/*! 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 19 Coherent light with wavelength 5... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Coherent light with wavelength 500 nm passes through narrow slits separated by 0.340 mm. At a distance from the slits large compared to their separation, what is the phase difference (in radians) in the light from the two slits at an angle of 23.0\(^\circ\) from the centerline?

Short Answer

Expert verified
The phase difference is approximately 4.22 radians.

Step by step solution

01

Convert Units

First, convert the given measurements into consistent units. The wavelength is given as 500 nm, which is equal to 500 x 10^{-9} meters. The slit separation is 0.340 mm, which is 0.340 x 10^{-3} meters.
02

Calculate Path Difference

The path difference between the two waves coming from the slits at angle \( \theta \) is given by \( d \sin \theta \), where \( d = 0.340 \times 10^{-3} \) meters and \( \theta = 23.0^\circ \). Calculate \( \sin \theta \):\[ \sin 23.0^\circ \approx 0.3907 \]So, the path difference, \( \Delta x = 0.340 \times 10^{-3} \times 0.3907 \approx 1.32838 \times 10^{-4} \) meters.
03

Calculate Phase Difference

The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) and the wavelength \( \lambda \) by\[ \Delta \phi = \frac{2 \pi}{\lambda} \Delta x \]Substitute the known values:\[ \Delta \phi = \frac{2 \pi}{500 \times 10^{-9}} \times 1.32838 \times 10^{-4} \approx \frac{2 \pi \times 1.32838 \times 10^{-4}}{500 \times 10^{-9}} \]Calculate the value:\[ \Delta \phi \approx 1670.7 \text{ radians} \]
04

Simplify the Phase Difference

Since the phase difference is typically considered within the range of 0 to \( 2\pi \), we take the modulus with \( 2\pi \):\[ \Delta \phi \approx 1670.7 \mod 2\pi \]This process results in a phase difference of approximately \( 4.22 \) radians, as we essentially want the equivalent angle within a single full cycle.

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.

Coherent Light
Coherent light is a fascinating phenomenon in the world of optics, essential for understanding interference patterns. Coherent light means that the light waves traveling together have a constant phase difference and the same frequency. This is important because it allows for constructive and destructive interference to take place. In simple terms, coherent light waves "line up" in phase, allowing them to "add" together predictably.
When two coherent light sources are used, like in the classic double-slit experiment, they generate an interference pattern of light and dark bands on a screen. This pattern occurs because at some points, the peaks of the light waves coincide (constructive interference), making those areas brighter. At other points, the peaks and troughs cancel out (destructive interference), resulting in dark areas.
• **Coherent Sources**: Two light sources are coherent if they emit waves that maintain a constant phase difference.
• **Importance in Interference**: Coherent sources are crucial for producing clear and stable interference patterns, as seen in many scientific and industrial applications, including holography and laser technology.
Phase Difference
Phase difference is a key concept when it comes to understanding light interference. It refers to the amount by which one wave is "ahead" or "behind" another wave measured in radians or degrees. This phase difference determines how waves interact with one another when they overlap.
In interference, the phase difference can dictate whether waves will interfere constructively or destructively. Constructive interference occurs when the phase difference is a multiple of 2π, making the waves align perfectly to increase their amplitude. On the other hand, destructive interference happens when the phase difference is an odd multiple of π, causing the waves to cancel each other out.
• **Calculating Phase Difference**: Phase difference \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2 \pi}{\lambda} \Delta x \]
where \( \Delta x \) is the path difference and \( \lambda \) is the wavelength.
• **Significance**: Determining the phase difference helps in predicting the resulting light pattern from two or more overlapping light waves, critical in experiments and technologies utilizing wave interference.
Wavelength
The concept of wavelength is fundamental in the study of light. Wavelength is the distance between two successive peaks (or troughs) of a wave. In the context of light, the wavelength determines the color we observe.
In applications like the interference pattern generated from slits, knowing the wavelength is crucial because it directly affects the pattern's spacing. Longer wavelengths (e.g., red light) have spaced out interference fringes compared to shorter wavelengths (e.g., blue light).
• **Units of Wavelength**: Wavelength is usually measured in nanometers (nm) for light waves. For example, visible light ranges approximately from 400 nm to 700 nm.
• **Relation with Frequency**: Wavelength is inversely related to frequency. This means that light waves with shorter wavelengths have higher frequencies and vice versa.
• **In Interference**: The wavelength used in calculating path and phase differences helps to predict where the light and dark fringes will appear in an interference pattern. Understanding wavelength is thus key to analyzing and interpreting any experiment involving light interference.

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

Young's experiment is performed with light from excited helium atoms (\(\lambda\) = 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference \(minima\) at \(\pm35.20^\circ\) 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 \(\pm19.46^\circ\) instead. What is the index of refraction of this liquid?

Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 cm. Point \(P\) is 4.86 m from \(A\) and 5.24 m from \(B\). What is the phase difference at \(P\) between these two waves?

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50. At room temperature (20.0\(^\circ\)C), this film is just thick enough for light with wavelength 582.4 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\(^\circ\)C, you find that the film cancels reflected light with wavelength 588.5 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.)

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 nm) of an incident white light beam in air can take place by reflection?

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.