## Section50.7Interference of Light Bootcamp

### Subsection50.7.4Miscellaneous

Two radio antennas are $0.5\text{ km}$ apart. They broadcast $4\text{ MHz}$ signals in phase. A car travels on a road parallel to the line joining the two antennas. The road is $20\text{ km}$ away from the antennas, i.e. the length of a line perpendicular to both the line joining the antennas and the road is $12\text{ km}\text{.}$ When the car moves on the road at a constant speed the net signal increases and decreases periodically. State why one would observe such variation in the intensity and determine five successive places on the road the strongest signals will be found and four successive places where the weakest signals will be found.

Hint

The car samples points on the “screen”.

$-3.6\text{ km}\text{,}$ $-1.8\text{ km}\text{,}$ $0\text{,}$ $1.8\text{ km}\text{,}$ $3.6\text{ km}\text{.}$

Solution

This is just Young's double-slit experiment, excpet that now, frequency is in rado wave, instead of visible light range. Let's calculate the wavelength since we can then use the condition from douoble-slit directly.

\begin{equation*} \lambda = \frac{c}{f} = \frac{3\times 10^{8}\text{ m/s}}{4.0\times 10^{6}\text{ Hz}} = 75\text{ m}. \end{equation*}

The directions of constructive interference are

\begin{equation*} d\sin\theta_m = m \lambda. \end{equation*}

Let $y$ axis is placed along the road with origin at the center. Let $L$ is the distance to the road. Then, we can write this in small angle approximation as

\begin{equation*} y_m = m \frac{\lambda L}{d}\ \ \ (m \text{ integer}). \end{equation*}

For five spots, I will pick $m=-2,\ -1,\ 0,\ 1,\ 2\text{.}$ Actually, we can just calculate $y_1$ and then use it to get the others. These give

\begin{align*} \amp y_0 = 0,\\ \amp y_1 = \frac{75\text{ m}\times 12\text{ km}}{0.5\text{ km}} = 1.8\text{ km},\\ \amp y_{-1} = - 1.8\text{ km},\\ \amp y_2 = 3.6\text{ km},\\ \amp y_{-1} = - 3.6\text{ km}. \end{align*}

A Lloyd's mirror is a set up that allows a double-slit experiment with only one source by placing a mirror between the slit and the screen as shown in Figure 50.7.20. Note that in this setup, the virtual image of the original source serves as the second source of light.

(a) Find the conditions for constructive and destructive interferences on the screen if the screen is a distance L away and the slit is a distance $D$ above the mirror.

(b) If $L = 2.5\text{ m}\text{,}$ $D = 20\ \mu\text{m}$ and wavelength of light used is the yellow sodium D line of wavelength $589\text{ nm}\text{,}$ find $y$ coordinates of two places on the screen you will find bright spots and two places where there will be dark spots. Use the simplification that comes with the approximation that $D\lt\lt L\text{.}$

Hint

(a) and (b) There is a phase changing event in one of the paths relative to the other path. Maky necessary modification to the condition for Young's double-slit formula for constructive and destructive interferences.

(a) $18.4\text{ mm}\text{,}$ $55.3\text{ mm}\text{,}$ (b) $36.8\text{ mm}\text{,}$ $73.6\text{ mm}\text{.}$

Solution 1 (a)

(a) Constructive Interferences. The geometry of the Lloyd's mirror is similar to Young's double slit experiment with separation between the slits $d = 2D$ with one exception: one of the paths has a additional phase shift of $pi$ rad due to the reflection in that path. Therefore, the condition for constructive interference will be what was the condition of destructive interference in the Young's double-slit experiment, viz.,

\begin{equation*} d\sin\theta = m\frac{\lambda}{2}, \ \ m\ \text{positive odd integer}. \end{equation*}

Let's calculate the sine of the directions for $m=1$ and $m=3\text{.}$

\begin{gather*} \sin\theta_1 = \frac{\lambda}{4D} = \frac{589\text{ nm}}{4\times 20\ \mu \text{m}} = 7.36 \times 10^{-3}. \\ \sin\theta_2 = 3\frac{\lambda}{4D} = 3\times 7.36\times 10^{-3} = 2.21\times 10^{-2} . \end{gather*}

These are such small angles that we can freely use small angle approximation. Using $\sin\theta\approx\tan\theta=y/L$ we get the $y$ corrdinates.

\begin{gather*} y_1 = 2.5 \text{m}\times 7.36\times 10^{-3} = 18.4\text{ mm}. \\ y_2 = 2.5 \text{m}\times 2.21\times 10^{-2}= 55.3\text{ mm}. \end{gather*}
Solution 2 (b)

(a) Destructive Interferences. The condition will be

\begin{equation*} d\sin\theta = m \lambda , \ \ m\ \text{positive integer}. \end{equation*}

Carrying through the calculations we did for (a), we get

\begin{gather*} y^\prime_1 = L\times \frac{\lambda}{2D} = 2.5 \text{m}\times 1.47\times 10^{-2}=36.8\text{ mm}. \\ y^\prime_2 = L\times \frac{2\lambda}{2D} = 2.5 \text{m}\times 2.94\times 10^{-2}= 73.6\text{ mm}. \end{gather*}

A radio transmitter atop a $50\text{-m}$ tall tower near a large lake emits a monochromatic electromagnetic wave of frequency $110\text{ MHz}\text{.}$ An airplane flying $20\text{ km}$ away and $6,000\text{ m}$ above ground finds the signal from the tower to vary in intensity with a period of $18\text{ sec}\text{.}$ How fast is the plane flying with respect to the ground?

Hint

Work out condition for $m^\text{th}$ order interference using Lloyd's mirror example, Problem 50.7.19. Cast the expression in terms of horizotal distance, calling it $l_m\text{.}$ Then equate $|l_\text{next order} - l_{this order}| = vT\text{.}$

$101\text{ m/s}\text{.}$

Solution

The varying signal occurs due to Lloyd effect addressed in the Problem 50.7.19. There are two signals that reach the airplane, one direct and the other reflected from the ground as shown in Figure 50.7.23.

The constructive interference condition for $m^\text{th}$ constructive interference will be

\begin{equation*} d \sin\theta_m = m \frac{\lambda}{2},\ \ m\text{ odd integer}. \end{equation*}

Here $d=2D$ with $D$ equal to the height of the tower. Since $H/L=0.3\text{,}$ the angles will be around $16^\circ\text{.}$ Therefore, we will use small angle approximations, such as $\sin\theta\approx\tan\theta=y/L\text{.}$

Let us denote the horizontal distance where $m^\text{th}$ order occurs by $l_m\text{.}$ We are going to write $l_{m+2}$ also, and then equate $|l_{m+2}-l_m$ to $vT$ with $T$ the period observed. From the interference condition for order $m$ we have

\begin{equation*} \frac{H}{l_m} = m \frac{\lambda}{4D},\ \ m\text{ odd integer}. \end{equation*}

Therefore, next order will be at $m+2$ since $m$ is restricted to odd integers.

\begin{equation*} \frac{H}{l_{m+2} = (m+2) \frac{\lambda}{4D},\ \ m\text{ odd integer}. \end{equation*}

Subtracting them, and setting |l_m -l_{m+2}| to $vT$ we get

\begin{equation*} v = \frac{\lambda}{2D}\: \frac{l_m \_{m+2}}{HT}. \end{equation*}

Now, we make assumption that $l_{m+2}\approx l_m = L$ to get

\begin{equation*} v = \frac{\lambda}{2D}\: \frac{L^2}{HT}. \end{equation*}

The data for the EM wave is given in frequence and the wave travels in air, where speed will be same as in vacuum. Therefore, \lambda=c/f and

\begin{equation*} v = \frac{c}{2fD}\: \frac{L^2}{HT}. \end{equation*}

Now we will put the numbers in.

\begin{align*} v \amp = \frac{3\times 10^8\text{ m/s}}{2\times 110\times 10^{6}\text{ Hz} \times 50\text{ m} }\: \frac{20,000^2\text{ m}^2}{6,000\text{ m} \times 18\text{ s} }\\ \amp = 101\text{ m/s}. \end{align*}