/*! 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 30 Two flat glass plates are presse... [FREE SOLUTION] | 91Ó°ÊÓ

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Two flat glass plates are pressed together at the top edge and separated at the bottom edge by a strip of tinfoil. The air wedge is examined in yellow sodium light (589 nm) reflected normally from its two surfaces, and 42 dark interference fringes are observed. Compute the thickness of the tinfoil.

Short Answer

Expert verified
The thickness of the tinfoil is approximately 12.27 micrometers.

Step by step solution

01

Understand the Problem

Two glass plates create an air wedge with a strip of tinfoil. We observe interference patterns using sodium light and need to find the thickness of the tinfoil that leads to 42 dark fringes.
02

Identify the Formula

The number of dark fringes in an air wedge is related to the thickness by the formula \( 2t = (m+1/2) \lambda \), where \( m \) is the fringe count, \( t \) is the thickness, and \( \lambda \) is the wavelength of the light.
03

Solve the Equation for Thickness

To find the thickness \( t \), we rearrange the formula to \( t = \left(m + \frac{1}{2}\right) \frac{\lambda}{2} \). Plug in \( m = 42 - 1 \) because the first fringe is at \( m = 0 \), and \( \lambda = 589 \times 10^{-9} \) m.
04

Calculate the Thickness

Substitute \( m = 41 \) and \( \lambda = 589 \times 10^{-9} \) into the equation: \[ t = \left(41 + \frac{1}{2}\right) \frac{589 \times 10^{-9}}{2} \]. Calculating this gives \( t \approx 1.227 \times 10^{-5} \) m or \( 12.27 \) micrometers.

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Key Concepts

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

Air Wedge
An air wedge is a simple optical device created by two flat surfaces separated by a thin layer of air. In the typical setup, two flat glass plates are used, with one end pressed together and the other end separated by a small object, like a strip of tinfoil. This arrangement causes the gap between the plates to gradually increase. The air gap acts as a medium where light can reflect and produce interference patterns when viewed using monochromatic light, such as sodium light at 589 nm. Observing these interference patterns helps in measuring the thickness of the object that created the air wedge, by counting the number of dark and bright fringes formed due to constructive and destructive interference. These visible bands of light, known as fringes, demonstrate the wave nature of light.
Thickness Measurement
Measuring the thickness of a thin object, like the tinfoil in the original exercise, involves understanding the interference pattern's relationship with the air wedge. By examining the pattern formed by the interference of light waves, we can estimate the thickness of the wedge. In the exercise, the air wedge is used in conjunction with the interference fringes to indirectly measure how thick the object separating the two plates is. The mathematical relationship used involves the counting of interference fringes, where each fringe corresponds to a specific change in the optical path length, directly relating to a change in thickness of the air wedge. By plugging the appropriate values, including the wavelength of the light used and the number of fringes observed, we can accurately determine the wedge's thickness.
Fringe Count
The fringe count is crucial in solving problems involving interference patterns. Each dark fringe corresponds to a condition known as destructive interference, where the path difference between the interfering waves is a half-integer multiple of the wavelength. In mathematical terms, the condition for dark fringes can be expressed as: \( 2t = (m + \frac{1}{2}) \lambda \). In the air wedge scenario, the total number of observable dark fringes helps in calculating the thickness of the wedge. Often, students may mistakenly begin counting from 1, while the correct start is from zero, as the first fringe corresponds to \( m = 0 \). In the provided exercise, after observing 42 dark fringes, it's essential to use \( m = 41 \) when applying the formula, resulting from initial enumeration starting from zero. This accuracy allows us to derive the correct thickness of objects like the tinfoil.
Wavelength
Wavelength is a fundamental property of waves, including light waves, and defines the distance between successive peaks. When discussing interference patterns, the wavelength of the light source becomes a vital factor. In the air wedge context, the wavelength of the light (\( \lambda \)) determines how light waves will interfere with each other and thus the spacing of the fringes. Sodium light with a wavelength of 589 nm is frequently used in these experiments because of its distinct color and coherent properties. Its value is essential in calculating the thickness of the tinfoil as it appears in the general formula for interference, where we're converting physical changes in light's path into measurable dimensions. Correct application of this wavelength in calculations ensures the precision and reliability of the thickness measurement, making it an indispensable part of analyzing interference patterns.

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Most popular questions from this chapter

A spectrum of the Sun's radiation in the infrared region is produced by a grating. What is the wavelength being studied if the infrared line in the first order occurs at an angle of \(25.0^{\circ}\) with the normal and the fourth-order image of the hydrogen line of wavelength \(656.3 \mathrm{~nm}\) occurs at \(30.0^{\circ}\) ?

A single slit of width \(0.140 \mathrm{~mm}\) is illuminated by monochromatic light, and diffraction bands are observed on a screen \(2.00 \mathrm{~m}\) away. If the second dark band is \(16.0 \mathrm{~mm}\) from the central bright band, what is the wavelength of the light?

Determine the ratio of the wavelengths of two spectral lines if the second- order image of one line coincides with the third-order image of the other line, both lines being examined by means of the same grating.

Red light falls normally on a diffraction grating ruled 4000 lines/cm, and the second-order image is diffracted \(34.0^{\circ}\) from the normal. Compute the wavelength of the light. From the grating equation \(m \lambda=a \sin \theta_{\mathrm{m}}\) $$ \lambda=\frac{a \sin \theta_{2}}{2}=\frac{\left(\frac{1}{4000} \mathrm{~cm}\right)(0.559)}{2}=6.99 \times 10^{-5} \mathrm{~cm}=699 \mathrm{~nm} $$

Red light of wavelength \(644 \mathrm{~nm}\), from a point source, passes through two parallel and narrow slits which are \(1.00 \mathrm{~mm}\) apart. Determine the distance between the central bright fringe and the third dark interference fringe formed on a screen parallel to the plane of the slits and \(1.00 \mathrm{~m}\) away.

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