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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

Expert verified
(a) 580 nm; (b) Intensity ratio is 0.1282.

Step by step solution

01

Understand Fraunhofer Diffraction Condition

In Fraunhofer diffraction, the condition for the first minima is given by the formula:\[ a \sin(\theta) = m\lambda \]where \(a\) is the slit width, \(\theta\) is the angle, \(m\) is the minima order (1 for first minima), and \(\lambda\) is the wavelength. For first diffraction minima at \(\theta = \pm 90^{\circ}\), \(m = \pm 1\).
02

Calculate the Width of the Slit

For \(\theta = 90^{\circ}\), we have:\[ a \sin(90^{\circ}) = \lambda \]Since \(\sin(90^{\circ}) = 1\), this simplifies to:\[ a = \lambda \]Substitute \(\lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m}\):\[ a = 580 \times 10^{-9} \text{ m}\]Hence, the width of the slit \(a\) is \(580 \text{ nm}\).
03

Determine Intensity Ratio Formula

In single-slit diffraction, the intensity as a function of angle \(\theta\) is given by:\[ I(\theta) = I_0 \left(\frac{\sin \beta}{\beta}\right)^2 \]where \(\beta = \frac{\pi a \sin\theta}{\lambda}\). The intensity at \(\theta = 0\) is \(I_0\).
04

Calculate \(\beta\) for \(\theta = 45^{\circ}\)

With \(a = 580 \times 10^{-9} \text{ m}\) and \(\lambda = 580 \times 10^{-9} \text{ m}\):\[ \beta = \frac{\pi \times 580 \times 10^{-9} \times \sin(45^{\circ})}{580 \times 10^{-9}} = \pi \times \frac{1}{\sqrt{2}} \]Thus, \(\beta \approx 2.221\).
05

Calculate Intensity Ratio

Substitute \(\beta\) into the intensity formula:\[ I(45^{\circ}) = I_0 \left( \frac{\sin(2.221)}{2.221} \right)^2 \]\(\sin(2.221) \approx 0.795\), so:\[ I(45^{\circ}) = I_0 \left( \frac{0.795}{2.221} \right)^2 \]\[ I(45^{\circ}) = I_0 \times 0.1282 \]Thus, the intensity ratio \( \frac{I(45^{\circ})}{I(0)} = 0.1282 \).

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Key Concepts

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

Monochromatic Light
Monochromatic light is crucial in studying diffraction phenomena. It refers to light of a single wavelength or color, which ensures that the diffraction pattern is as clear and defined as possible. Without the presence of different wavelengths or colors, there's no interference due to variations in path length. This homogeneity is vital in achieving precise observations, as variations can blur the distinct bands of the diffraction pattern.
Using monochromatic light allows researchers to study phenomena like diffraction more accurately. It's commonly used in experiments involving diffraction and interference. This light source is typically produced using lasers or by passing white light through a filter to isolate a single wavelength.
In this exercise, the light used has a wavelength of 580 nm, which ensures that the observed diffraction pattern is due solely to this specific wavelength, simplifying calculations and increasing accuracy.
Wavelength
Wavelength is a fundamental concept in understanding light properties, especially when discussing diffraction. It is the distance between consecutive crests of a wave. For light, this means the distance in which the wave repeats itself.
In this context, the wavelength of monochromatic light used in the experiment is 580 nm, or 580 x 10^-9 meters. This small unit reflects the tiny nature of light waves, which are often measured in nanometers due to their minuscule size.
The wavelength is a key factor in the Fraunhofer diffraction equation. Knowing it allows for the calculation of slit width and positions of minima. Specifically, it helps to predict where dark bands (minima) and bright bands will appear in the diffraction pattern.
Diffraction Minima
Diffraction minima are crucial points in a diffraction pattern. They are the dark bands where wave interference causes a reduction in light intensity, essentially nullifying it at these angles. In Fraunhofer diffraction, the condition for diffraction minima is described by the relation:\[ a \sin(\theta) = m\lambda\]where \(a\) is the slit width, \(\theta\) is the angle, \(m\) indicates the minima order, and \(\lambda\) is the wavelength.
Typically, the first minima occur at \(m=\pm1\), and their positions help in determining the width of the slit used. For this problem, the first diffraction minima appear at \(\theta = \pm90^{\circ}\), leading to the width of the slit being equal to the wavelength (580 nm).
Understanding diffraction minima is essential for calculating slit dimensions and analyzing light behavior in experiments concerning wave interference.
Intensity Ratio
The intensity ratio in diffraction patterns explores how light's brightness varies across different angles. It is specifically the comparison of light intensity at a certain angle with that at the center of the diffraction pattern. Calculating this ratio involves understanding how light interferes as it passes through a slit and diffracts.
For a single-slit diffraction, light intensity at any angle \(\theta\) is given by:\[ I(\theta) = I_0 \left(\frac{\sin \beta}{\beta}\right)^2\]where \(\beta\) links angle, slit width, and wavelength by:\[ \beta = \frac{\pi a \sin\theta}{\lambda}\]At \(\theta = 0\), \(\beta = 0\), and the brightness \(I(0) = I_0\), which is the maximum. Comparing intensity at \(\theta = 45^{\circ}\) revealed an intensity ratio of approximately 0.1282, indicating the brightness at this angle is significantly less than at the center. This highlights the effect of diffraction and wave interference in altering perceived light intensity.

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Most popular questions from this chapter

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 \(\mathrm{mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 \(\mathrm{W} / \mathrm{m}^{2} .\) (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits/cm? (b) What is the second order deviation of this wavelength? Assume normal incidence.

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Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm} .\) The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is 36.5 \(\mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

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