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A fl at screen is located 0.60 m away from a single slit. Light with a wavelength of 510 nm (in vacuum) shines through the slit and produces a diffraction pattern. The width of the central bright fringe on the screen is 0.050 m. What is the width of the slit?

Short Answer

Expert verified
The width of the slit is approximately 0.0001224 meters.

Step by step solution

01

Identify Key Variables

We are given a few important pieces of information:1. The distance from the slit to the screen, \[ L = 0.60 \text{ m} \].2. The wavelength of the light, \[ \lambda = 510 \text{ nm} = 510 \times 10^{-9} \text{ m} \].3. The width of the central bright fringe, \[ W = 0.050 \text{ m} \].Our task is to find the width of the slit, \( a \).
02

Understand the Relationship Between Variables

The width of the central bright fringe \( W \) on the screen is approximately related to the slit width \( a \) by the formula:\[ W = \frac{2 \lambda L}{a} \]This relationship derives from the diffraction angle and the geometry of the pattern.
03

Rearrange the Equation

We want to find \( a \), the width of the slit. Rearranging the formula for \( W \) gives:\[ a = \frac{2 \lambda L}{W} \].
04

Plug in the Values

Substitute the known values into the rearranged formula:\[ a = \frac{2 \times 510 \times 10^{-9} \text{ m} \times 0.60 \text{ m}}{0.050 \text{ m}} \].Calculating this gives the width of the slit.
05

Calculate the Slit Width

Perform the calculation:\[ a = \frac{2 \times 510 \times 10^{-9} \times 0.60}{0.050} \approx 1.224 \times 10^{-4} \text{ m} = 0.0001224 \text{ m} \].

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Key Concepts

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

single slit diffraction
Single slit diffraction occurs when a wave, such as light, passes through a narrow opening and spreads out on the other side, creating a distinctive pattern of light and dark bands on a screen. This pattern arises from the wave nature of light and results in areas of constructive and destructive interference. When the light encounters the slit, portions of the wavefront are obstructed, leading to differences in the path lengths the light travels, which causes this pattern.

In a single slit diffraction setup, the central bright fringe is typically the largest and brightest band seen on the pattern. It appears directly in line with the slit, while smaller, less intense bright fringes can be observed on either side, interspersed with dark fringes. These results from fitting different numbers of wavefronts into the path difference across the slit opening.
wavelength
Wavelength is the distance between successive crests of a wave, particularly in relation to electromagnetic waves such as light. It plays a crucial role in determining the diffraction pattern produced. The wavelength is crucial in diffraction because it determines how much light bends when passing through the slit.

If the slit width is comparable to the wavelength, significant diffraction occurs, strongly influencing the resulting pattern. For instance, in our exercise, the wavelength of the light used is 510 nm, which is critical for calculating how the light bends and interferes with itself after passing through the slit, causing the observable diffraction pattern.
central bright fringe
The central bright fringe, also known as the principal maximum, is the brightest part of the diffraction pattern created on the screen. This fringe occurs at the center and is symmetric on both sides. It captures most of the light's intensity that passes through the slit, making it visually dominant compared to the adjacent fringes.

The width of this fringe is often measured to determine other characteristics of the diffraction setup, like the slit width. The width is critical since it directly correlates to the geometry of the diffraction pattern, the distance to the screen, and the wavelength of light used. In our case, the central bright fringe is 0.050 m wide and is crucial for calculating the slit width using the appropriate formula.
slit width calculation
Calculating the slit width involves understanding the geometric relationship between the central bright fringe, the light’s wavelength, and its distance to the screen. The formula to find the slit width \( a \) is given by \[ a = \frac{2\lambda L}{W} \]where \( \lambda \) is the wavelength,\( L \) is the distance from the slit to the screen,and \( W \) is the width of the central bright fringe.

By substituting our known values into this equation, we can calculate the slit width accurately. In the given exercise, with \( \lambda = 510 \times 10^{-9} \text{ m} \),\( L = 0.60 \text{ m} \),and \( W = 0.050 \text{ m} \),the slit width \( a \) becomes approximately \( 0.0001224 \text{ m} \).

This calculation is key to understanding how light behaves when passing through openings of different sizes.

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Most popular questions from this chapter

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?

Orange light \(\left(\lambda_{\text {vacuum }}=611 \mathrm{nm}\right)\) shines on a soap film \((n=1.33)\) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

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 \(540 \mathrm{nm} \leq \lambda \leq 560 \mathrm{nm} ?\)

Two stars are \(3.7 \times 10^{11} \mathrm{m}\) apart and are equally distant from the earth. A telescope has an objective lens with a diameter of \(1.02 \mathrm{m}\) and just detects these stars as separate objects. Assume that light of wavelength 550 nm is being observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the telescope. Find the maximum distance that these stars could be from the earth.

The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen and the slit. Find the ratio \(\lambda / W\) of the wavelength \(\lambda\) of the light to the width \(W\) of the slit.

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