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Chapter 36: Problem 71

A glass sheet is covered by a very thin opaque coating. In the middle of this sheet there is a thin scratch 0.00125 \(\mathrm{mm}\) thick. The sheet is totally immersed beneath the surface of a liquid. Parallel rays of monochromatic coherent light with wavelength 612 \(\mathrm{nm}\) in air strike the sheet perpendicular to its surface and pass through the scratch. A screen is placed in the liquid a distance of 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on the screen are 22.4 \(\mathrm{cm}\) apart. What is the refractive index of the liquid?

Short Answer

Expert verified
The refractive index of the liquid is approximately 1.64.

Step by step solution

01

Understand the Problem

We are given a thin scratch on a glass sheet immersed in a liquid. Light passes through this scratch, creating an interference pattern, and we need to find the refractive index of the liquid.
02

Define Variables and Formulas

Let \(d\) be the distance between the first dark fringes (22.4 \(\text{cm}\) in our case), \(D\) be the distance from the sheet to the screen (30.0 \(\text{cm}\)), \(n\) be the refractive index of the liquid, \(\lambda\) be the wavelength of light in air (612 \(\text{nm}\)), and \(\lambda'\) be the wavelength of light in the liquid. The formula for the fringe separation \(d\) in terms of \(n\) and \(D\) is: \[d = \frac{\lambda' D}{t} = \frac{\left( \frac{\lambda}{n} \right) D}{t}\] where \(t\) is the thickness of the scratch (0.00125 \(\text{mm} = 1.25 \times 10^{-6} \text{m}\)).
03

Substitute Known Values and Simplify

Rearrange to solve for \(n\): \[n = \frac{\lambda D}{d t}\] Substitute the known values: \(\lambda = 612 \times 10^{-9} \text{ m}\), \(D = 0.3 \text{ m}\), \(d = 0.224 \text{ m}\), and \(t = 1.25 \times 10^{-6} \text{ m}\).
04

Calculate the Refractive Index

Use the rearranged formula: \[ n = \frac{(612 \times 10^{-9})(0.3)}{(0.224)(1.25 \times 10^{-6})} \] After calculating, you will find: \[ n \approx 1.64 \] Thus, the refractive index of the liquid is about 1.64.

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Key Concepts

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

Interference Pattern
When light waves meet, they can superimpose on each other, creating an interference pattern. In this exercise, light beams pass through a thin scratch on a glass, resulting in bright and dark fringes on a distant screen. This pattern arises due to the phenomenon of interference, where the light waves either constructively or destructively combine.
  • Constructive interference occurs when waves are in phase, reinforcing each other to produce bright fringes.
  • Destructive interference happens when waves are out of phase, canceling each other and creating dark fringes.
Understanding these patterns is crucial in optics, as it reveals properties of both the light and the medium it travels through.
Optics
Optics is the study of light and its behaviors. It encompasses various phenomena, including reflection, refraction, and interference. In our scenario, we explore how light behaves when it travels through different mediums, like air and liquid, before forming an interference pattern.
  • Reflection refers to the bouncing back of light waves upon encountering a surface.
  • Refraction involves the bending of light waves as they pass through different materials, due to changes in speed.
  • Interference, as seen in this exercise, results when waves overlap, varying in intensity.
Grasping these core concepts is essential in understanding how devices, lenses, and many optical instruments function.
Monochromatic Light
Monochromatic light consists of waves with a single wavelength. In this exercise, the light used has a wavelength of 612 nm in air. Monochromatic light is crucial in creating clear, consistent interference patterns.
  • It eliminates variations caused by multiple wavelengths, enabling precise analysis.
  • Lasers often provide monochromatic light, due to their high coherence.
Analysis with monochromatic light helps in calculating variables like the refractive index more accurately, as seen in this problem.
Fringe Separation
Fringe separation refers to the distance between consecutive light or dark bands in an interference pattern. Calculating this separation helps in determining properties of the medium or the light itself. In this problem, the calculated fringe separation is 22.4 cm for the first dark fringes.
  • This separation depends on several factors: the wavelength of the light, the distance between the light source and the screen, and the medium's properties, like the refractive index.
  • Using the formula provided, parameters such as the path difference, serve to calculate the refractive index of the liquid.
Accurate measurements of fringe separation aid in diverse applications, from optical measurements to engineering projects.

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Most popular questions from this chapter

In a large vacuum chamber, monochromatic laser light passes through a narrow slit in a thin aluminum plate and forms a diffraction pattern on a screen that is 0.620 \(\mathrm{m}\) from the slit. When the aluminum plate has a temperature of \(20.0^{\circ} \mathrm{C},\) the width of the central maximum in the diffraction pattern is 2.75 \(\mathrm{mm}\) . What is the change in the width of the central maximum when the temperature of the plate is raised to \(520.0^{\circ} \mathrm{C}\) ? Does the width of the central diffraction maximum increase or decrease when the temperature is increased?

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda=587.9782 \mathrm{nm}\) and \(\lambda=\) 587.8002 \(\mathrm{nm} .\) You wish to resolve these spectral lines in first order using a grating 1.20 \(\mathrm{cm}\) in length. What minimum number of slits per centimeter must the grating have?

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 \(\mathrm{mm}\) wide. At the point in the pattern \(3.25^{\circ}\) . from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maxi-mum is \(I_{0} ?\)

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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