Help with two multiple choice questions related to light and interference

AI Thread Summary
In the discussion, the first question addresses the behavior of light reflecting off a thin film of glass and plastic when separated by an air gap. It is concluded that the air gap does not significantly change the destructive interference condition, leading to minimal reflection from both sides, suggesting answer d) as correct. The second question focuses on the diffraction pattern produced by red and blue light passing through a single slit, with the central spot appearing purple and gradually shifting to red as one moves outward, indicating answer b) is correct. The analysis emphasizes the importance of phase shifts and path differences in determining interference patterns. Overall, the responses highlight key principles of light behavior in thin films and diffraction.
whatislifehue
Messages
4
Reaction score
0

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.
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top