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A slit of width \(0.50 \mathrm{~mm}\) is illuminated with light of wavelength \(500 \mathrm{~nm}\), and a screen is placed \(120 \mathrm{~cm}\) in front of the slit. Find the widths of the first and second maxima on each side of the central maximum.

Short Answer

Expert verified
The widths of the first and second maxima on each side of the central maximum are both \(1.2 \, mm\).

Step by step solution

01

Understand the problem and gather data

In this problem, we need to find the widths of the first and second maxima on each side of the central maximum. The data given are: wavelength of light \(\lambda = 500 \, nm = 500 \times 10^{-9} \, m\), width of the slit \(a = 0.5 \, mm = 0.5 \times 10^{-3} \, m\), and the distance of the screen from the slit \(L = 120 \, cm = 1.2 \, m\).
02

Use the formula for minima

The formula for the positions of minima (dark fringes) in a single-slit diffraction pattern is \(x_{min} = L \cdot \frac{m \lambda}{a}\), where \(m=1\) for the first minimum, \(m=2\) for the second minimum, etc.
03

Find the widths of the maxima

The width of a maxima is the distance between two adjacent minima. Therefore, the width of the first maxima on either side of the central maximum is \(x_{1, width} = x_{min}(m=2) - x_{min}(m=1) = L \cdot \frac{\lambda}{a} = 1.2 \cdot \frac{500 \times 10^{-9}}{0.5 \times 10^{-3}} \, m = 1.2 \, mm\). The width of the second maxima is \(x_{2, width} = x_{min}(m=3) - x_{min}(m=2) = L \cdot \frac{\lambda}{a} = 1.2 \, mm\).

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Key Concepts

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

Diffraction Patterns
When a wave encounters an obstacle or a slit that is comparable in size to its wavelength, it bends around it, a phenomenon known as diffraction. This bending creates a pattern of alternating dark and bright regions on a screen, known as a diffraction pattern. In the context of the textbook problem, we talk about light, a form of electromagnetic radiation, diffracting through a single slit and forming a pattern on the screen.

The diffraction pattern from a single-slit includes a central bright region known as the central maximum, which is flanked on both sides by a series of progressively dimmer bright regions (maxima) separated by dark regions (minima). These patterns can be predicted using the principles of wave optics and are characterized by their unique features depending on the slit's width and the wavelength of light used.
  • The central maximum is the brightest and widest because it receives light diffracted from the entire width of the slit.
  • The positions of the minima are determined by the destructive interference of light waves, which occurs when the path difference between waves is equal to a half-integer multiple of the wavelength.
Wave Optics
Wave optics, also known as physical optics, is the branch of optics that studies interference, diffraction, polarization, and other phenomena for which the ray approximation of geometric optics is not valid. These phenomena occur when the size of the obstacles or apertures through which the wave passes is comparable to the wavelength of the wave.

In our problem, wave optics allows us to understand how light waves spread out after passing through a slit and interfere with one another to form the observed diffraction pattern on a screen. The calculations in the solution rely on principles from wave optics, such as the path difference between waves and conditions for constructive and destructive interference. Using wave optics, we can predict and calculate the positions and widths of the diffraction pattern's maxima and minima with the formula for the positions of minima being pivotal for finding the maxima widths as well.
Physical Optics
Physical optics is a field that provides a detailed description of the behavior of light waves. Unlike geometric optics, which considers light as rays that travel in straight lines, physical optics considers light as waves that can exhibit behaviors such as diffraction and interference mentioned in the example problem.

Understanding physical optics principles is crucial to solving problems related to the diffraction patterns produced by a single slit. The exercise provided demonstrates how physical optics can be applied to determine the widths of diffraction maxima. Through the application of the single-slit diffraction formula, which derives from the wave nature of light, we achieve an accurate depiction of how light behaves when it encounters an object of a size comparable to its wavelength. This knowledge not only allows us to calculate the pattern on a screen but also to comprehend the fundamental nature of light as a wave.

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Most popular questions from this chapter

A thin film of \(\mathrm{MgF}_{2}(n=1.38)\) with thickness \(1.00 \times\) \(10^{-5} \mathrm{~cm}\) is used to coat a camera lens. Are any wave- lengths in the visible spectrum intensified in the reflected light?

Light of wavelength \(587.5 \mathrm{~nm}\) illuminates a slit of width \(0.75 \mathrm{~mm}\). (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be \(0.85 \mathrm{~mm}\) from the central maximum? (b) Calculate the width of the central maximum.

Ecp Astronomers observe the chromosphere of the Sunt with a filter that passes the red hydrogen spectral line of wavelength \(656.3 \mathrm{~nm}\), called the \(\mathrm{H}_{\alpha}\) line. The filter consists of a transparent dielectric of thickness \(d\) held between two partially aluminized glass plates. The filter is kept at a constant temperature. (a) Find the minimum value of \(d\) that will produce maximum transmission of perpendicular \(\mathrm{H}_{a}\) light if the dielectric has an index of refraction of \(1.378 .\) (b) If the temperature of the filter increases above the normal value increasing its thickness, what happens Lo the transmitted wavelength? (c) The dielectric will also pass what near-visible wavelength? One of the glass plates is colored red to absorb this light.

A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of \(3.00 \mathrm{~cm}\) and the index of refraction of the polymer is \(n=1.50\), how thick would you make the coating?

A pair of slits, separated by \(0.150 \mathrm{~mm}\), is illuminated by light having a wavelength of \(\lambda=643 \mathrm{~nm}\). An interference pattern is observed on a screen \(140 \mathrm{~cm}\) from the slits. Consider a point on the screen located at \(y=1.80 \mathrm{~cm}\) from the central maximum of this pattern. (a) What is the path difference \(\delta\) for the two slits at the location \(y^{2}\) (b) Express this path difference in terms of the wavelength. (c) Will the interference correspond to a maximum, a minimum, or an intermediate condition?

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