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A single slit has a width of \(2.1 \times 10^{-6} \mathrm{~m}\) and is used to form a diffraction pattern. Find the angle that locates the second dark fringe when the wavelength of the light is (a) 430 \(\mathrm{nm}\) and (b) \(660 \mathrm{~nm}\).

Short Answer

Expert verified
For \( \lambda = 430 \mathrm{nm} \), \( \theta \approx 24.2^\circ \); for \( \lambda = 660 \mathrm{nm} \), \( \theta \approx 38.9^\circ \).

Step by step solution

01

Understanding the Problem

In a single-slit diffraction pattern, dark fringes occur at angles \( \theta \) where the path difference causes destructive interference. The condition for dark fringes is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( m \) is the order of the fringe, and \( \lambda \) is the wavelength. We need to find \( \theta \) for the second dark fringe \( m = 2 \).
02

Applying the formula for \( \lambda = 430 \mathrm{nm} \)

For the second dark fringe \( m = 2 \) and \( \lambda = 430 \mathrm{nm} = 430 \times 10^{-9} \mathrm{~m} \). Substitute these values into the formula: \( a \sin \theta = m \lambda \). This gives \( 2.1 \times 10^{-6} \sin \theta = 2 \times 430 \times 10^{-9} \).
03

Solving for \( \sin \theta \) with \( \lambda = 430 \mathrm{nm} \)

Rearrange the equation to find \( \sin \theta \): \( \sin \theta = \frac{2 \times 430 \times 10^{-9}}{2.1 \times 10^{-6}} \). Calculate this to find \( \sin \theta \approx 0.4095 \).
04

Finding \( \theta \) for \( \lambda = 430 \mathrm{nm} \)

Now find \( \theta \) by taking the inverse sine (arc sine) of 0.4095: \( \theta = \arcsin(0.4095) \), which yields \( \theta \approx 24.2^\circ \).
05

Applying the formula for \( \lambda = 660 \mathrm{nm} \)

Use the same initial setup: \( a \sin \theta = m \lambda \) for \( m = 2 \) and \( \lambda = 660 \mathrm{nm} = 660 \times 10^{-9} \mathrm{~m} \). This becomes \( 2.1 \times 10^{-6} \sin \theta = 2 \times 660 \times 10^{-9} \).
06

Solving for \( \sin \theta \) with \( \lambda = 660 \mathrm{nm} \)

Rearrange to find \( \sin \theta \): \( \sin \theta = \frac{2 \times 660 \times 10^{-9}}{2.1 \times 10^{-6}} \). Calculate this to get \( \sin \theta \approx 0.6286 \).
07

Finding \( \theta \) for \( \lambda = 660 \mathrm{nm} \)

Finally, find \( \theta \) using the inverse sine: \( \theta = \arcsin(0.6286) \), resulting in \( \theta \approx 38.9^\circ \).

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Key Concepts

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

Diffraction Pattern
When light passes through a single narrow slit, it spreads out and creates a series of bright and dark areas on a screen placed behind the slit. This spread is known as a diffraction pattern. In this pattern, the bright areas are called bright fringes, and they form where waves constructively interfere. The dark areas, or dark fringes, occur where light waves cancel each other out through destructive interference. The central bright fringe is the most intense, and as you move away from it, the brightness of the fringes decreases.

Diffraction patterns are visual proof of the wave nature of light, showing how light bends around obstacles. The pattern results because light acts as a wave, and when waves pass through small openings, like a slit, they spread out on the other side, creating interference patterns based on the wave interaction.
Wavelength
A crucial factor in determining how a diffraction pattern looks is the wavelength of the light used. Wavelength is the distance between consecutive crests of a wave. Different colors of light have different wavelengths. For example, blue light has a shorter wavelength (~430 nm) compared to red light (~660 nm).

The wavelength of light affects the diffraction pattern by influencing the spacing of the fringes. Longer wavelengths, such as red light, have larger spacing between the fringes compared to shorter wavelengths like blue light, which creates closer fringes. When you increase the wavelength, the angle at which the dark fringes appear also increases, resulting in a more spread-out diffraction pattern.
Destructive Interference
Destructive interference occurs when two light waves meet in such a way that their crests align with the troughs of the other wave. This causes the waves to cancel each other out, leading to minimal or no light intensity at those points. In a diffraction pattern, destructive interference is responsible for the dark fringes.

The mathematical condition for destructive interference, specifically in a single-slit diffraction, is given by the equation \( a \sin \theta = m \lambda \), where:
  • \( a \) represents the width of the slit
  • \( \theta \) is the angle relative to the original direction of the light
  • \( m \) is the order number of the dark fringe (e.g., 1 for the first, 2 for the second, etc.)
  • \( \lambda \) is the wavelength of the incident light
By using this condition, one can precisely determine where dark spots will appear in the diffraction pattern.
Dark Fringes
Dark fringes in a diffraction pattern are the locations where light waves completely cancel each other out due to destructive interference. These dark bands are evenly spaced, making it possible to predict their exact position.

The formula \( a \sin \theta = m \lambda \) helps calculate the angle \( \theta \) where these dark regions appear. The integer \( m \) signifies the order of the fringe, which are the series of bands formed on either side of the central bright fringe:
  • For the first dark fringe (\( m = 1 \)), position is closest to the bright central fringe.
  • For the second dark fringe (\( m = 2 \)), farther from the center.
For example, when using a wavelength of 430 nm and 660 nm, the angles for the second dark fringe differ due to the wavelength difference. Using the slit width of 2.1 x 10-6 m, calculations show these angles as approximately 24.2° and 38.9°, respectively.

Understanding dark fringes is essential to mastering how light behaves when encountering small apertures, which is fundamental in optics and various practical and scientific applications.

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Most popular questions from this chapter

Concept Questions (a) In a single-slit diffraction pattern the width of the central bright fringe is defined by the location of the first dark fringe that lies on either side of it. For a given slit width, does the width of the central bright fringe increase, decrease, or remain the same as the wavelength of the light increases? (b) For a given wavelength, does the width of the central bright fringe increase, decrease, or remain the same as the slit width increases? (c) When both the wavelength and the slit width change, it is possible for the width of the central bright fringe to remain the same. What condition must be satisfied for this to happen? In each case, give your reasoning. Problem A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{~m}\). When light with a wavelength of \(\lambda_{1}=510\) tim passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ) and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\).

(a) Two diffraction gratings are located at the same distance from observation screens. Light with the same wavelength \(\lambda\) is used for each. The principal maxima of grating \(\mathrm{A}\) are observed to be closer together on the screen than the principal maxima of grating \(\mathrm{B}\). Which grating diffracts the light to a greater extent? (b) Which grating has the smaller slit separation \(d ?(\mathrm{c})\) Which grating has the greater number of lines permeter? Justify each of your answers. The separation between adjacent principal maxima for grating \(\mathrm{A}\) is \(2.7 \mathrm{~cm}\), and for grating \(\mathrm{B}\) it is \(3.2 \mathrm{~cm}\). Grating \(\mathrm{A}\) has 2000 lines per meter. How many lines per meter does grating B have? The diffraction angles are small enough that \(\sin \theta \approx \tan \theta .\) Be sure that your answer is consistent with your answers to the Concept Questions.

Light that has a wavelength of \(668 \mathrm{~nm}\) passes through a slit \(6.73 \times 10^{-6} \mathrm{~m}\) wide and falls on a screen that is \(1.85 \mathrm{~m}\) away. What is the distance on the screen from the center of the central bright fringe to the third dark fringe on either side?

A mix of red light \(\left(\lambda_{\text {vacuum }}=661 \mathrm{nm}\right)\) and green light \(\left(\lambda_{\text {vacuum }}=551 \mathrm{nm}\right)\) is directed perpendicularly onto a soap film \((n=1.33)\) that has air on either side. What is the minimum nonzero thickness of the film, so that destructive interference causes it to look red in reflected light?

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is \(675 \mathrm{~nm}\). Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{~m}\) and (b) \(1.8 \times 10^{-6} \mathrm{~m}\).

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