Section 50.7 Interference of Light Bootcamp
Subsection 50.7.1 Young's Double-Slit Experiment
Problem 50.7.1. Interference of Laser Light in a Double-Slit Experiment.
Follow the link: Checkpoint 50.2.7.
Problem 50.7.2. Location of Interference Maxima and Minima in Young's Double-Slit Experiment.
Follow the link: Checkpoint 50.2.8.
Problem 50.7.3. Separation of Red and Blue Colors by Interference.
Follow the link: Checkpoint 50.2.9.
Problem 50.7.4. Finding Wavelength from Interference Pattern in Young's Double-Slit Experiment.
Follow the link: Checkpoint 50.2.10.
Subsection 50.7.2 Interference From Dielectric Film
Problem 50.7.5. Thickness of a Soap Film from Wavelength of the Reflected Light.
Follow the link: Checkpoint 50.3.3.
Problem 50.7.6. Interference of Laser Light Reflected from a Wedge.
Follow the link: Checkpoint 50.3.8.
Problem 50.7.7. Change in Interference When Glass Plate is Inserted in one Path.
Follow the link: Checkpoint 50.3.11.
Problem 50.7.8. Thicknesss of an Oil Film on Glass from Wavelength of Light Observed at Normal.
Follow the link: Checkpoint 50.3.12.
Problem 50.7.9. Color of Very Thin Soap Film When Observed from Normal Direction.
Follow the link: Checkpoint 50.3.13.
Problem 50.7.10. Radius of Curvature of a Lens from Number of Newtonian Rings.
Follow the link: Checkpoint 50.3.16.
Subsection 50.7.3 Interferometers
Problem 50.7.11. Precise Distance Measurement by Michelson Interferometer.
Follow the link: Checkpoint 50.5.3.
Problem 50.7.12. Movement of Fringes in a Michelson's Interferometer with Moving Mirror.
Follow the link: Checkpoint 50.5.4.
Problem 50.7.13. Resolving Power Required in a Fabry Perot Instrument for Separating Sodium Yellow Doublet.
Follow the link: Checkpoint 50.6.9.
Problem 50.7.14. Minimum Wavelength Difference Resolvable in a Fabry-Perot Interferometer.
Follow the link: Checkpoint 50.6.10.
Problem 50.7.15. Width of Transmittance Peaks of a Fabry-Perot Interferometer.
Follow the link: Checkpoint 50.6.11.
Problem 50.7.16. Reflectivity of Mirrors of Fabry-Perot Cavity Required to Separate Mercury Yellow-Orange Doublet.
Follow the link: Checkpoint 50.6.12.
Subsection 50.7.4 Miscellaneous
Problem 50.7.17. First Two Angles From Normal Direction of Bright Fringes from a Plastic Film.
Follow the link: Checkpoint 50.3.14.
Problem 50.7.18. Interference on Road Trip for Electromagnetic Wave.
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.
The car samples points on the “screen”.
\(-3.6\text{ km}\text{,}\) \(-1.8\text{ km}\text{,}\) \(0\text{,}\) \(1.8\text{ km}\text{,}\) \(3.6\text{ km}\text{.}\)
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.
The directions of constructive interference are
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
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
Problem 50.7.19. Lloyd's Mirror.
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 \(D\lt\lt L\text{.}\)

(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) \(36.8\text{ mm}\text{,}\) \(73.6\text{ mm}\text{.}\)
(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.,
Let's calculate the sine of the directions for \(m=1\) and \(m=3\text{.}\)
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.
(a) Destructive Interferences. The condition will be
Carrying through the calculations we did for (a), we get
Problem 50.7.21. Signal Intensity Picked up by Airplane Varying Intensity due to Interference.
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?

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 \(|l_\text{next order} - l_{this order}| = vT\text{.}\)
\(101\text{ m/s}\text{.}\)
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
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
Therefore, next order will be at \(m+2\) since \(m\) is restricted to odd integers.
Subtracting them, and setting |l_m -l_{m+2}| to \(vT\) we get
Now, we make assumption that \(l_{m+2}\approx l_m = L\) to get
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
Now we will put the numbers in.