/*! 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 17 A single-slit diffraction patter... [FREE SOLUTION] | 91Ó°ÊÓ

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A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 \(\mathrm{mm}\) wide. At the point in the pattern \(3.25^{\circ}\) from the center of the central maximum, the total phase difference between wavelets from the tup and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_{0} ?\)

Short Answer

Expert verified
(a) The wavelength is 667 nm. (b) The intensity is 1.9 x 10^{-4} I_0.

Step by step solution

01

Convert Angle to Radians

Given that the angle is \(3.25^{\circ}\), we first need to convert it from degrees to radians for further calculations, as the phase difference is given in radians. The conversion formula is \(\theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180}\). So:\[\theta = 3.25 \times \frac{\pi}{180} \approx 0.0567 \text{ radians}\]
02

Using the Phase Difference Formula

The phase difference between the top and bottom of the slit in a single-slit diffraction pattern is given by:\[ \Delta \varphi = \frac{2\pi}{\lambda} \cdot a \cdot \sin(\theta) \]where \(\Delta \varphi = 56.0\) rad, \(a = 0.105\) mm, and \(\sin(\theta)\) is the sine of the converted angle. We can solve for the wavelength \(\lambda\):\[ \lambda = \frac{2\pi \cdot a \cdot \sin(\theta)}{\Delta \varphi} \approx \frac{2\pi \cdot 0.105 \cdot \sin(0.0567)}{56.0} \]
03

Calculate the Wavelength

Calculate \(\sin(\theta)\) using the angle in radians from Step 1:\[ \sin(0.0567) \approx 0.0567 \]Now plug this back into the wavelength formula from Step 2:\[ \lambda \approx \frac{2\pi \cdot 0.105 \cdot 0.0567}{56.0} \approx 6.67 \times 10^{-7} \text{ m} \approx 667 \text{ nm} \]
04

Determine the Intensity at the Given Point

Intensity in a single-slit diffraction pattern can be described by:\[ I = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \]where \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \). Calculate \(\beta\) using the calculated \(\lambda\) from Step 3:\[ \beta = \frac{\pi \cdot 0.105 \cdot 0.0567}{667 \times 10^{-9}} \approx 27.87 \]Thus, the intensity ratio \((\sin(\beta) / \beta)^2\) needs to be computed:\[ \left(\frac{\sin(27.87)}{27.87}\right)^2 \approx \left(\frac{-0.121}{27.87}\right)^2 \approx 1.9 \times 10^{-4} \]The intensity at this point is:\[ I \approx I_0 \times 1.9 \times 10^{-4} \]
05

Express the Final Answers

For part (a), the wavelength \(\lambda\) is approximately \(667\text{ nm}\). For part (b), the intensity at the point is approximately \(1.9 \times 10^{-4} I_0\).

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Key Concepts

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

Monochromatic Electromagnetic Radiation
Monochromatic electromagnetic radiation refers to light that consists of waves with only one specific wavelength. In simpler terms, all the light waves are uniform in color and frequency. As a result, this type of light is ideal for experiments involving interference and diffraction.
This uniformity enables clear patterns, allowing us to study phenomena like diffraction through narrow slits. For this exercise, the consistency in the light's wavelength is crucial to trailing and measuring parameters such as wavelength and intensity under defined conditions.
Examples:
  • Lasers - produce almost purely monochromatic light.
  • Sodium vapor lamps - often used in street lighting, also known for producing almost monochromatic light.
In our exercise, the monochromatic nature of the light allows for precise calculations using formulas that assume no wavelength variation. This ensures accurate diffraction pattern analysis, aiding in the determination of various optical properties, such as wavelength and intensity.
Phase Difference
The phase difference in optics is a concept that helps us understand how waves, originating from different points, interact with each other. This interaction can lead to either constructive or destructive interference. In the context of single-slit diffraction, it is the difference in phase between the wavelets from the top and bottom of the slit.
When using single-slit diffraction, each part of the slit is considered a source of wavelets that interfere with one another at a given point on a screen. The phase difference ()d) can be calculated using the formula:\[∆\varphi = \frac{2\pi}{\lambda} \cdot a \cdot \sin(\theta)\]where:
  • \(\lambda\) is the wavelength of the light.
  • \(a\) is the slit width.
  • \(\theta\) is the angle relative to the central maximum.
In the exercise, this phase difference is given as 56.0 radians. This value provides critical information that allows us to solve for the wavelength and understand the structure of the interference pattern. Understanding phase difference is key to realizing how these interference patterns form and how they can change with different setups and inputs.
Intensity Calculation
The concept of intensity in optics is crucial when analyzing diffraction patterns. Intensity relates to the brightness of the light at a given point on a screen. In single-slit diffraction, intensity varies, creating bright and dark regions.The formula used to determine intensity at a point in the diffraction pattern is:\[I = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2\]where:
  • ) is the primary intensity at the central maximum.
  • \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\)
This equation tells us how light spreads and dims as we move away from the center of the diffraction pattern. In the given example, the calculated intensity is approximately 0.019% of the initial intensity \(I_0\). This dramatic decrease in intensity is typical for points away from the central maximum.
Intensity calculations provide insight into how light energy is distributed across a diffraction pattern, which has applications in various fields, including optics and photonics.
Wavelength Determination
Determining the wavelength of electromagnetic radiation can be done through careful analysis of diffraction patterns. In the exercise, where a single-slit diffraction setup is used, the wavelength of the radiation is crucial. Calculations are necessary to relate the observed phase differences and intensities.To find the wavelength \(\lambda\), the formula used is:\[\lambda = \frac{2\pi \cdot a \cdot \sin(\theta)}{\Delta\varphi}\]This relationship relies on accurately measuring factors such as:
  • The slit width \(a\)
  • The angle \(\theta\)
  • The phase difference \(\Delta\varphi\)
In the example, after computations, the wavelength was found to be approximately 667 nm. This value falls into the visible spectrum and helps explain the color and characteristics of the light being analyzed in the experiment.
The process of determining wavelength is fundamental in optics and is key in applications requiring precision measurements.

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Most popular questions from this chapter

Laser light of wavelength 500.0 \(\mathrm{nm}\) illuminates two identical slits, producing an interference pattern on a screen 90.0 \(\mathrm{cm}\) from the slits. The bright bands are 1.00 \(\mathrm{cm}\) apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

If the planes of a crystal are 3.50 \(\mathrm{A}\left(\mathrm{i} \mathrm{A}=10^{-10} \mathrm{m}=\right.\) 1 Angstrom unit) apart, (a) what wavelength of electromagnetic waves is necded so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of \(15.0^{\circ}\) , and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

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Measuring Wavelengths with a CD. A laser beasm of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pits in which information is coded onto the CD are 1.60\(\mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum?

The Very Long Baseline Array can resolve (by Rayleigh's criterion) signals from sources separated by \(1.0 \times 10^{-8} \mathrm{rad} .\) If the effective diameter of the receiver is 8000 \(\mathrm{km}\) , what is the wavelength of these signals?

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