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A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{m} .\) When light with a wavelength of \(\lambda_{1}=510 \mathrm{nm}\) 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 \(W_{2}\)

Short Answer

Expert verified
The second slit width must be 3.34 x 10鈦烩伓 m.

Step by step solution

01

Understanding the Problem

We are given two light configurations producing identical central bright fringes in a single-slit diffraction experiment. The goal is to calculate the width of a new slit, given the condition that the width of the central bright fringe remains constant despite changes in light wavelength and slit used.
02

Identify Formula for Fringe Width

The width of the central bright fringe in a single slit diffraction is determined by the formula: \( W = \frac{2 \lambda L}{a} \), where \( W \) is the fringe width, \( \lambda \) is the wavelength of light, \( L \) is the distance from the slit to the screen, and \( a \) is the slit width.
03

Set Up Proportionality

Since the fringe width remains unchanged for both slits, the expressions \( \frac{\lambda_1}{W_1} = \frac{\lambda_2}{W_2} \) must hold, as \(2L\) cancels out being a constant in both configurations.
04

Solve for New Slit Width

Rearrange the equation \( \frac{\lambda_1}{W_1} = \frac{\lambda_2}{W_2} \) to solve for \( W_2 \): \[ W_2 = W_1 \frac{\lambda_2}{\lambda_1} \]
05

Substitute Known Values and Compute

Substitute \( W_1 = 2.3 \times 10^{-6} \text{ m} \), \( \lambda_1 = 510 \text{ nm} = 510 \times 10^{-9} \text{ m} \), and \( \lambda_2 = 740 \text{ nm} = 740 \times 10^{-9} \text{ m} \) into the equation: \[ W_2 = \left( 2.3 \times 10^{-6} \right) \left( \frac{740 \times 10^{-9}}{510 \times 10^{-9}} \right) \]
06

Final Calculation

Compute the expression: \[ W_2 = 2.3 \times 10^{-6} \cdot \frac{740}{510} = 3.34 \times 10^{-6} \text{ m} \]Resulting in \( W_2 = 3.34 \times 10^{-6} \text{ m} \).

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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 slit, it bends around the edges, creating patterns of light and dark regions on a screen 鈥 this is known as a diffraction pattern. A diffraction pattern consists of bright fringes called maxima and dark fringes called minima. These patterns result from the constructive and destructive interference of light waves. The central maxim, which is the brightest and widest part of the pattern, is flanked by alternating minima and smaller maxima.

Understanding diffraction patterns is essential because it helps us analyze how waves behave when they encounter obstructions. The principles governing these patterns are applicable not only to light waves but also to other types of waves, such as sound and water waves.

In a single-slit scenario, how light waves constructively and destructively interfere determines the pattern's appearance. A narrow slit produces a broader diffraction pattern, while a wider slit leads to a more confined one. This relationship between slit width and pattern structure is crucial for understanding wave behavior in optics.
Wavelength of Light
Light has wave-like properties, and one important characteristic is its wavelength. Wavelength is the distance between successive crests (or troughs) of the wave and is typically measured in nanometers (nm) for light waves. Wavelength directly influences the color of light we perceive; for example, shorter wavelengths correspond to blue light, while longer wavelengths correspond to red light.

The wavelength of light affects how it diffracts through a slit. In a single-slit diffraction, a change in the wavelength alters the width of the central bright fringe. Longer wavelengths, like red light, tend to produce wider fringes, whereas shorter wavelengths, like blue light, create narrower ones.

In the given exercise, two different wavelengths are used: 510 nm (green light) and 740 nm (red light). The challenge arises from maintaining the same fringe width despite the difference in wavelengths, showcasing the direct impact wavelength has on diffraction and interference patterns.
Fringe Width
Fringe width is a crucial aspect of single-slit diffraction patterns. It refers to the size of the central bright band, often the largest and most prominent feature of the pattern. The width of this central fringe is determined by the formula:\[ W = \frac{2 \lambda L}{a} \]Where:
  • \(W\) is the fringe width,
  • \(\lambda\) is the wavelength of light,
  • \(L\) is the distance from the slit to the screen,
  • \(a\) is the width of the slit.
This formula shows that fringe width is directly proportional to both the wavelength of the light and the distance to the screen, but inversely proportional to the slit width.

In practical terms, when the wavelength increases or the slit width decreases, the central fringe widens. Conversely, a decrease in wavelength or an increase in slit width results in a narrower fringe. The exercise described explains how, despite changing these variables, keeping the central fringe width constant involves adjusting the slit width according to the new wavelength, affirming the interplay between these variables in wave physics.

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Most popular questions from this chapter

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is \(296 \mathrm{nm}\). What is the vacuum wavelength of the light in \(\mathrm{nm} ?\)

The light shining on a diffraction grating has a wavelength of 495 nm (in vacuum). The grating produces a second-order bright fringe whose position is defined by an angle of \(9.34^{\circ} .\) How many lines per centimeter does the grating have?

An Optical Monochromator. You and your team are designing a device that inputs a beam of white light (i.e., a continuous spectrum of visible light spanning all wavelengths from \(410 \mathrm{nm}\) to \(660 \mathrm{nm}\) ), and outputs a nearly monochromatic beam (i.e., a single color). Such a device is called an optical monochromator and is used in a wide variety of instruments and scientific experiments. In the instrument you are building, white light impinges upon the backside of a diffraction grating that has 1200 lines/cm. A movable rectangular aperture (a slit) is located on the opposite side of the grating, and can translate along a circular arc of radius \(20.0 \mathrm{cm},\) the center of which is located at the grating. (a) At what angle relative to the normal of the grating should the center of the slit be located in order to pass green light \((\lambda=550 \mathrm{nm})\) from first order \((m=1)\) diffracted light? (b) How wide should the slit be so that the wavelengths passing through the slit fall in the range \(540 \mathrm{nm} \leq \lambda \leq 560 \mathrm{nm} ?\)

Violet light (wavelength \(=410 \mathrm{nm}\) ) and red light (wavelength \(=\) \(660 \mathrm{nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the thirdorder maximum. (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe is \(5.4^{\circ} .\) Find the ratio \(d / \lambda\) of the slit separation \(d\) to the wavelength \(\lambda\) of the light.

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