Help with two multiple choice questions related to light and interference

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SUMMARY

The discussion focuses on two multiple-choice questions regarding light interference and diffraction patterns. For the first question, the conclusion is that when a thin film of glass is separated from a plastic film by an air gap, both sides will still reflect very little light due to destructive interference, making option d) the correct answer. The second question addresses the diffraction pattern created by monochromatic red and blue light, where the central bright spot will appear purple, transitioning to red as one moves outward, confirming option b) as the correct choice.

PREREQUISITES
  • Understanding of light interference principles
  • Knowledge of phase shifts in wave optics
  • Familiarity with diffraction patterns and their formation
  • Basic equations of wave optics, including \(\Phi = \Phi_{incoming} + \frac{2\pi\Delta x}{\lambda} + \phi\)
NEXT STEPS
  • Study the principles of thin film interference in optics
  • Learn about phase shifts in wave reflections and their impact on interference
  • Explore diffraction patterns produced by single slits and multiple wavelengths
  • Investigate the mathematical modeling of light behavior using wave equations
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Students of physics, educators teaching optics, and anyone interested in understanding light behavior in thin films and diffraction phenomena.

whatislifehue
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Homework Statement


A thin film of glass is in flush contact with a thin film of transparent plastic that has a higher index of refraction than the glass. Monochromatic light of a certain frequency is shown on both sides of this combination, and there is very little light reflected from either side. If the two slices are now separated slightly to allow for a small air gap between them, and we repeat the process with the same light, which of the following will we see?
a. Light will be reflected from the glass side and the plastic side will still reflect very little light.
b. Light will be reflected from the plastic side and the glass side will still reflect very little light.
c. Light will be reflected from both sides.
d. Both sides will still reflect very little light.
e. More information is needed in order to draw a conclusion here.

Monochromatic red (low frequency) and monochromatic blue (higher frequency) light are shown simultaneously through a single slit, making a diffraction pattern on a screen behind it. What will be seen in the central bright spot? a. a uniformly purple spot all the way out to a dark fringe
b. a purple spot in the middle which gradually turns redder as you move out from the center
c. a purple spot in the middle which gradually turns bluer as you move out from the center
d. a purple spot in the middle which gradually turns bluer as you move out from the center in one direction, and redder as you move out from the center in the other direction.
e. darkness - the two light waves cancel each other in the center

Homework Equations


\Phi = \Phi_{incoming} + \frac{2\pi\Delta x}{\lambda} + \phi (1)\\<br /> <br /> asin(\theta) = m\lambda (2)<br /> <br />

The Attempt at a Solution


For the first question, because very little light is reflected from either side, there is destructive interference going on. As light passes from air to glass, there is a phase shift (phi) of pi. Then as it travels through the glass and reflects off the plastic, there is another phase shift of pi, but also a path difference (second term in first equation) equal to 2*pi*(2*t/lambda) (since delta x = 2*t). The difference in phase is simply 2*pi*(2*t/lambda), and for destructive interference to occur, 2*t/lambda has to equal n/2 for n = 1,2,3... Now, if we allow a gap of air inside, the same logic follows as it travels from air to glass, but from glass to air, there is no phase shift since n_air < n_glass. Then from air to plastic again, there is a phase shift, so ultimately the phase difference is the same as without the gap of air, so I would think the answer is d) both sides will still reflect very little light.

For the second question, the central bright spot will be a blend of the two monochromatic lights (blue + red --> purple). As we move away from the center, theta increases, which means that the wavelength increases, and increased wavelength means the wave is becoming more red, so I would think b) is the answer.
 
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whatislifehue said:

Homework Statement


For the first question, because very little light is reflected from either side, there is destructive interference going on. As light passes from air to glass, there is a phase shift (phi) of pi. Then as it travels through the glass and reflects off the plastic, there is another phase shift of pi, but also a path difference (second term in first equation) equal to 2*pi*(2*t/lambda) (since delta x = 2*t). The difference in phase is simply 2*pi*(2*t/lambda), and for destructive interference to occur, 2*t/lambda has to equal n/2 for n = 1,2,3... Now, if we allow a gap of air inside, the same logic follows as it travels from air to glass, but from glass to air, there is no phase shift since n_air < n_glass. Then from air to plastic again, there is a phase shift, so ultimately the phase difference is the same as without the gap of air, so I would think the answer is d) both sides will still reflect very little light.
The thickness of the air gap also counts. Think, if it is much thicker than the wavelength, the reflected intensity is about the sum of the intensities reflected from the separate glass and plastic films, both of them showing constructive interference with the air gap between them.
There are lot of ways the rays go, and all contribute to the reflected light. There are reflections from each interface and phase shifts in each films.
reflections.JPG
 

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