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Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.15 \(\times 10^{-5} \mathrm{m}\) and strike a screen 1.20 \(\mathrm{m}\) from the slit. Two diffraction patterns are formed on the screen. What is the distance (in \(\mathrm{cm}\) ) between the common center of the diffraction patterns and the first occurrence of \(\mathrm{a}\) dark fringe from one pattern falling on top of a dark fringe from the other pattern?

Short Answer

Expert verified
The distance is approximately 10.6 cm.

Step by step solution

01

Understanding the Problem

We are given two light waves with wavelengths 632 nm and 474 nm passing through a single slit of width \( w = 7.15 \times 10^{-5} \text{ m} \). The screen is placed at a distance \( L = 1.20 \text{ m} \) from the slit. We need to find the distance on the screen where the first dark fringe of one pattern coincides with the dark fringe of the other pattern.
02

Concept of Diffraction Minima

For a single slit, the condition for minima (dark fringes) is given by: \[ a \sin(\theta) = m \lambda \] where \( a \) is the slit width, \( \theta \) is the angle, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum (an integer).
03

Calculating Position of Dark Fringes

The position of the dark fringe on the screen is given by: \[ y_m = \frac{m \lambda L}{a} \] We will use this formula to calculate the position of the dark fringes for both wavelengths.
04

Finding Common Dark Fringe Position

To find the position where dark fringes coincide for both wavelengths, we set up the equation: \[ m_{632} \times 632 \text{ nm} = m_{474} \times 474 \text{ nm} \] where \( m_{632} \) and \( m_{474} \) are integers representing the order of the minima for the respective wavelengths.
05

Calculate Common Integer Multiples

We need the least common multiple (LCM) of 632 and 474 as the two minima must have \( m \cdot \lambda \) equal. The LCM is 6320, meaning \( m_{632} = 10 \) and \( m_{474} = 13 \).
06

Determine the Position Distance

Using \( m_{632} = 10 \), calculate the position using the equation: \[ y_{10} = \frac{10 \times 632 \times 10^{-9} \times 1.20}{7.15 \times 10^{-5}} \] Calculate this value and convert it to cm.
07

Conversion and Final Result

Calculate the value: \[ y_{10} \approx \frac{10 \times 632 \times 1.20}{7.15} \times 10^{-4} \approx 10.6 \text{ cm} \] This is the distance from the center to the common dark fringe position.

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Key Concepts

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

Diffraction Patterns
When light passes through a single slit, it spreads out and creates a specific pattern on a screen, known as a 'diffraction pattern'. This is due to the bending of light waves around the edges of the slit, which interferes with itself and produces regions of maximum and minimum intensity. This pattern consists of a central bright fringe, surrounded by alternating dark and bright fringes.
The intensity of the light decreases as you move away from the central maximum. The resulting pattern gives us insight into the wave nature of light. It highlights how light can behave like a wave, bending and spreading out when it encounters an obstacle or opening.
Dark Fringes
Dark fringes in diffraction patterns are spots on the screen where light waves cancel each other out, leading to minima or complete darkness. This happens because of destructive interference, where the peaks of one wave align with the troughs of another.
To determine the positions of these dark fringes for a single slit, we use the formula for diffraction minima: \[ a \sin(\theta) = m \lambda \] where \( a \) is the slit width, \( \theta \) is the angle relative to the slit, \( \lambda \) is the wavelength of light, and \( m \) is an integer indicating the order of the minima.
  • These dark fringes appear at regular intervals.
  • The order \( m \) starts at 1 for the first dark fringe, 2 for the second, and so on.
This method allows us to calculate exactly where on the screen these dark interference positions occur.
Wavelength
The wavelength of a light wave is the distance between successive peaks or troughs. It determines the color of the light in the visible spectrum and influences how light diffracts when passing through a slit.
In the context of diffraction, varying wavelengths create different interference patterns because each wavelength will have different path differences that result in constructive or destructive interference.
  • A longer wavelength (e.g., red light at 632 nm) will spread out more than a shorter wavelength (e.g., blue light at 474 nm), leading to broader patterns.
  • Shorter wavelengths tend to create narrower patterns with closely spaced fringes.
Therefore, two different wavelengths will form overlapping but distinguishable patterns on the screen.
Slit Width
The slit width is the physical measure across which the light wave travels; it significantly affects the diffraction pattern produced. A narrow slit causes more pronounced diffraction, causing waves to spread out more after passing through.
To predict the intensity pattern, we consider how the slit width \( a \) interacts with the wavelength \( \lambda \). The first dark fringe occurs where the angle \( \theta \) satisfies \[ a \sin(\theta) = m \lambda \] for \( m = 1 \). Thus, for a given wavelength, the slit width determines the angles of diffraction and how wide or narrow the fringes will appear on the screen.
  • A smaller slit (relative to wavelength) results in a wider spread of the diffraction pattern.
  • A larger slit confines the light more, producing narrower diffraction patterns with more closely spaced fringes.
Understanding this helps in designing experiments where controlling diffraction is essential.

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Most popular questions from this chapter

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18 000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W,\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\)

Light shines through a single slit whose width is \(5.6 \times 10^{-4} \mathrm{m} .\) A diffraction pattern is formed on a flat screen located 4.0 \(\mathrm{m}\) away. The distance between the middle of the central bright fringe and the first dark fringe is 3.5 \(\mathrm{mm}\) . What is the wavelength of the light?

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about 7.0 mm. The taillights of this car are separated by a distance of 1.2 m and emit red light (wavelength \(=660 \mathrm{nm}\) in vacuum. How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

You are standing in air and are looking at a flat piece of glass \((n=1.52)\) on which there is a layer of transparent plastic \((n=1.61)\) . Light whose wavelength is 589 nm is vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

A nonreflective coating of magnesium fluoride \((n=1.38)\) covers the glass \((n=1.52)\) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum \(=565 \mathrm{nm}\) ), determine the minimum nonzero thickness that the coating can have.

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