/*! 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 55 Measuring Refractive Index. A th... [FREE SOLUTION] | 91Ó°ÊÓ

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Measuring Refractive Index. A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm 38.2^{\circ}\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm 21.6^{\circ} .\) Find the refractive index of the liquid.

Short Answer

Expert verified
The refractive index of the liquid is approximately 1.68.

Step by step solution

01

Understanding the problem

The problem involves finding the refractive index of a liquid in which a diffraction pattern's first dark band angle changes. We will use the relation for the angle of the first dark band formed in single-slit diffraction.
02

Use the diffraction condition

The condition for the first dark band in single-slit diffraction is given by the equation: \( a \sin(\theta) = m \lambda \), where \(a\) is the slit width, \(\theta\) is the angle, \(m\) is the order number (1 for first dark band), and \(\lambda\) is the wavelength.
03

Relate wavelengths in air and liquid

In air, the wavelength \(\lambda_{air}\) is the speed of light divided by frequency \(f\): \(\lambda_{air} = c/f\). In the liquid, the wavelength \(\lambda_{liquid}\) is \(\lambda_{air}/n\), where \(n\) is the refractive index of the liquid.
04

Setup equations in air and liquid

Using the diffraction condition: \( a \sin(38.2^\circ) = \lambda_{air} \) in air, and \( a \sin(21.6^\circ) = \lambda_{liquid} \) in liquid.
05

Solve for the refractive index

Since \(\lambda_{liquid} = \lambda_{air}/n\), equate the two conditions: \( a \sin(38.2^\circ) = n \cdot a \sin(21.6^\circ) \). Simplify to find \(n = \frac{\sin(38.2^\circ)}{\sin(21.6^\circ)}\).
06

Calculate the refractive index

Compute the value of \(n\): \(n = \frac{\sin(38.2^\circ)}{\sin(21.6^\circ)}\). Using a calculator, find \(\sin(38.2^\circ) \approx 0.6196\) and \(\sin(21.6^\circ) \approx 0.3686\). Therefore, \(n \approx \frac{0.6196}{0.3686} \approx 1.68\).

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Key Concepts

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

Single-Slit Diffraction
Single-slit diffraction is a fundamental concept in optics that describes how light spreads out, or diffracts, when it passes through a narrow slit. In this context, light behaves much like a wave. When it encounters an obstacle, such as a slit, it bends around it and creates a pattern of alternating bright and dark bands on a screen placed behind the slit. This pattern is due to the interference of the light waves. The bands, usually called fringes, result from constructive and destructive interference. When considering single-slit diffraction, the central maximum, which is the brightest band, is flanked by progressively dimmer maxima and intervening minima (dark bands), where destructive interference occurs. The position of these fringes depends on the wavelength of the light, the width of the slit, and the angle from the central axis. Knowing these factors allows us to predict the pattern's characteristics. This knowledge is crucial for experiments involving refractive index measurements, as seen in our exercise.
Wavelength
Wavelength is a vital concept in understanding light and its interaction with different medium. It is defined as the distance between two successive peaks of a wave. The wavelength of light determines its color when observed. In physics, especially optics, wavelength (\(lambda\) is fundamental because it affects how light interacts with materialsWavelength is inversely related to frequency. This means light with a higher frequency has a shorter wavelength and vice versa. In the context of single-slit diffraction, wavelength plays a key role in determining the diffraction pattern. When light moves from one medium to another, its speed changes, altering its wavelength but not its frequency. This change is important when calculating the refractive index, as seen in the exercise above, where the wavelength of the light was affected by the change in medium from air to liquid.
Angle of Diffraction
The angle of diffraction is the angle between the original direction of an incoming wave and the direction of the diffracted wave. It is a critical concept in analyzing diffraction patterns. This angle determines the position of light and dark bands in diffraction experiments.When light passes through a single slit, the angle where the first minimum or dark band occurs can be calculated using the diffraction condition: \(a \sin(\theta) = m \lambda\). Here,\(a\) is the slit width,\(\theta\) is the angle of diffraction,\(m\) is the order of the dark band, and\(\lambda\) is the wavelength of the light.In our exercise, angles of 38.2° and 21.6° represent the diffraction in air and in the liquid. These angles are essential to determine the refractive index, which indicates how much the light bends when entering the liquid compared to air.
Optics
Optics is a branch of physics dealing with light and its properties, including its interactions with matter, reflection, refraction, and diffraction. Studying optics involves understanding how light behaves in different environments, which is essential for applications in eyeglasses, cameras, and scientific instruments.One key aspect of optics is the study of refractive index, which is a measure of how much light bends, or refracts, when entering a different medium. This bending occurs due to a change in the speed of light between different materials. The refractive index is calculated using the formula: \(n = \frac{c}{v}\), where \(n\) is the refractive index,\(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the material. Understanding optics and refractive indices is vital in accurately describing how light behaves in experiments, like the measurement of refractive index in the given textbook exercise.

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Most popular questions from this chapter

A converging lens 7.20 \(\mathrm{cm}\) in diameter has a focal length of 300 \(\mathrm{mm} .\) If the resolution is diffraction limited, how far away can an object be if points on it 4.00 \(\mathrm{mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{nm} .\)

Monochromatic \(\mathrm{x}\) rays are incident on a crystal for which the spacing of the atomic planes is 0.440 \(\mathrm{nm} .\) The first-order maximum in the Bragg reflection occurs when the incident and reflected \(x\) rays make an angle of \(39.4^{\circ}\) with the crystal planes. What is the wavelength of the \(\mathrm{x}\) rays?

An interference pattern is produced by light of wave- length 580 \(\mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 \(\mathrm{mm}\) . (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit, interference maxima? (b) Let the slits have width 0.320 \(\mathrm{mm}\) . In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

A loudspeaker having a diaphragm that vibrates at 1250 \(\mathrm{Hz}\) is traveling at 80.0 \(\mathrm{m} / \mathrm{s}\) directly toward a pair of holes in a very large wall in a region for which the speed of sound is 344 \(\mathrm{m} / \mathrm{s} .\) You observe that the sound comind through the openings first cancels at \(\pm 11.4^{\circ}\) with respect to the original direction of the speaker when observed far from the wall. (a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 \(\mathrm{cm} / \mathrm{s}\) on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 \(\mathrm{m}\) away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm 61.3 \mathrm{cm}\) from the point directly opposite the hole, but waves do reach the shore every- where within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

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