/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Many cells are transparent and c... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The number of dark fringes that appear when water replaces the air between the plates is 113.

Step by step solution

01

Identify the Relationship between Variables

The number of dark fringes is directly proportional to the refractive index of the medium between the plates. Mathematically, we can express this relation as \(F_n = k \cdot n\), where \(F_n\) is the number of dark fringes observed, \(n\) is the refractive index and \(k\) is a constant of proportionality.
02

Find the Constant of Proportionality

In the initial scenario, where air is the medium between the plates, we have the refractive index \(n_{air} = 1\) and the fringe count \(F_{air} = 85\). Substituting these values into the formula, we find that the constant \(k = F_{air}/n_{air} = 85/1 = 85\).
03

Use the Constant to Find the Number of Fringes with Water

We now want to find the number of fringes when the medium has been changed to water. The refractive index of water is given as \(n_{water} = 1.33\). We can use the formula from Step 1 with our constant \(k = 85\) to find the new fringe count: \(F_{water} = k \cdot n_{water} = 85 \cdot 1.33 = 113.05\). Since we cannot have a fractional number of fringes, we round this to the nearest whole number, which gives us 113.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index, denoted as 'n', is a measure of how much the speed of light is reduced within a particular medium compared to its speed in a vacuum. When light passes from one medium to another, it changes speed, which results in a change in direction—a phenomenon known as refraction.

From a practical standpoint, every transparent material has its own refractive index, and understanding this property is essential in fields like optics and microscopy. For example, the refractive index of air is approximately 1, which means light travels through air almost as fast as it does through a vacuum. In contrast, water has a refractive index of about 1.33, indicating light travels slower in water than in air.

Lenses utilize the refractive index to bend light and create images, whereas in an interference microscope, differences in refractive index between various cellular structures can be translated into an observable pattern of interference fringes. This is an invaluable tool for visualizing cells and tissues that are otherwise difficult to see with conventional microscopes.
Monochromatic Light
Monochromatic light is composed of waves that are all the same wavelength and, in turn, all the same color. This uniformity is critical when creating interference patterns, as it ensures that waves combine predictably when they overlap, based on their phase. These patterns are indicative of the optical path differences that result from changes in medium or wavelength, which are harnessed in an interference microscope.

Common monochromatic light sources include lasers and filtered lights. In educational experiments or analytical applications like thin-film measurement and refractometry, the use of monochromatic light allows for precise calculation since any observed interference can be attributed to changes in path length rather than wavelength variances. This precision is what makes the calculation of dark fringes in the given problem possible, assuming that a single wavelength is responsible for the observed fringes.
Dark Fringes
Dark fringes are a result of destructive interference of light waves, where two waves cancel each other out due to being out of phase by half a wavelength. In the context of an interference microscope, when monochromatic light reflects off different layers or structures, the path lengths traveled by the light waves can result in them arriving at the observer in phase (constructive interference) or out of phase (destructive interference).

The dark fringes visualized in an interference pattern directly correspond to areas where the path differences lead to destructive interference. These fringes provide a wealth of information about the material, including thickness, refractive index, and topography. For instance, the problem given describes how an air wedge between two glass plates creates a series of dark fringes. By counting these fringes and understanding their relationship with the refractive index, as demonstrated in the steps provided, one can infer qualitative and quantitative details about the substance between the plates.

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Most popular questions from this chapter

Light of wavelength \(546 \mathrm{~nm}\) (the intense green line from a mercury source) produces a Young's interference pattern in which the second minimum from the central maximum is along a direction that makes an angle of \(18.0\) min of arc with the axis through the central maximum. What is the distance between the parallel slits?

Intense white light is incident on a diffraction grating that has 600 lines/mm. (a) What is the highest order in which the complete visible spectrum can be seen with this grating? (b) What is the angular separation between the violet edge \((400 \mathrm{~nm})\) and the red edge \((700 \mathrm{~nm})\) of the first-order spectrum produced by the grating?

Light with a wavelength in vacuum of \(546.1 \mathrm{~nm}\) falls perpendicularly on a biological specimen that is \(1.000 \mu \mathrm{m}\) thick. The light splits into two beams polarized at right angles, for which the indices of refraction are \(1.320\) and \(1.333\), respectively. (a) Calculate the wavelength of each component of the light while it is traversing the specimen. (b) Calculate the phase difference between the two beams when they emerge from the specimen.

A thin film of oil \((n=1.25)\) is located on smooth, wet pavement. When viewed from a direction perpendicular to the pavement, the film reflects most strongly red light at \(640 \mathrm{~nm}\) and reflects no green light at \(512 \mathrm{~nm}\). (a) What is the minimum thickness of the oil film? (b) Let \(m_{1}\) correspond to the order of the constructive interference and \(m_{2}\) to the order of the destructive interference. Obtain a relationship between \(m_{1}\) and \(m_{2}\) that is consistent with the given data.

A helium-neon laser \((\lambda=632.8 \mathrm{~nm})\) is used to calibrate a diffraction grating. If the first-order maximum occurs at \(20.5^{\circ}\), what is the spacing between adjacent grooves in the grating?

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