Chapter 11: Problem 2
Call the irradiance at the center of the central Fraunhofer diffraction maximum of a single slit \(I_{0}\) and the irradiance at some other point in the pattern \(I .\) Obtain the ratio \(I / I_{0}\) for a point on the screen that is \(3 / 4\) of a wavelength farther from one edge of the slit than the other.
Short Answer
Step by step solution
Understanding the Fraunhofer Diffraction Pattern
Defining the Scenario
Formula Transformation Using Path Difference
Calculating Intensity Ratio
Final Result
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.
Single Slit Diffraction
The diffraction pattern consists of a central bright fringe called the central maximum, alongside several dimmer fringes, alternately dark and bright. These fringes are due to the interference of waves emerging from different points along the slit.
Overall, each wave travels different distances before converging on a point on the screen. This variation in travel distance is responsible for the interference pattern, including the characteristic central brightness and diminishing intensity side fringes.
Path Difference
At any given point where the waves meet, constructive or destructive interference can occur based on this path difference. Constructive interference happens if the path difference is a whole number of wavelengths, while destructive interference occurs for half-wave multiples. This interference effect produces the distinctive bright and dark bands in a diffraction pattern.
Wavelength
The wavelength, in combination with slit width and wave interference, determines the angles and distances at which bright and dark fringes appear. Hence, knowing the wavelength allows precise predictions and calculations regarding the diffraction pattern, such as the intensity at various points.