/*! 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 20 Monochromatic green light, of wa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Monochromatic green light, of wavelength \(550 \mathrm{~nm}\), illuminates two parallel narrow slits \(7.70 \mu \mathrm{m}\) apart. Calculate the angular deviation \((\theta\) in Fig. \(35-10\) ) of the third-order \((m=3)\) bright fringe (a) in radians and (b) in degrees.

Short Answer

Expert verified
(a) \( \theta \approx 0.216 \) radians, (b) \( \theta \approx 12.37 \) degrees.

Step by step solution

01

Understand the Problem

We need to calculate the angular deviation of the third-order bright fringe in a double-slit experiment. The separation between the slits is given as \( d = 7.70 \mu \mathrm{m} \) and the wavelength of the light is \( \lambda = 550 \mathrm{~nm} \).
02

Calculate the Path Difference

The path difference for bright fringes in a double-slit experiment must be a multiple of the wavelength. It can be expressed as \( d \sin \theta = m\lambda \) where \( m \) is the order of the fringe (\( m = 3 \)).
03

Solve for Angular Deviation in Radians

We rearrange the formula to solve for \( \theta \): \[ \sin \theta = \frac{m\lambda}{d} = \frac{3 \times 550 \times 10^{-9} \text{ m}}{7.70 \times 10^{-6} \text{ m}} \]. Calculate \( \sin \theta \) and then use the inverse sine function to find \( \theta \) in radians.
04

Calculate Angular Deviation in Degrees

Convert the angle obtained in radians to degrees using the conversion \( 1 \, \text{rad} = \frac{180}{\pi} \, \text{degrees} \). Use the formula: \( \theta (\text{degrees}) = \theta (\text{radians}) \times \frac{180}{\pi} \).
05

Perform Numerical Calculation

Calculate \( \sin \theta = \frac{3 \times 550 \times 10^{-9}}{7.70 \times 10^{-6}} = 0.2143 \). Using a calculator, find \( \theta \approx 0.216 \) radians. Convert to degrees: \( \theta \approx 0.216 \times \frac{180}{\pi} \approx 12.37 \) degrees.

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
In the double-slit experiment, light is shone upon two closely spaced slits, creating an interference pattern on a screen. This pattern consists of a series of bright and dark fringes, or bands, forming due to the superposition of light waves.
The light waves emanate from the two slits and overlap, leading to regions of constructive and destructive interference.
  • Constructive interference occurs when the waves are in phase, meaning their peaks and troughs align, resulting in bright fringes.
  • Destructive interference happens when the waves are out of phase, causing their amplitudes to cancel each other out, leading to dark fringes.
The spacing and intensity of these fringes depend on the wavelength of light and the separation distance between the slits. This phenomenon beautifully demonstrates the wave nature of light.
Angular deviation
Angular deviation in a double-slit experiment refers to the angle at which a particular order of fringe appears off the central maximum (directly in front of the slits). It describes how far the bright or dark fringes are deviated from this central bright line.
Mathematically, the angle of the bright fringe is determined using the formula: \[ d \sin \theta = m\lambda \] where \( d \) is the distance between the slits, \( \theta \) is the angular deviation, \( \lambda \) is the wavelength, and \( m \) is the fringe order number (0, 1, 2, ...).
This angular deviation gives insight into the positions where constructive interference creates bright fringes. In our exercise, this calculation explains both in radians and degrees how light is diverted from its direct path.
Third-order bright fringe
The term "third-order bright fringe" refers to the third bright band on either side of the central maximum in an interference pattern. This is a result of the waves constructively interfering three wavelengths path difference apart.
For any bright fringe of order \( m \), the condition \( d \sin \theta = m\lambda \) must be satisfied, where \( m = 3 \) for the third-order.
  • The higher the order of the fringe, the greater the path difference and hence the larger the angular deviation.
  • These fringes show how light waves can still add up even after traveling different distances.
Conceptually, each subsequent bright fringe represents light waves that have traveled further relative to the others, maintaining the coherence to form distinct fringe orders.
Wavelength
The wavelength of light is a key factor in forming the interference pattern observed in the double-slit experiment. It signifies the distance between consecutive wave peaks. This length defines how and where the interference fringes will appear.
In this context, a monochromatic wavelength, meaning light of a single color (or frequency), ensures a clear and organized interference pattern.
The slit separation \( d \) and wavelength \( \lambda \) dictate the spacing of the fringes. A larger wavelength relative to slit separation will generally lead to more widely spaced fringes.
  • A typical laser might provide a consistent wavelength, seen in our problem as 550 nm, which aligns with green light.
  • Changes in wavelength will shift the position of bright and dark bands, altering the interference pattern dramatically.
Understanding wavelength helps predict the behavior of light in various interference setups.

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

Two rectangular glass plates \((n=1.60)\) are in contact along one edge and are separated along the opposite edge (Fig. 35-44). Light with a wavelength of \(600 \mathrm{~nm}\) is incident perpendicularly onto the top plate. The air between the plates acts as a thin film. Nine dark fringes and eight bright fringes are observed from above the top plate. If the distance between the two plates along the separated edges is increased by \(600 \mathrm{~nm}\), how many dark fringes will there then be across the top plate?

The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at \(600 \mathrm{~nm}\) and a minimum at \(450 \mathrm{~nm}\), with no minimum in between. If \(n=1.33\) for the film, what is the film thickness, assumed uniform?

-30 Find the sum \(y\) of the following quantities: $$ y_{1}=10 \sin \omega t \text { and } y_{2}=8.0 \sin \left(\omega t+30^{\circ}\right) \text { . } $$

A \(600-\mathrm{nm}\) -thick soap film \((n=1.40)\) in air is illuminated with white light in a direction perpendicular to the film. For how many different wavelengths in the 300 to \(700 \mathrm{~nm}\) range is there (a) fully constructive interference and (b) fully destructive interference in the reflected light?

A thin film of acetone \((n=1.25)\) coats a thick glass plate \((n=1.50)\). White light is incident normal to the film. In the reflections, fully destructive interference occurs at \(600 \mathrm{~nm}\) and fully constructive interference at \(700 \mathrm{~nm}\). Calculate the thickness of the acetone film.

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.