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Light of wavelength \(420 \mathrm{~nm}\) is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is \(2.00^{\circ}\). What is the width of the slit?

Short Answer

Expert verified
The width of the slit is approximately 24.1 µm.

Step by step solution

01

Identify the relevant equation

The formula to determine the position of the diffraction minima in a single-slit diffraction pattern is given by \(a \sin \theta = m \lambda\), where \(a\) is the slit width, \(\theta\) is the diffraction angle, \(m\) is the order of the minimum (integer, not zero), and \(\lambda\) is the wavelength of the light.
02

Understand the problem conditions

The angle given is the total angle between the first diffraction minimum on one side and the first diffraction minimum on the other side, which is \(2.00^{\circ}\). Therefore, each minimum is located at an angle of \(1.00^{\circ}\) from the central maximum.
03

Rearrange the equation for slit width

For the first diffraction minimum, \(m=1\). The equation becomes \(a \sin \theta = \lambda\). We need to find \(a\), so rearrange the equation to \(a = \frac{\lambda}{\sin \theta}\).
04

Convert angle to radians

Convert the angle from degrees to radians because the sine function in physics usually requires angles in radians: \[ \theta = 1.00^{\circ} \times \frac{\pi}{180^{\circ}} = 0.01745 \, \text{radians} \].
05

Substitute numbers and solve for slit width

Substitute the given wavelength \(\lambda = 420 \times 10^{-9} \text{ m}\) and \(\theta = 0.01745\) radians into the equation \(a = \frac{\lambda}{\sin \theta}\): \[ a = \frac{420 \times 10^{-9} \text{ m}}{\sin(0.01745)} \approx \frac{420 \times 10^{-9} \text{ m}}{0.01745} \approx 24.1 \times 10^{-6} \text{ m} \].
06

State the final result

The width of the slit is approximately \(24.1 \text{ µm}\) or \(24.1 \times 10^{-6} \text{ m}\).

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Key Concepts

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

Single-slit diffraction
Single-slit diffraction occurs when a wave encounters a narrow slit, causing the wave to spread out or 'diffract'. This effect is most noticeable when the width of the slit is comparable to the wavelength of the light being used. In a typical lab setup, light is shone through a single narrow slit, creating a pattern of bright and dark regions on a screen. These dark and bright areas are caused by the interaction of waves interfering constructively and destructively with one another.

In single-slit diffraction:
  • The central bright region, or the central maximum, is the brightest and widest part of the diffraction pattern.
  • Each side of the central maximum has alternating dark and bright regions, with the dark areas known as diffraction minima.
  • The intensity of light decreases as you move away from the central maximum.
Understanding this concept helps comprehend how light behaves when forced through a narrow passage, providing greater insight into the wave nature of light.
Diffraction minima
Diffraction minima are the dark regions in the pattern produced by single-slit diffraction. In these areas, the light waves cancel each other out due to destructive interference. This happens because waves from different parts of the slit arrive out of phase when they recombine on the other side.

For single-slit diffraction, the condition for destructive interference, or minima, is given by the equation:
  • \[a \sin \theta = m \lambda\]
where:
  • \(a\) is the slit width,
  • \(\theta\) is the angle of diffraction minima,
  • \(m\) is the order of the minima (integer, except zero), and
  • \(\lambda\) is the wavelength of the incident light.
For the first minima, which we often need to calculate, \(m=1\), making it possible to find the width of the slit if the wavelength and diffraction angle are known.
Wavelength
Wavelength is a fundamental property of waves, describing the distance between consecutive points of the same phase, such as from peak to peak. It is denoted by the Greek letter \(\lambda\), and it plays a crucial role in diffraction phenomena.

In the context of light, the wavelength determines the color of the light perceived by humans and directly influences the diffraction pattern created when it passes through a slit. For example, shorter wavelengths result in narrower diffraction patterns, whereas longer wavelengths cause wider patterns.

In experiments similar to the given exercise, precise knowledge of the wavelength allows for accurate calculations of slit width using the formula for diffraction minima. When you know the light's wavelength and the angle at which minima occur, you can compute the slit width necessary to produce that specific diffraction pattern.

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Most popular questions from this chapter

a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if \(\lambda=550 \mathrm{~nm}, d=0.180 \mathrm{~mm}\), and \(a=30.0 \mu \mathrm{m}\) ? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

A diffraction grating \(24.0 \mathrm{~mm}\) wide has 5000 rulings. Light of wavelength \(589 \mathrm{~nm}\) is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of \(\theta\) at which maxima appear on a distant viewing screen?

The wall of a large room is covered with acoustic tile in which small holes are drilled \(6.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm}\), and the wavelength of the room light to be \(550 \mathrm{~nm}\) ?

Sound waves with frequency \(2500 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\) diffract through the rectangular opening of a speaker cabinet and into a large auditorium of length \(d=100 \mathrm{~m}\). The opening, which has a horizontal width of \(30.0 \mathrm{~cm}\), faces a wall \(100 \mathrm{~m}\) away (Fig. 36-35). Along that wall, how far from the central axis will a listener be at the first diffraction minimum and thus have difficulty hearing the sound? (Neglect reflections.)

The two headlights of an approaching automobile are \(1.4 \mathrm{~m}\) apart. At what (a) angular separation and (b) maximum distance will the eye resolve them? Assume that the pupil diameter is \(4.5 \mathrm{~mm}\), and use a wavelength of \(550 \mathrm{~nm}\) for the light. Also assume that diffraction effects alone limit the resolution so that Rayleigh's criterion can be applied.

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