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In a single-slit diffraction pattern on a flat screen, the central bright fringe is 1.2 \(\mathrm{cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{m} .\) When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of 1.9 \(\mathrm{cm} .\) What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta\)

Short Answer

Expert verified
The width of the second slit is 2.56 x 10^-5 m.

Step by step solution

01

Understanding Diffraction Formula

In a single-slit diffraction pattern, the angular position of the first minimum is given by the formula \( a \sin \theta = m\lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength, and \( m \) is the order number of the minimum, with \( m = 1 \) for the first minimum.
02

Relating Fringe Width to Geometry

For small angle approximation, \( \sin \theta \approx \tan \theta \approx \frac{y}{D} \), where \( y \) is the fringe width and \( D \) is the distance to the screen. For the central fringe, \( y = 2 \frac{\lambda D}{a} \), since it spans from \( -\lambda D/a \) to \( \lambda D/a \).
03

Calculating Wavelength

Using the central fringe width of 1.2 cm for the first slit: \( \frac{1.2}{2} = \frac{\lambda D}{3.2 \times 10^{-5}} \). Solving for \( \lambda D \), we find \( \lambda D = 1.92 \times 10^{-5} \).
04

Applying to Second Slit

With the central fringe width of 1.9 cm for the second slit: \( \frac{1.9}{2} = \frac{\lambda D}{a_2} \). Using the previously calculated \( \lambda D \), we solve for the new slit width \( a_2 \).
05

Finding the Second Slit Width

From \( \frac{0.95}{1.92 \times 10^{-5}} = \frac{1.92 \times 10^{-5}}{a_2} \), solve for \( a_2 \). We find that \( a_2 = 2.56 \times 10^{-5} \) 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, it doesn't just go straight ahead but spreads out in a pattern known as a diffraction pattern. This pattern consists of a series of alternating bright and dark bands on a screen placed behind the slit. The central bright band, or fringe, is usually the brightest and broadest, flanked by successively dimmer fringes on either side.
The cause of these patterns is the wave nature of light. As the light waves encounters the slit, each point inside the slit acts as a new source of light. These new wave fronts interfere with each other, creating zones of constructive and destructive interference that appear as bright and dark bands. This is why we see a diffraction pattern instead of a single spot of light. Understanding this concept is crucial for explaining phenomena like resolving power in optical instruments and other wave-related applications.
Slit Width
The slit width plays a significant role in determining the characteristics of the diffraction pattern. Specifically, it influences the size and intensity of the central and subsequent fringes. A wider slit will typically produce a narrower diffraction pattern, while a narrower slit results in a broader pattern. This is because the angle of deviation caused by diffraction increases as the slit width decreases, spreading light over a larger area.
In mathematical terms, the slit width is denoted as "a" in the diffraction formula: \[ a \sin \theta = m\lambda \]\where \(a\) is the slit width, \(\theta\) is the angle of diffraction, \(\lambda\) is the wavelength, and \(m\) is the order of the minimum. By manipulating the slit width, we can affect how wavefronts interfere, altering the pattern that appears on the screen.
Central Fringe
The central fringe in a single-slit diffraction pattern is the bright band directly opposite the slit on the screen. Its prominence is due to the constructive interference of light waves that are in phase when they travel paths of equal length through the slit.
Its width is inversely proportional to the slit width—if the slit becomes narrower, the central fringe becomes broader. This is given by the relationship:\[ y = 2 \frac{\lambda D}{a} \]
where \(y\) is the fringe width, \(\lambda\) is the wavelength, \(D\) is the distance from the slit to the screen, and \(a\) is the slit width. Since the energy is concentrated in the central fringe due to wave interference, understanding this concept helps in fields like spectroscopy and the analysis of wave behavior in physics.
Wavelength
Wavelength is a fundamental property of waves, including light waves, and it significantly affects the diffraction pattern. It is the distance between successive crests of a wave and is denoted by \(\lambda\) in equations.
In single-slit diffraction, the wavelength helps to determine the angles and positions of the diffraction minima and maxima. The relation of wavelength to the central fringe’s width is crucial:\[ \lambda = \frac{y a}{2D} \]
Given other known values, this expression can be used to estimate the wavelength of light when observing the width of the central fringe, combined with the slit width and screen distance. Understanding how wavelength interacts with other factors in diffraction is important for fields like wave optics and communication technologies.

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Most popular questions from this chapter

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 hunter who is a bit of a braggart claims that from a distance of 1.6 km he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What’s more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of 498 nm (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to 8 mm, the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of 498 nm.

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 \(585 \mathrm{nm}\).

Light that has a wavelength of 668 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?

Two gratings \(A\) and \(B\) have slit separations \(d_{A}\) and \(d_{B},\) respectively. They are used with the same light and the same observation screen. When grating \(A\) is replaced with grating \(B,\) it is observed that the first order maximum of \(A\) is exactly replaced by the second-order maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating A and the principal maxima of \(\mathrm{B}\) that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

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