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Chapter 35: Problem 1

(I) If 680 -nm light falls on a slit \(0.0365 \mathrm{~mm}\) wide, what is the angular width of the central diffraction peak?

Short Answer

Expert verified
The angular width of the central diffraction peak is approximately 2.128 degrees.

Step by step solution

01

Understanding the Problem

We have a 680 nm wavelength light falling on a slit with width 0.0365 mm, and we need to find the angular width of the central diffraction peak.
02

Converting Units

First, convert the slit width from millimeters to meters for standard SI units. Given slit width is 0.0365 mm. Convert it to meters: \(0.0365\, \text{mm} = 0.0365 \times 10^{-3} \, \text{m} = 3.65 \times 10^{-5} \, \text{m}\).
03

Formula for Angular Width

The angular width \( \theta \) of the central diffraction peak can be found using the formula for single-slit diffraction: \( \lambda = a \sin(\theta) \), where \(\lambda\) is the wavelength of light, \(a\) is the slit width, and \(\theta\) is the angle.
04

Solving for Angular Width

Calculate \(\sin(\theta)\) using the formula for the first minimum of diffraction, which occurs at \(\theta = \arcsin\left(\frac{\lambda}{a}\right)\). The central peak extends from \(-\theta\) to \(\theta\), so its angular width is \(2\theta\).
05

Calculate and Solve

Substitute values into the formula \(\theta = \arcsin\left(\frac{680 \times 10^{-9}}{3.65 \times 10^{-5}}\right)\). Solve for \(\theta\) and then multiply by 2 to get the full angular width of the central peak.

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Key Concepts

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

Diffraction
Diffraction is a phenomenon that occurs when a wave encounters an obstacle or a slit that is comparable in size to its wavelength. In the case of light waves, when they pass through a small slit, they spread out, creating a pattern of light and dark regions, known as a diffraction pattern. This pattern is characterized by a central bright fringe, surrounded by alternating dark and bright fringes on either side.
The spreading or bending of the waves occurs because each point on a wavefront acts as a source of secondary wavelets. As these secondary wavelets overlap and interfere with each other, they produce the observable diffraction pattern.
  • For single-slit diffraction, the width of the slit and the wavelength of the light are crucial in determining the pattern.
  • A narrower slit or a longer wavelength results in a wider diffraction pattern.
Understanding diffraction is essential for analyzing how waves behave in various contexts, such as in optical instruments and technologies.
Wavelength
Wavelength is a key concept in wave physics, representing the distance between two consecutive points that are in phase on a wave, such as crest to crest or trough to trough. In the electromagnetic spectrum, wavelength is inversely proportional to frequency, meaning longer wavelengths correspond to lower frequencies.
Light typically has wavelengths in the range of hundreds of nanometers. For example, in the given problem, the wavelength of the light is 680 nm (nanometers), where 1 nm equals 10-9 meters. This wavelength is part of the visible spectrum, specifically towards the red end of the spectrum.
  • Wavelength determines many properties of the light, including its color when in the visible spectrum.
  • In diffraction, the wavelength is vital in determining the pattern produced after passing through a slit.
The relationship between wavelength and slit width ultimately determines the angles at which destructive interference occurs, shaping the diffraction pattern.
Angular Width
The angular width in the context of single-slit diffraction refers to the range of angles over which the central bright fringe extends. It is an important measurable component of the diffraction pattern. To find it, it's necessary to understand how light behaves as it passes through the slit and interferes constructively and destructively.
The angular width of the central peak is specifically the angle between the first minima, or the points where the intensity of light first drops to zero on either side of the central maximum. This is given by calculating the sine of the angle considering the ratio between the wavelength and the slit width:
  • The calculation involves \( \theta = \arcsin\left(\frac{\lambda}{a}\right) \), where \( \lambda \) is the wavelength, and \( a \) is the slit width.
  • The central peak's angular width is then \( 2\theta \), encompassing the spread both to the left and right of the central maximum.
Understanding angular width is key to examining how light intensity varies with angle and is crucial in applications where precise measurements of light spread are necessary.

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Most popular questions from this chapter

(II) A student shined a laser light onto a single slit of width \(0.04000 \mathrm{~mm} .\) He placed a screen at a distance of \(1.490 \mathrm{~m}\) from the slit to observe the diffraction pattern of the laser light. The accompanying Table shows the distances of the dark fringes from the center of the central bright fringe for different orders. $$ \begin{array}{lccccccc} \hline \text { Order } & & & & & & & & \\ \text { number, } & \text { : }: 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \text { Distance } & & & & & & & & \\ \text { (m) } & 0.0225 & 0.0445 & 0.0655 & 0.0870 & 0.1105 & 0.1320 & 0.1540 & 0.1775 \\ \hline \end{array} $$ Determine the angle of diffraction, \(\theta\), and \(\sin \theta\) for each order. Make a graph of \(\sin \theta\) vs. order number, \(m\), and find the wavelength, \(\lambda\), of the laser from the best-fit straight line.

(II) If you double the width of a single slit, the intensity of the light passing through the slit is doubled. (a) Show, however, that the intensity at the center of the screen increases by a factor of \(4 .\) (b) Explain why this does not violate conservation of energy.

(II) Monochromatic light falls on a transmission diffraction grating at an angle \(\phi\) to the normal. (a) Show that Eq. 13 for diffraction maxima must be replaced by \(d(\sin \phi+\sin \theta)=\pm m \lambda . \quad m=0,1,2, \cdots\) (b) Explain the \(\pm\) sign. (c) Green light with a wave- length of 550 \(\mathrm{nm}\) is incident at an angle of \(15^{\circ}\) to the normal on a diffraction grating with 5000 lines/cm. Find the angles at which the first- order maxima occur. \(\begin{aligned} \sin \theta &=\frac{m \lambda}{d}, \\ m &=0,1,2, \cdots \end{aligned} \quad \left[ \begin{array}{l}{\text { diffraction grating }} \\\ {\text { principal maxima }}\end{array}\right]\)

The Earth and Moon are separated by about \(400 \times 10^{6} \mathrm{~m}\). When Mars is \(8 \times 10^{10} \mathrm{~m}\) from Earth, could a person standing on Mars resolve the Earth and its Moon as two separate objects without a telescope? Assume a pupil diameter of \(5 \mathrm{~mm}\) and \(\lambda=550 \mathrm{nm}\).

(II) Two stars 16 light-years away are barely resolved by a 66-cm (mirror diameter) telescope. How far apart are the stars? Assume \(\lambda=550 \mathrm{nm}\) and that the resolution is limited by diffraction.
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