Phase differences of light and multiples of lambda

[h00di3]

Hello All,
My new physics professor literally talks about nothing but his childhood and women during lecture halls. We seriously have taken half of a page of notes throughout this entire semester, and our book doesn't explain the content very well. I was wondering if any of you could help me out here, because even after talking to my professor, he still doesn't teach any relevant information. I would really appreciate your input for either this problem or what to do about this professor. He's the only one who teaches all of my physics classes until I'm a junior in college, and right now I haven't even finished my freshman semester... Suggestions?

Now that my rambling is complete, here's this homework problem:

Two waves of light in air, of wavelength [lambda]= 6.00 nm, are initially in phase. Each wave then travels through a different plastic layer, the thickness of layer 1 being L1=4.00 micrometers, and layer to being L2=3.50 micrometers. The indices of refraction are n1=1.40 and n2=1.60, respectively.

a) What multiple of lambda gives their phase difference after they both have emerged from
the layers?
b) If the waves later arrive at some common point with the same amplitude, is their
intereference fully constructive, fully destructive, intermediate but closer to fully
constructive, or intermediate but closer to fully destructive?

 

[lippyka]

To begin answering this question, you will first need to calculate the velocities of the two waves in each layer. Since the two waves have different indices of refraction, they will travel at different speeds. The equation for the speed of light in a medium is v = c/n, where c is the speed of light in a vacuum and n is the index of refraction. Using this equation, you can calculate the speed of the waves in each layer. Once you have the speeds of the waves, you can then use the equation for the wavelength of a wave in a medium, which is [lambda] = v/f, where v is the velocity of the wave and f is the frequency of the wave. This equation can be used to calculate the wavelength of the waves in each layer. Finally, you can use the equation for the phase change of a wave, which is [phi] = 2[pi](d/[lambda]), where d is the distance traveled by the wave and [lambda] is the wavelength of the wave. Using this equation, you can calculate the phase difference between the two waves after they have both emerged from the layers. For part b, you can use the equation for the interference of two waves, which is I = I1 + I2 + 2[square root of I1I2]cos([phi]), where I1 and I2 are the amplitudes of the two waves, [phi] is the phase difference between the two waves, and I is the total intensity of the interference pattern. Using this equation, you can determine if the interference is fully constructive, fully destructive, or somewhere in between. I hope this helps! Good luck with your problem.

 

[kerimek]

As a scientist, it is important to have a clear understanding of the concepts being taught in class. It seems like your professor may not be effectively communicating the material, which can be frustrating for students. My suggestion would be to seek out additional resources, such as online tutorials or study groups, to help supplement your understanding of the subject.

Now, regarding the homework problem, let's break it down. The first thing to note is that the two waves of light have the same wavelength, but travel through different mediums (air and plastic) with different indices of refraction. This means that their velocities will be different, and thus, they will experience a phase difference.

To calculate the phase difference, we can use the equation phi = 2pi*d/lambda, where phi is the phase difference, d is the distance the wave travels, and lambda is the wavelength. In this case, the distance traveled by each wave is the thickness of the plastic layers, L1 and L2.

a) To determine the multiple of lambda that gives the phase difference, we need to divide the total phase difference by 2pi. So, phi/2pi = (n1*L1 - n2*L2)/lambda. Plugging in the given values, we get phi/2pi = (1.40*4.00 micrometers - 1.60*3.50 micrometers)/6.00 nm = 0.167. This means that the phase difference is equal to 0.167 times the wavelength of the light.

b) To answer the second part of the question, we need to consider the interference between the two waves. Since they are initially in phase, their interference will depend on the phase difference after they have traveled through the plastic layers. If the phase difference is a multiple of 2pi, the interference will be fully constructive. If it is an odd multiple of pi, the interference will be fully destructive. In this case, the phase difference is 0.167 times the wavelength, which is not a multiple of 2pi or pi. Therefore, the interference will be somewhere in between fully constructive and fully destructive, but closer to fully constructive.

In conclusion, it is important to seek out additional resources to supplement your understanding of the subject if your professor is not effectively communicating the material. As for the homework problem, we can use the equation phi = 2pi*d/lambda to calculate the phase difference and determine the interference