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Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

Short Answer

Expert verified
The intensity at \( \theta = 1.20^\circ \) is approximately \( 1.33 \times 10^{-8} \text{ W/m}^2 \).

Step by step solution

01

Understand the Diffraction Equation

The intensity of light in a single-slit diffraction pattern is given by the equation: \[ I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]where \( I_0 \) is the intensity at the center, \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), and \( a \) is the slit width. We are tasked with finding \( I(1.20^\circ) \).
02

Convert Units and Calculate \( \beta \)

Convert wavelength \( \lambda = 592 \text{ nm} = 592 \times 10^{-9} \text{ m} \) and slit width \( a = 0.029 \text{ mm} = 0.029 \times 10^{-3} \text{ m} \). For \( \theta = 1.20^\circ \), \( \sin(\theta) \approx \sin(1.20^\circ) = 0.0209 \). Calculate:\[ \beta = \frac{\pi \times 0.029 \times 10^{-3} \times 0.0209}{592 \times 10^{-9}} \]
03

Simplify and Insert Values

Calculate \( \beta \) using the given values:\[ \beta = \frac{\pi \times 0.029 \times 10^{-3} \times 0.0209}{592 \times 10^{-9}} \approx 3.20 \]This will be used to find the new intensity at \( \theta = 1.20^\circ \).
04

Compute the Intensity Ratio

Using the intensity equation:\[ \frac{I(\theta)}{I_0} = \left( \frac{\sin(3.20)}{3.20} \right)^2 \]Calculate \( \sin(3.20) \) and divide by \( 3.20 \). Then square the result to find the intensity ratio.
05

Determine the Intensity at \( 1.20^\circ \)

Calculate \( \sin(3.20) \approx -0.0584 \) (note the negative value, but the square is taken):\[ \frac{I(1.20^\circ)}{4.00 \times 10^{-5}} = \left( \frac{-0.0584}{3.20} \right)^2 \approx \left( -0.01825 \right)^2 \approx 0.0003335 \]Therefore, \( I(1.20^\circ) = 4.00 \times 10^{-5} \times 0.0003335 \approx 1.33 \times 10^{-8} \text{ W/m}^2 \).

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Key Concepts

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

Diffraction Pattern
When light encounters an obstacle, such as a slit, it tends to bend and spread out. This is known as diffraction. In a single-slit diffraction setup, when monochromatic light passes through a narrow slit, it produces a characteristic pattern of alternating bright and dark bands on a screen, known as a diffraction pattern.
This pattern results from the constructive and destructive interference of light waves. The central bright band, called the central maximum, is the brightest part of the pattern. Flanking it are alternating bands of light (maxima) and darkness (minima), whose intensity gradually decreases with distance from the center.
Intensity Calculation
The intensity of light in a diffraction pattern is not uniform. It varies based on the angle \(\theta\)\, which measures the position on the screen relative to the central maximum. The intensity at any point in the pattern can be calculated using:
  • \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2\)
Where \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\)\, \(I_0\) is the intensity at the central maximum, and \(\lambda\) is the wavelength. This equation accounts for the wave nature of light and demonstrates how intensity diminishes as one moves away from the central axis. Calculating \(\beta\) provides insight into intensity changes due to different angles and slit dimensions.
Monochromatic Light
Monochromatic light is light consisting of a single wavelength and frequency. This type of light is crucial for diffraction experiments because it ensures clear and distinguishable patterns.
In the given exercise, the light used has a wavelength of 592 nm. As all photons have the same energy and wavelength, the pattern produced is coherent and precise, aiding in accurate determination of diffraction characteristics such as maxima and minima positions.
Wavelength
The wavelength \(\lambda\) of light is the distance between consecutive peaks (or troughs) in a wave. It’s a vital parameter in understanding many optical phenomena, including diffraction. The wavelength determines the scale of the diffraction pattern.
  • Longer wavelengths create more spread-out patterns,
  • Shorter wavelengths yield narrower bands.
In our context, the light's wavelength is 592 nm, translating to 592 \(\times\) 10^{-9}\ m. This small scale emphasizes the microscopic nature of diffraction patterns.
Slit Width
Slit width is the width (denoted as \(a\)) of the opening through which light passes. It significantly influences the diffraction pattern's appearance. As the width changes, so does the interference pattern's distribution:
  • Narrower slits cause more pronounced diffraction, leading to wider separation between maxima.
  • Wider slits result in less diffraction and narrower spacing of the bands.
In this exercise, the slit width is 0.0290 mm, which defines the initial condition for the diffraction phenomena observed.

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Most popular questions from this chapter

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longerwavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

Light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. What is the distance between the two dark fringes on either side of the central bright fringe?

Monochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4 \(\mu\)m. The resulting diffraction pattern is observed on a screen that is 4.5 m from the aperture. What is the diameter of the Airy disk on the screen?

Monochromatic light of wavelength \(\lambda\) = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity \(I_0\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) 1.00 mm; (b) 3.00 mm; (c) 5.00 mm?

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at \(\pm\)17.8\(^\circ\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

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