/*! 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 43 Suppose you illuminate two thin ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

Short Answer

Expert verified
The index of refraction of the liquid is approximately 1.724.

Step by step solution

01

Understand the Problem

We are dealing with a double-slit interference pattern where the position of minima is given by the formula \(m \lambda = d \sin \theta\), where \(m\) is the order of the minimum, \(\lambda\) is the wavelength of light, \(d\) is the distance between slits, and \(\theta\) is the angle of the minima. We need to find the index of refraction when the slits are in a liquid, given the change in the angle of the minima.
02

Define the Relation for Minima in Air

For the slits in air, the first-order minima occur when \(m = \pm 1\). The angle of the minima is given as \(\theta = 35.20^{\circ}\). Thus, the equation becomes:\[d \sin 35.20^{\circ} = \lambda\]This equation provides a reference for the wavelength of light in air.
03

Define the Relation for Minima in Liquid

When the slits are in the liquid, the angle of the first-order minima changes to \(\theta' = 19.46^{\circ}\). The relation becomes:\[d \sin 19.46^{\circ} = n \lambda\]where \(n\) is the index of refraction of the liquid.
04

Set Up the Equation for the Index of Refraction

We know the wavelength remains the same, so equating \(\lambda\) in both conditions, we get:\[\frac{d \sin 35.20^{\circ}}{d \sin 19.46^{\circ}} = n\]This equation can be simplified to:\[n = \frac{\sin 35.20^{\circ}}{\sin 19.46^{\circ}}\]
05

Calculate the Index of Refraction

Using the above equation, compute \(n\):Convert angles to radians or use a calculator with degree mode.\[n = \frac{\sin 35.20^{\circ}}{\sin 19.46^{\circ}} \approx \frac{0.576}{0.334} \approx 1.724\]Thus, the index of refraction of the liquid is approximately 1.724.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Interference Pattern
An interference pattern arises when waves overlapping create a distinctive pattern of maxima and minima due to constructive and destructive interference, respectively. In the context of double-slit experiments, an interference pattern manifests as a series of bright and dark fringes on a screen.
The key components that contribute to this pattern are the wavelength of light and the slit separation. When monochromatic light—light with a single wavelength—passes through two closely spaced slits, it diffracts and the waves overlap.
  • Bright spots, or maxima, occur where the waves constructively interfere.
  • Dark spots, or minima, occur where the waves destructively interfere.
By understanding where these interference minima (dark spots) occur, we can gain insights into various properties of the light and the medium through which it travels.
Index of Refraction
The index of refraction, often denoted by the symbol \( n \), is a measure of how much the speed of light is reduced inside a medium compared to a vacuum. This property influences how light waves travel through different substances.
When light enters a material from air or another medium, its speed changes, which in turn affects several characteristics, such as the wavelength and the direction of the light. The index of refraction is determined by comparing the sine of the angle of incidence in the original medium to the sine of the angle of refraction in the new medium, according to Snell's law.
For the double-slit experiment:
  • The change in the angle of minima from when the slits are in air to when they are in a liquid helps determine the index of refraction of the liquid.
  • Mathematically expressed as \( n = \frac{\sin \theta_{air}}{\sin \theta_{liquid}} \), it allows calculation of how the medium affects wavefronts.
Understanding this concept clarifies how light interacts with different materials and why angles of interference sites shift when conditions change.
Monochromatic Coherent Light
Monochromatic coherent light refers to light that consists of a single wavelength and has a consistent phase relationship, which is crucial for producing stable interference patterns. Using monochromatic light in experiments ensures that the interference patterns remain constant and predictable, as there is no variation in the wavelength that could lead to pattern blurring.
  • This light ensures that all wavefronts from the slits are synchronized.
  • Using sources like lasers can produce monochromatic and coherent light.
In the context of this double-slit experiment, using the same monochromatic coherent light both in air and while submerged in a liquid ensures that changes in the interference pattern can be attributed to the refractive effects of the liquid rather than variations in the light source itself.
Angle of Minima
The angle of minima is a key aspect of analyzing interference patterns in double-slit experiments. It is the specific angle at which destructive interference causes a reduction in light intensity, resulting in a dark band or minimum in the pattern.The angle of minima is calculated using the formula \(m \lambda = d \sin \theta \), where \(m\) is the order of the minima, \(\lambda \) is the wavelength, \(d \) is the distance between slits, and \(\theta \) is the angle. This formula tells us how far the light spreads based on its interaction with the slits.
  • In a medium of different refractive index, the angle of minima changes, illustrating how the medium affects the wave's path.
  • Tracking these angles helps reveal properties of the medium, like the index of refraction.
Understanding the angle of minima helps in calculating critical optical properties of mediums through experiments. This comprehension aids in practical applications like optical coating designs and understanding material properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

What is the thinnest film of a coating with \(n=1.42\) on glass \((n=1.52)\) for which destructive interference of the red component \((650 \mathrm{nm})\) of an incident white light beam in air can take place by reflection?

Coherent light with wavelength 450 nm falls on a double slit. On a screen 1.80 \(\mathrm{m}\) away, the distance between dark fringes is 4.20 \(\mathrm{mm} .\) What is the separation of the slits?

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) this film is just thick enough for light with wavelength 582.4 \(\mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength 588.5 \(\mathrm{nm}\) . What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.