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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 1.6

Step by step solution

01

Write down the formula for the angular position of minima.

The formula is given by \[\sin(\theta)=m\lambda/a\]
02

Write down the values

The first dark band occurs at \(38.2^{\circ}\) in air and at \(21.6^{\circ}\) when immersed in the liquid. Thus \(\theta_{air}=38.2^{\circ}\) and \(\theta_{liquid}=21.6^{\circ}\)
03

Set up equation using the formula.

Since \(\sin(\theta_{air})/ \sin(\theta_{liquid}) = v_{air}/v_{liquid}=n_{liquid}\), where \(v_{air}, v_{liquid}\) are the speeds of the light in the mediums and \(n_{liquid}\) is the refractive index, inserting our values gives \(\sin(38.2^{\circ})/ \sin(21.6^{\circ}) = n_{liquid}\)
04

Solve for refractive index.

Evaluate \(\sin(38.2^{\circ})/ \sin(21.6^{\circ}) = n_{liquid}\). Therefore, \( n_{liquid}\approx 1.6 \)

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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 is a measure of how much the speed of light is reduced inside a medium compared to the speed of light in a vacuum. This crucial concept in optics determines how light propagates through different substances.

  • It is denoted as "n" and is defined as the ratio of the speed of light in a vacuum (c) to that in the medium (v): \( n = \frac{c}{v} \).
  • For instance, water has a refractive index of 1.33, meaning light travels slower in water than in air.
  • In the context of the provided exercise, the refractive index helps determine how light behaves when it transitions from air to a liquid medium.
When light enters a material other than a vacuum or air, it slows down proportionately to the refractive index of the new medium.

If you've ever wondered why objects appear bent or distorted underwater, it's due to the refractive index creating this optical illusion. By calculating the refractive index of the unknown liquid, we ascertain its optical properties, which help determine many things like its identity or purity.
Diffraction
Diffraction is the phenomenon where waves, such as light, bend around obstacles or spread as they pass through narrow openings.

  • This effect is particularly significant when the aperture or obstacle size is comparable to the wavelength of the light.
  • The bending and spreading of light encompass the creation of patterns, typically consisting of dark and light bands or fringes.
  • Important in many technologies, diffraction explains why we can hear sounds around corners, but with light, it becomes crucial in studying the wave nature of light.
In optics, diffraction plays a pivotal role in understanding how light interacts with materials.

When light diffracts through a slit, it creates interference patterns where the light overlaps, known as minima (dark bands) and maxima (bright bands). This phenomenon helps scientists understand wave properties and is used in various applications, like spectrometry.

It's important to note that while diffraction is a natural occurrence with many types of waves, its noticeable impact makes it crucial in finely understanding the behavior of visible light in distinct settings.
Thin Slit
A thin slit refers to a narrow opening through which light passes. This concept is central in many physics experiments demonstrating wave behaviors, especially diffraction.

  • In our exercise, the slit is thin, meaning its width is a small fraction of the light's wavelength. This setup is ideal for observing clear diffraction patterns.
  • As light passes through the slit, it spreads out and creates patterns that can be used to measure wavelengths or investigate wave-particle duality.
  • Thin slits are used to explore foundational principles in optics. Through designs like Young's double-slit experiment, these principles show how light waves interfere to create bands of light and dark.
Such simplicity—just a light and a slit—holds profound information about light's nature and behavior.

In practical applications, thin slits help refine optical devices, like cameras and microscopes, enabling them to achieve better precision and clarity in imaging.

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Most popular questions from this chapter

Laser light of wavelength \(632.8 \mathrm{nm}\) falls normally on a slit that is \(0.0250 \mathrm{~mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is \(8.50 \mathrm{~W} / \mathrm{m}^{2}\). (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

Monochromatic light of wavelength \(580 \mathrm{nm}\) passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

If a diffraction grating produces a third-order bright spot for red light (of wavelength \(700 \mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength \(400 \mathrm{nm}\) )?

Coherent electromagnetic waves with wavelength \(\lambda\) pass through a narrow slit of width \(a\). The diffraction pattern is observed on a tall screen that is \(2.00 \mathrm{~m}\) from the slit. When \(\lambda=500 \mathrm{nm}\), the width on the screen of the central maximum in the diffraction pattern is \(8.00 \mathrm{~mm}\). For the same slit and screen, what is the width of the central maximum when \(\lambda=0.125 \mathrm{~mm} ?\)

Light of wavelength \(633 \mathrm{nm}\) from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide, and the resulting diffraction pattern is observed on a screen \(3.50 \mathrm{~m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

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