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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 width of the second slit \( W_2 \) is approximately \( 3.34 \times 10^{-6} \text{ m}.\)

Step by step solution

01

Understand the Situation

Given two slits with widths \( W_1 = 2.3 \times 10^{-6} \) m and \( W_2 \), we're dealing with a single-slit diffraction pattern. Initially, light of wavelength \( \lambda_1 = 510 \) nm passes through the first slit creating a central bright fringe of a certain width on the screen. The same fringe width is obtained for a second slit with width \( W_2 \) when the wavelength is changed to \( \lambda_2 = 740 \) nm.
02

Recall Diffraction Fringe Width Formula

The width of the central bright fringe in single-slit diffraction is given by:\[ 2 \cdot L \cdot \frac{\lambda}{W}\]where \( L \) is the distance from the slit to the screen, \( \lambda \) is the wavelength of light, and \( W \) is the slit width. The factor 2 accounts for the fringe extending equally on both sides of the center.
03

Set Equal Fringe Widths for Two Cases

Since the central bright fringe width is the same in both cases:\[ 2 \cdot L \cdot \frac{\lambda_1}{W_1} = 2 \cdot L \cdot \frac{\lambda_2}{W_2}\]This equation shows \[ \frac{\lambda_1}{W_1} = \frac{\lambda_2}{W_2}. \]
04

Solve for Unknown Slit Width \( W_2 \)

Rearrange the equation from Step 3 to solve for \( W_2 \):\[W_2 = W_1 \cdot \frac{\lambda_2}{\lambda_1}.\]Plug in the known values:\[ W_2 = 2.3 \times 10^{-6} \cdot \frac{740 \times 10^{-9}}{510 \times 10^{-9}}.\]Calculate \( W_2 \).
05

Calculate and Confirm

Perform the calculation:\[W_2 = 2.3 \times 10^{-6} \cdot \frac{740}{510} \approx 3.34 \times 10^{-6} \text{ m}.\]Therefore, the width of the second slit \( W_2 \) is approximately \( 3.34 \times 10^{-6} \text{ m} \).

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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 single slit, the waves spread out, creating a distinct pattern of dark and light bands on a screen. This is known as a diffraction pattern. It happens because light behaves like a wave, bending around the edges of the slit.
The most visible part of this pattern is the central bright fringe, which is the wide bright band in the middle. It is surrounded by alternating dark and lighter fringes. The central fringe is brighter and wider than the others because more of the light waves constructively interfere here. This pattern helps us understand wave behavior more deeply and is a key component when designing devices that use light, such as lenses and diffraction gratings.
Wavelength
Wavelength is a fundamental property of waves, especially in this context of light. It represents the distance between two consecutive points on a wave that are in phase, such as from crest to crest. In terms of light, it determines the color visible to us. For example, light with a wavelength of 510 nm appears green, while one with 740 nm appears red.
These differences in wavelength affect how light behaves when it encounters obstacles like slits. Specifically, longer wavelengths (like 740 nm) will spread out more when passing through a slit than shorter wavelengths (like 510 nm). This is why the wavelength is crucial in the formula that calculates the fringe width in a diffraction pattern.
Fringe Width
The width of the central bright fringe in a single-slit diffraction pattern is directly related to both the wavelength of light and the width of the slit. This fringe width is the distance between the positions of the first minimum on either side of the central maximum.
  • Formula: The fringe width can be estimated by the equation: \(2 \cdot L \cdot \frac{\lambda}{W}\), where \(L\) is the distance from the slit to the screen, \(\lambda\) is the light's wavelength, and \(W\) is the slit width.
  • Implications: This formula shows that a larger wavelength or a smaller slit width results in a wider fringe.
This information can be applied to calculate unknown slit widths when other variables are kept constant, as in the exercise.
Slit Width
The width of the slit in a diffraction experiment significantly impacts the resulting pattern. A narrower slit causes the light waves to spread out more, leading to a broader central diffraction pattern. This is because as the slit gets narrower, the points where the light can pass through are closer together, enhancing diffraction effects.
In solving for an unknown slit width as seen in the exercise, we use the relationship \(\frac{\lambda_1}{W_1} = \frac{\lambda_2}{W_2}\). This formula helps us maintain the same fringe width even when the wavelength changes. Adjusting the slit width compensates for the change in light's behavior, ensuring the observed diffraction pattern remains consistent.

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Most popular questions from this chapter

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?

In a Young's double-slit experiment, the wavelength of the light used is 520 \(\mathrm{nm}\) (in vacuum), and the separation between the slits is \(1.4 \times 10^{-6} \mathrm{m}\) . Determine the angle that locates \((\mathrm{a})\) the dark fringe for which \(m=0, \quad\) (b) the bright fringe for which \(m=1\) (c) the dark fringe for which \(m=1,\) and \(\quad\) (d) the bright fringe for which \(m=2\)

There are 5620 lines per centimeter in a grating that is used with light whose wavelength is 471 nm. A flat observation screen is located at a distance of 0.750 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen?

In a Young's double-slit experiment the separation \(y\) between the second- order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of \(425 \mathrm{nm} .\) Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta .\) Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{nm}\).

In a Young's double-slit experiment, the seventh dark fringe is located 0.025 \(\mathrm{m}\) to the side of the central bright fringe on a flat screen, which is 1.1 \(\mathrm{m}\) away from the slits. The separation between the slits is \(1.4 \times 10^{-4} \mathrm{m} .\) What is the wavelength of the light being used?

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