/*! 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 29 \(\bullet$$\bullet\) Red light o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

Short Answer

Expert verified
Central bright fringe width: 0.01084 m; first bright fringe width: 0.01086 m.

Step by step solution

01

Understand the Problem

We are asked to find the width of the central bright fringe and the first bright fringe on either side in a single-slit diffraction pattern. We use the formula for minima: \(a \sin \theta = m \lambda\), where \(a\) is the slit width, \(m\) is the order of the minimum, \(\lambda\) is the wavelength, and \(\theta\) is the diffraction angle. The width of a bright fringe is the distance between two adjacent minima on the screen.
02

Find the Angle for the Minima

For the central bright fringe, we need the angles for the first minimum on each side: \(m = \pm 1\). Using small angle approximation, \(\theta \approx \sin \theta \approx \tan \theta\): \(a \sin \theta = 1 \times \lambda\) \(0.350 \times 10^{-3} \sin \theta = 633 \times 10^{-9}\) \(\sin \theta = \frac{633 \times 10^{-9}}{0.350 \times 10^{-3}}\)
03

Simplify and Solve for Sin(θ)

Calculate the right side of the equation to find \(\sin \theta\):\[ \sin \theta = \frac{633 \times 10^{-9}}{3.5 \times 10^{-4}} = 1.8086 \times 10^{-3} \]
04

Calculate Distance for Central Bright Fringe

Using the calculated \(\sin \theta\), find \(\theta\) using small angle approximation \(\theta \approx \sin \theta\). Calculate the position on the screen with \(y = L \tan \theta \approx L \sin \theta\):\[ y = 3.00 \, \mathrm{m} \times 1.8086 \times 10^{-3} \approx 0.00542 \, \mathrm{m} \]This is one side, so the total width is:\[ 2y = 2 \times 0.00542 \, \mathrm{m} = 0.01084 \, \mathrm{m} \]
05

Calculate Width of the First Non-Central Bright Fringe

For the first bright fringe, we need angles for the second and third minima \(m = 1\) and \(m = 2\) outside the central fringe:\[ a \sin \theta = m \lambda \ \text{for other minima; repeat similar calculations} \] For \(m=2\) \(\sin \theta_2 = \frac{2 \times 633 \times 10^{-9}}{0.350 \times 10^{-3}} = 3.6171 \times 10^{-3} \)Now, \[ y_2 = L \sin \theta_2 \approx 3.00 \, \mathrm{m} \times 3.6171 \times 10^{-3} \approx 0.01085 \, \mathrm{m} \]Thus, total width:\[ 2(y_2 - y_1) \] where \(y_1\) corresponds to the first minimum used in central fringe.
06

Calculate First Bright Fringe Dimensions

Calculate the width of the first bright fringe using the positions calculated for \(y_1\) and \(y_2\):\[ \text{Fringe Width} = 2(y_2 - y_1) = 2(0.01085 \, \mathrm{m} - 0.00542 \, \mathrm{m}) \approx 0.01086 \, \mathrm{m} \]

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.

Central Bright Fringe Width
In the single-slit diffraction pattern produced when light of wavelength 633 nm passes through a slit that is 0.350 mm wide, the central bright fringe is particularly significant. This fringe is the result where the light waves interfere constructively, creating a bright band in the middle of the pattern.
To determine the width of this bright fringe, we calculate the distance between the minima, essentially where the bright fringe begins to taper off on either side. Using the formula for minima, \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum, we can then find these positions on the screen.
For the central fringe, the minima occur at \( m = \pm 1 \). Given the small angle approximation, \( \theta \approx \sin \theta \approx \tan \theta \), the position \( y \) on the screen can be simplified to \( y = L \sin \theta \). This calculation gives the position across the screen for one side of the bright central fringe. Doubling this result gives the entire width of the central bright fringe. Through the calculations provided, one can see that the central bright fringe is approximately 0.01084 meters wide.
Diffraction Minima
In a diffraction pattern, minima are crucial in understanding the pattern's structure. The minima represent points where constructive interference transitions into destructive interference, resulting in darker bands adjacent to the brighter ones.
For a single slit, minima are calculated using the equation \( a \sin \theta = m \lambda \), with \( m \) representing the order of the minima. For the central bright fringe, the most relevant minima are those on either side, marked by \( m = \pm 1 \).
The calculation involves finding the angle \( \theta \) for each respective order of the minima. By solving for \( \sin \theta \) using the slit width and wavelength, each minimum angle is identified. This process shows that the entire set of minima and maxima creates the ordered fringe pattern observed on the screen.
These calculations predict the locations on the screen where no light reaches, helping define where the bright fringes end.
Diffraction Pattern Calculation
Calculating a diffraction pattern involves understanding both the constructive and destructive interference of light waves as they pass through a narrow opening. Here, we particularly focus on how to mathematically describe the appearance of bright and dark bands on a screen.
The step-by-step process of calculating the diffraction pattern begins with determining positions for maxima and minima using the formula \( a \sin \theta = m \lambda \). These calculations are essential, as they provide a framework to predict where each fringe will appear on the screen.
Using the concept of small angles, where \( \sin \theta \approx \tan \theta \approx \theta \), positions \( y \) on the screen can be derived: \( y = L \times \sin \theta \). With this information, the locations of bright fringes can be pinpointed, enhancing our understanding of the diffraction pattern.
Ultimately, knowing how to calculate the exact positions of diffraction elements allows anyone observing the experiment to anticipate the order and width of fringes, crucial for studies in optics and wave behavior.

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

\(\bullet\) 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.

\(\bullet\) A thin film of polystyrene of refractive index 1.49 is used as a nonreflecting coating for Fabulite (strontium titanate) of refractive index \(2.409 .\) What is the minimum thickness of the film required? Assume that the wavelength of the light in air is 480 \(\mathrm{nm}\) .

\( (f… # \)\bullet$$\bullet\( Two small loudspeakers that are 5.50 \)\mathrm{m}\( apart are emitting sound in phase. From both of them, you hear a singer singing \)\mathrm{C} \\#\( (frequency 277 \)\mathrm{Hz} )\( , while the speed of sound in the room is 340 \)\mathrm{m} / \mathrm{s}\( . Assuming that you are rather far from these speak- ers, if you start out at point \)P\( equidistant from both of them and walk around the room in front of them, at what angles (measured relative to the fine from \)P$ to the midpoint between the speakers) will you hear the sound (a) maximally enhanced, (b) cancelled? Neglect any reflections from the walls.

\(\bullet$$\bullet\) Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illumi- nated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), includ- ing the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

\(\bullet$$\bullet\) Sensitive eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 \(\mathrm{nm}\) has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

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.