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Problem 68 Hard Difficulty

Many cells are transparent and colorless. Structures of great interest in biology and medicine can be practically invisible to ordinary microscopy. An interference microscope reveals a difference in refractive index as a shift in interference fringes to indicate the size and shape of cell structures. The idea is exemplified in the following problem: An air wedge is formed between two glass plates in contact along one edge and slightly separated at the opposite edge. When the plates are illuminated with monochromatic light from above, the reflected light has 85 dark fringes. Calculate the number of dark fringes that appear if water $(n=1.33)$ replaces the air between the plates.

Answer

113

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Top Physics 103 Educators
Elyse G.

Cornell University

Andy C.

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Marshall S.

University of Washington

Jared E.

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Video Transcript

our car. Our question wants us to calculate the number of dark friends. Is that appear if the water, which has an index of refraction in sub w, replaces the air, which has an index of refraction inside a between the place. So in order to do this, we're gonna consider the fact that as the refractive index of glass is greater than the refractive index of air and water, there will be a phase change of 100 80 degrees when the light undergoes refraction at the top surface of the lower glass plate. And no face change takes place when my under grows reflection at the bottom surface of the top plate. So the total face change for the two reflected waves of 180 degrees in the condition for destructive interference. For these reflective waves, eyes given by two times the index of refraction times the thickness T is equal to, um, in times the wavelength lambda. So, for a maximum value here, we can go ahead and calculate the maximum thickness of the film. When the film is made out of heirs were called his team Max. This is equal to em times. Linda, divided by two, and we're considering it when it's airs. This is gonna be in some A and we're told that there are a total of 85 dark fringes from the reflected light. So 85 dark fringes means that M is equal to 85 minus one, which is 84. So you click that value in as well as the value for our index every fraction of air. We can put everything in terms of the wavelength Lambda, since we don't know that this is 42 times lambda. So for the water film, we can now find the maximum value of the M or the total number of fringe is that that are gonna take place when we replace the air with water. So in order to do that, we can use the team Max value that we just found and the equation right up here. But now we're going to solve for M. So we're gonna call this in next, and it's equal to to kind of the index of refraction of the medium. And in this case, it's water times. The team max value we just found divided by Lambda. So it's okay that we didn't know Lambda when we calculated to T. Max because the lander's air gonna cancel. So we end up with two times in some W T. Max was 42 times lambda, divided by Lambda. Now, as I mentioned, those Lambda values cancel so we can cross those out. So we're left with two times index of refraction of water times 42 which is equal to approximately 112. Um, but so since M is equal to 112 the number of dark fringes is going to be equal to him. Max plus one. So because we have to consider the, uh, sources in next plus one, which is equal 113 we have to consider that zeroth quarter fringe as well. Counting the number of fringe is making box set in as the solution to our quest.

University of Kansas
Top Physics 103 Educators
Elyse G.

Cornell University

Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Jared E.

University of Winnipeg