Chapter 36: Problem 13
Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm\)90.0\(^\circ\), so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta\) = 45.0\(^\circ\) to the intensity at \(\theta\) = 0?
Short Answer
Step by step solution
Understanding the Problem
Using Formula for First Minima in Single-Slit Diffraction
Calculating the Slit Width
Apply Intensity Formula for Single-Slit Diffraction
Calculate Intensity at \( \theta = 45.0° \)
Calculate Intensity Ratio
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Single-slit diffraction
Single-slit diffraction is a key example of Fraunhofer diffraction. This type of diffraction occurs when both the light source and the observation screen are far enough from the slit. In such cases, the light waves can be treated as parallel rays. The central bright area on the screen is flanked by a series of dark and bright fringes.
These patterns arise because parts of the light waves overlap and interfere. Where they overlap constructively, we see brightness. Where they interfere destructively, darkness appears. Understanding single-slit diffraction is fundamental in optics and helps in explaining various concepts like the resolving power of optical instruments.
Diffraction pattern
The position of these dark and bright fringes depends on the wavelength of the light and the width of the slit. For instance, the condition for the first minima occurs when the path difference between light from the edges of the slit is exactly one wavelength. This is described by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle from the center of the pattern, \( m \) is the order of the minima, and \( \lambda \) is the wavelength. This equation helps determine the angles where darkness appears, thus mapping out the diffraction pattern.
Wavelength of light
For light of wavelength 580 nm, as given in the example, the diffraction pattern's features—such as how the light and dark bands (minima and maxima) are spaced—are also dictated by this wavelength. A longer wavelength results in a more spread-out pattern, with bands further apart. A shorter wavelength causes a more compressed pattern.
Understanding the wavelength's role helps in predicting and analyzing optical phenomena. It also aids in designing experiments and interpreting the results. Wavelength is not only crucial for visible light but also for understanding various applications ranging from telecommunications to spectroscopy.
Intensity distribution
For a single slit, the intensity at any angle \( \theta \) is given by the formula: \( I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \), where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( I_0 \) is the maximum intensity at \( \theta = 0 \). This means that the light is brightest at the center and gradually diminishes as you move away, forming the distinct pattern of peaks and troughs.
At specific angles such as 45 degrees, the formula allows us to calculate the exact intensity relative to the central maximum. For example, in the provided exercise, at \( \theta = 45.0^\circ \), the intensity is about 40.5% of the center. These calculations are crucial for applications that rely on precise control and manipulation of light, such as optical engineering and beam shaping.