/*! 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 55 35.55. A source \(S\) of monochr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 35: Problem 55

35.55. A source \(S\) of monochrounatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass, and are separated by a horizontal distance \(x\) . Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

Short Answer

Expert verified
The longest wavelength for constructive interference is 720 nm.

Step by step solution

01

Understanding Path Difference

The path difference between the light directly from source \(S\) to detector \(D\) and the light reflected off the glass is expressed as \( \sqrt{x^2 + 4h^2} - x \). This expression results from the geometry of the setup, where \( \sqrt{x^2 + 4h^2} \) is the hypotenuse of the triangle formed by \( x \) and \( 2h \).
02

Condition for Constructive Interference

Constructive interference occurs when the path difference is equal to an integer multiple of the wavelength plus half the wavelength due to the \( \pi \) phase shift upon reflection. Thus, the condition for constructive interference is given by: \[ \sqrt{x^2 + 4h^2} - x = \left(m + \frac{1}{2}\right) \lambda \] where \( m \) is an integer.
03

Condition for Destructive Interference

Destructive interference occurs when the path difference is an integer multiple of the wavelength without additional phase factors. Hence, the condition for destructive interference is: \[ \sqrt{x^2 + 4h^2} - x = m \lambda \] where \( m \) is an integer.
04

Substitute Values for Constructive Interference

To find the longest wavelength for which there is constructive interference, substitute \(h = 24\) cm and \(x = 14\) cm into the constructive interference condition:\[ \sqrt{14^2 + 4(24)^2} - 14 = \left(m + \frac{1}{2}\right) \lambda \] Simplifying the left side gives: \[ \sqrt{196 + 2304} - 14 = \left(m + \frac{1}{2}\right) \lambda \] \[ \sqrt{2500} - 14 = \left(m + \frac{1}{2}\right) \lambda \] \[ 50 - 14 = \left(m + \frac{1}{2}\right) \lambda \] \[ 36 = \left(m + \frac{1}{2}\right) \lambda \]
05

Solve for Longest Wavelength

We need the longest wavelength, which corresponds to the smallest value of \( m \). For this, set \( m = 0 \): \[ 36 = \left(0 + \frac{1}{2}\right) \lambda \] \[ 36 = \frac{1}{2} \lambda \] Solving for \( \lambda \):\[ \lambda = 72 \text{ cm} \] or 720 nm once converted to nanometers.

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 91影视!

Key Concepts

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

Constructive Interference
Constructive interference is a beautiful phenomenon observed when two light waves overlap in such a way that their peaks (or crests) align perfectly. This alignment maximizes the combined amplitude of the waves, creating brighter light. In the context of our exercise, constructive interference occurs when the path difference between a direct light wave and one reflecting off a surface satisfies a specific condition involving the wavelength.
The formula given is:
  • Path Difference = \( \sqrt{x^2 + 4h^2} - x \)
  • Condition for Constructive Interference = \( \left(m + \frac{1}{2}\right) \lambda \)
To understand why, remember that a half-wavelength phase shift happens when light reflects off a medium with a higher refractive index, such as from air to glass. This phase shift introduces a condition which is why we add half a wavelength (\(\frac{1}{2}\lambda \)).
For achieving constructive interference, this path difference should equal an integer multiple plus half the wavelength, where \(m\) is any whole number (0, 1, 2, etc.). When these conditions are met, the waves amplify one another, resulting in a stronger signal observed at the detector.
Destructive Interference
Destructive interference is the opposite of constructive interference. Here, the overlapping waves have peaks and troughs that do not match, causing them to cancel each other out, resulting in dimmer or no light at all. In terms of our exercise, destructive interference is described by a condition where the path difference matches exactly the wavelength.
The formula used in this scenario is:
  • Path Difference = \( \sqrt{x^2 + 4h^2} - x \)
  • Condition for Destructive Interference = \( m \lambda \)
Here, the integer \(m\) adjusts so that exact multiples of wavelengths are considered. When the condition is satisfied, the crest of one wave nullifies the trough of another, leading to a minimal or zero amplitude at the detector.
It's fascinating how these conditions live up to the wave nature of light, manipulating it simply by adjusting path lengths and reflecting off surfaces.
Phase Change on Reflection
Phase change on reflection is a crucial concept that affects how interference patterns occur. In our setup, the light wave reflecting from the glass surface undergoes a phase change of half a wavelength (\(\pi\) radians). This shift can significantly alter whether waves amplify (constructively interfere) or cancel out (destructively interfere).
The concept is simple: - When light reflects off a medium with a higher refractive index (like air to glass), a phase shift of \(\pi\) occurs. - This phase change effectively adds \(\frac{1}{2} \lambda\) to the path difference.
When considering interference, correctly accounting for this phase change is essential. The constructive interference condition adjusts to include \(\frac{1}{2}\lambda\), while the destructive condition remains unchanged because it naturally fits a whole wavelength multiple. Understanding these shifts allows us to predict and manipulate patterns like those in the exercise.

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

35.32. A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraetion 1.52 . (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manu- facture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

35.22. GPSTransmission. The GPS (Global Positioning System) satellites are approximately 5.18 \(\mathrm{m}\) across and transmit two low-power signals, one of which is at 1575.42 \(\mathrm{MHz}\) (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42-MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (a) At how many other angles in the range \(0^{\circ}<\theta<90^{\circ}\) is the intensity also 2.00 \(\mathrm{W} / \mathrm{m}^{2} ?\) (b) Find the four smallest angles in the range \(0^{\circ}<\theta<90^{\circ}\) for which the intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (c) What is the intensity at a point on the circle at an angle of \(4.65^{\circ}\) from the centerline?

35.17. Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated 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), including 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?

35.1. Two coherent sources \(A\) and \(B\) of radio waves are 5.00 \(\mathrm{m}\) apart. Each source emits waves with wavelength 6.00 \(\mathrm{m}\) . Consider points along the line between the two sources. At what distances, if any, from \(A\) is the interference (a) constructive and (b) destructive?

35.2. Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q\) ?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Classical Mechanics

Read Explanation

Physics of Motion

Read Explanation

Particle Model of Matter

Read Explanation

Space Physics

Read Explanation

Oscillations

Read Explanation
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.