A microscope slide 10. cm long is separated from a glass plate at one end by a sheet of paper. As shown below, the other end of the slide is in contact with the plate. The slide is illuminated from above by light from a sodium lamp (λ=589nm�=589nm). The dimensions of the figure are distorted for visualization. In reality, the air gap is comparable to a small multiple of the wavelength, and the glass pieces are very thick compared to the wavelength. If 16 fringes per centimeter are seen along the slide. What is the thickness of the piece of paper, ΔyΔ�, in micrometers? Suppose that the setup of the preceding problem is immersed in an unknown liquid. If 19 fringes per centimeter are now seen along the slide, what is the index of refraction, n� of the liquid? Is the first fringe near the edge where the plates are in contact a bright fringe or a dark fringe? Select the best answer and explanation. The first fringe is dark because reflection from each the upper and the lower interfaces of the air gap causes phase change π�, but the path difference causes the two reflected waves to be maximally out of phase. The first fringe is bright because reflection from each the upper and the lower interfaces of the air gap causes phase change 00 while the path difference is approximately zero. The first fringe is bright because reflection from each the upper and the lower interfaces of the air gap causes phase change π� while the path difference is approximately zero. The first fringe is dark because reflection from the upper interface of the air gap causes phase change 00 while reflection from the lower interface causes phase change π� while the path difference is approximately zero. The first fringe is dark because reflection from the upper interface of the air gap causes phase change π� while reflection from the lower interface causes phase change 00 while the path difference is approximately zero. The first fringe is bright because, while reflection from the upper interface of the air gap causes phase change π� and reflection from the lower interface causes phase change 00, the path difference brings the two reflected waves back into phase. The first fringe is dark because reflection from each the upper and the lower interfaces of the air gap causes phase change 00, but the path difference causes the two reflected waves to be maximally out of phase. The first fringe is bright because, while reflection from the upper interface of the air gap causes phase change 00 and reflection from the lower interface causes phase change π�, the path difference brings the two reflected waves back into phase.

A microscope slide 10. cm long is separated from a glass plate at one end by a sheet of paper. As shown below, the other end of the slide is in contact with the plate. The slide is illuminated from above by light from a sodium lamp (λ=589nm�=589nm). The dimensions of the figure are distorted for visualization. In reality, the air gap is comparable to a small multiple of the wavelength, and the glass pieces are very thick compared to the wavelength. If 16 fringes per centimeter are seen along the slide. What is the thickness of the piece of paper, ΔyΔ�, in micrometers? Suppose that the setup of the preceding problem is immersed in an unknown liquid. If 19 fringes per centimeter are now seen along the slide, what is the index of refraction, n� of the liquid? Is the first fringe near the edge where the plates are in contact a bright fringe or a dark fringe? Select the best answer and explanation. The first fringe is dark because reflection from each the upper and the lower interfaces of the air gap causes phase change π�, but the path difference causes the two reflected waves to be maximally out of phase. The first fringe is bright because reflection from each the upper and the lower interfaces of the air gap causes phase change 00 while the path difference is approximately zero. The first fringe is bright because reflection from each the upper and the lower interfaces of the air gap causes phase change π� while the path difference is approximately zero. The first fringe is dark because reflection from the upper interface of the air gap causes phase change 00 while reflection from the lower interface causes phase change π� while the path difference is approximately zero. The first fringe is dark because reflection from the upper interface of the air gap causes phase change π� while reflection from the lower interface causes phase change 00 while the path difference is approximately zero. The first fringe is bright because, while reflection from the upper interface of the air gap causes phase change π� and reflection from the lower interface causes phase change 00, the path difference brings the two reflected waves back into phase. The first fringe is dark because reflection from each the upper and the lower interfaces of the air gap causes phase change 00, but the path difference causes the two reflected waves to be maximally out of phase. The first fringe is bright because, while reflection from the upper interface of the air gap causes phase change 00 and reflection from the lower interface causes phase change π�, the path difference brings the two reflected waves back into phase.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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If 16 fringes per centimeter are seen along the slide. What is the thickness of the piece of paper, ΔyΔ�, in micrometers?

Suppose that the setup of the preceding problem is immersed in an unknown liquid. If 19 fringes per centimeter are now seen along the slide, what is the index of refraction, n� of the liquid?

Is the first fringe near the edge where the plates are in contact a bright fringe or a dark fringe? Select the best answer and explanation.

The first fringe is dark because reflection from each the upper and the lower interfaces of the air gap causes phase change π�, but the path difference causes the two reflected waves to be maximally out of phase.

The first fringe is bright because reflection from each the upper and the lower interfaces of the air gap causes phase change 00 while the path difference is approximately zero.

The first fringe is bright because reflection from each the upper and the lower interfaces of the air gap causes phase change π� while the path difference is approximately zero.

The first fringe is dark because reflection from the upper interface of the air gap causes phase change 00 while reflection from the lower interface causes phase change π� while the path difference is approximately zero.

The first fringe is dark because reflection from the upper interface of the air gap causes phase change π� while reflection from the lower interface causes phase change 00 while the path difference is approximately zero.

The first fringe is bright because, while reflection from the upper interface of the air gap causes phase change π� and reflection from the lower interface causes phase change 00, the path difference brings the two reflected waves back into phase.

The first fringe is dark because reflection from each the upper and the lower interfaces of the air gap causes phase change 00, but the path difference causes the two reflected waves to be maximally out of phase.

The first fringe is bright because, while reflection from the upper interface of the air gap causes phase change 00 and reflection from the lower interface causes phase change π�, the path difference brings the two reflected waves back into phase.

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