\lambda \\ \sin \theta =\frac{m\lambda }{a} \\ \theta ={\sin }^{-1}\left(\frac{m\lambda }{a}\right) \end{gathered}$$ (3)

The angular spread is double of the angle of diffraction.

Multiply the equation (3) by 2.

$2\theta ={\sin }^{-1}\left(\frac{m\lambda }{a}\right)$

Substitute $1$for $m$, $10\mathrm{mm}$for $\lambda$, and $5\mathrm{m}$for $a$into the above equation.

localid="1664271363740" $$\begin{gathered} 2\theta =2{\sin }^{-1}\left(\frac{1\times 10\times {10}^{-3}m}{5m}\right) \\ 2\theta =2{\sin }^{-1}\left(2\times {10}^{-3}\right) \\ 2\theta =2\times 0.114° \\ 2\theta =0.23° \end{gathered}$$

Hence the angular spread is $0.23°$.