/*! 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 7 In a Young's double-slit experim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a Young's double-slit experiment the separation \(y\) between the second- order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of 425 \(\mathrm{nm}\). Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta\). Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{~nm}\).

Short Answer

Expert verified
The new separation \(y_2\) is approximately 0.0248 m.

Step by step solution

01

Understand the Problem and Given Data

We are tasked with finding the separation \(y\) between the second-order bright fringe and the central bright fringe when the wavelength is changed. The given data is:- Initial separation \(y_1 = 0.0180 \text{ m}\)- Initial wavelength \(\lambda_1 = 425 \text{ nm} = 425 \times 10^{-9} \text{ m}\)- New wavelength \(\lambda_2 = 585 \text{ nm} = 585 \times 10^{-9} \text{ m}\)We need to find the new \(y_2\).
02

Relationship between Fringe Separation, Wavelength, and Order

In Young's double-slit experiment, the distance \(y\) between the \(m\)-th order bright fringe and the center is given by the formula:\[y = \frac{m \cdot \lambda \cdot L}{d}\]where \(m\) is the fringe order, \(\lambda\) is the wavelength, \(L\) is the distance from the slits to the screen, and \(d\) is the slit separation. Since \(L\) and \(d\) remain constant for both wavelengths, the ratio of the separations is proportional to the ratio of the wavelengths.
03

Set Up Proportional Equation

Since the angles are small, \(\tan \theta \approx \sin \theta\), making the expression for \(y\) proportional to \(\lambda\). The equation becomes:\[\frac{y_2}{y_1} = \frac{\lambda_2}{\lambda_1}\]We can use this to solve for \(y_2\).
04

Calculate the New Separation

Substitute the known values into the proportional equation:\[\frac{y_2}{0.0180} = \frac{585 \times 10^{-9}}{425 \times 10^{-9}}\]Solve for \(y_2\):\[y_2 = 0.0180 \times \frac{585}{425}\]\[y_2 \approx 0.0248 \text{ m}\]
05

Verify and Interpret the Result

The new separation \(y_2\) is approximately \(0.0248 \text{ m}\). Since the wavelength increased, the fringe separation also increases, consistent with the direct proportionality.

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.

Fringe Separation
Fringe separation in Young's double-slit experiment refers to the distance between consecutive bright (or dark) fringes on the observation screen. It helps us understand how the interference pattern is distributed across the screen. The formula for calculating fringe separation is:\[ y = \frac{m \cdot \lambda \cdot L}{d} \]Here's what each term signifies:- **\(y\)** is the fringe separation.- **\(m\)** is the order number of the fringe (1st, 2nd, etc.).- **\(\lambda\)** is the wavelength of the light used.- **\(L\)** is the distance from the slits to the screen.- **\(d\)** is the distance between the two slits.Understanding fringe separation is crucial because it lets us predict how changing different parameters (like the wavelength) affects the observed pattern. For example, if the wavelength increases, as in the problem above, the fringe separation also increases.
Wavelength
Wavelength is a measure of the distance between successive crests of a wave, typically used in the context of electromagnetic waves like light. It's denoted by the Greek letter \(\lambda\), and its units are usually meters (m) or nanometers (nm).In the context of optical experiments, the wavelength determines several characteristics:- **Color of the Light**: Different wavelengths correspond to different colors in the visible spectrum.- **Pattern Visibility**: Wavelength changes affect how interference patterns appear on the screen. Longer wavelengths result in wider fringe separations.In Young's experiment, knowing the wavelength is essential as it directly influences the outcome and appearance of the fringes. When comparing wavelengths of 425 nm and 585 nm, as mentioned in the solution, the latter leads to a greater fringe separation due to its longer length.
Optical Interference
Optical interference occurs when two light waves superpose, creating a resultant wave of greater, lesser, or the same amplitude. In Young's double-slit experiment, interference is what causes the fringe pattern. When light from two slits meets, it can interfere constructively or destructively: - **Constructive Interference**: Occurs when waves are in phase, meaning their crests and troughs align, resulting in bright fringes. - **Destructive Interference**: Occurs when waves are out of phase, leading to cancellation and dark fringes. The essence of optical interference is that the path difference between the two waves determines the type of interference, subsequently forming the characteristic bright and dark bands on the screen. This fundamental property of light provides insights into wave behavior and underpins technologies such as lasers and even noise-canceling headphones.
Bright Fringes
Bright fringes are the visible result of constructive interference in Young's double-slit experiment. As light from both slits meets and aligns correctly in phase, the intensity increases, forming these bright regions. The distance between these bright fringes, known as fringe separation, is influenced by several factors: - **Order of Fringe**: Higher order fringes appear further from the center. - **Wavelength**: A longer wavelength increases fringe separation. - **Slit Distance and Screen Distance**: Consistent in the formula, they help determine the precise separation. The appearance of bright fringes is essential not just for aesthetic reasons but for scientific measurements. By studying the patterns, scientists can deduce information about the light's wavelength or the medium through which it travels. Thus, bright fringes aren't just an optical curiosity but a tool for discovery and analysis.

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

Review Conceptual Example 2 before attempting this problem. Two slits are \(0.158 \mathrm{~mm}\) apart. A mixture of red light (wavelength \(=665 \mathrm{~nm}\) ) and yellow-green light (wavelength \(=565 \mathrm{~nm}\) ) falls on the slits. A flat observation screen is located \(2.24 \mathrm{~m}\) away. What is the distance on the screen between the third-order red fringe and the thirdorder yellow-green fringe?

Violet light (wavelength \(=410 \mathrm{~nm}\) ) and red light (wavelength \(=660 \mathrm{~nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines \(/ \mathrm{cm}\). This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the third-order maximum, (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{~m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{~m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{~nm}\), what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

Light of wavelength \(691 \mathrm{nm}\) (in vacuum) is incident perpendicularly on a soap film \((n=1.33)\) suspended in air. What are the two smallest nonzero film thicknesses (in \(\mathrm{nm})\) for which the reflected light undergoes constructive interference?

A flat observation screen is placed at a distance of \(4.5 \mathrm{~m}\) from a pair of slits. The separation on the screen between the central bright fringe and the first-order bright fringe is \(0.037 \mathrm{~m}\). The light illuminating the slits has a wavelength of \(490 \mathrm{~nm}\). Determine the slit separation.

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.