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A thin sheet of transparent material has an index of refraction of 1.40 and is 15.0$\mu \mathrm{m}$ thick. When it is inserted in the light path along one arm of an interferometer, how many fringe shifts occur in the pattern? Assume the wavelength (in a vacuum) of the light used is 600 $\mathrm{nm}$ . Hint: The wavelength will change within the material.

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Rutgers, The State University of New Jersey

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Hope College

So we have the refractive index of a transparent material as one point for, and the thickness of that material is 15 micrometer. The web linked off light used in the interferometer is 600 nanometer. Now the mitral is inserted into the interferometer and the fridge shift occurs because of the change in the web length of delight inside the material, and that causes the change in pad difference. So let's find the web link inside the material first. Okay, so by definition of the refractive index, the refractive index is the ratio of velocity off light in vacuum to the velocity off light in that material. So remember velocities frequency timesweb licked over frequency times web link in material. I'm just gonna give it Lambda Prime. There you can see that f is constant and book both sides because the frequency does not change whether the light is in vacuum or any material. So it's just gonna be constant. So this gives us Lambda over Lambda Prime. And from here, Lambda Prime is actually Lambda over end. So I'm just gonna put this as equation one now for the vacuum for the thickness t the friendship If you remember friendship in vacuum. The total friendship is going to be an equals four times thickness over wavelength of the light and similarly friendship and material in, uh, material that is given the total friendship I'm gonna call this and Prime is going to be four t over Lambda Prime and Lambda Prime is Lambda over and that is going to give us for tea and over Lambda. Okay, so let's go to the next page. That means the total change infringes which is delta and is going to be the total French shift in the material, minus the total friendship in the vacuum. And this is going to be a four t over Lambda and minus one. And this is four times thicknesses, 15 metre micrometer, and the Web link that 600 nanometers and and is 1.4 so and minus one is basically is 1.4 minus one, and this will give us 40

University of Wisconsin - Milwaukee