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Chapter 24: Problem 32

Light of wavelength \(600 \mathrm{~nm}\) falls on a \(0.40\) -mm-wide slit and forms a diffraction pattern on a screen \(1.5 \mathrm{~m}\) away. (a) Find the position of the first dark band on each side of the central maximum. (b) Find the width of the central maximum.

Short Answer

Expert verified
The positions of the first dark band on each side of the central maximum is equal to \(L \cdot \tan(\theta)\) m and the width of the central maximum is twice this value i.e. \(2 \cdot L \cdot \tan(\theta)\) m.

Step by step solution

01

Calculating Angular Position

First, calculate the angular position of the first dark band. Plugging \(m = 1\), \(\lambda = 600 \times 10^{-9}\) m (converted from nm to m), and \(w = 0.40 \times 10^{-3}\) m (converted from mm to m) into the formula: \(\theta = \arcsin(\frac{m\lambda}{w}) = \arcsin(\frac{(1) (600 \times 10^{-9} \, m)}{0.40 \times 10^{-3} \, m})\). Calculate this angle.
02

Calculating Position

The actual position of the first dark band on the screen can be found using the small angle approximation (for small angles, sin(胃) 鈮� 胃), and the formula \(L \cdot \tan(\theta)\), where \(L\) is the distance to the screen. Plugging \(\theta\) from previous step and \(L = 1.5\) m into the formula, calculate the position. If the value of \(\theta\) found in step 1 is not a small angle, then it would be necessary to use the full form of the formula \(L \cdot \tan(\theta)\) without the small angle approximation.
03

Calculating Width of the Central Maximum

The width of the central maximum is twice the position of the first-order dark band. Hence, multiply the position calculated in the previous step by 2. This will provide the width of the central maximum in meters.

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Key Concepts

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

Diffraction Pattern
When a wave encounters an obstacle or slit, it bends around the edges, a phenomenon known as diffraction. In the case of light, this results in a pattern of bright and dark fringes called a diffraction pattern.

A diffraction pattern occurs because light waves interfere with one another after passing through the slit. The bright bands, known as maxima, appear where the waves are in phase and constructively interfere. Alternatively, the dark bands, or minima, occur where the waves are out of phase and destructively interfere.

Understanding diffraction patterns is crucial in applications like resolving details in microscopes or telescopes, where wave interference can impact the clarity of the image.
Central Maximum
In a diffraction pattern, the central maximum is the brightest band found directly in the middle. This occurs because the wave paths emanating from the center of the slit interfere constructively, resulting in maximum intensity at this point.

The position and width of the central maximum are critical components when analyzing diffraction. It is usually the widest and most intense band. The width of the central maximum can be calculated by doubling the position of the first dark band on either side, as we find the dark bands symmetric around the center. This aspect is essential because it defines how much the wave has spread after passing through the slit.
Wavelength
Wavelength is the distance between successive crests of a wave. In the case of light waves, it is a measure of color in visible light, or type of electromagnetic wave for other wavelengths.

Wavelength plays a significant role in determining the diffraction pattern. Longer wavelengths result in wider diffraction patterns, causing more spread out intensity bands.
  • Analyzing wavelength's impact helps us understand how light behaves when encountering obstacles.
  • In calculating diffraction parameters, the wavelength is typically one of the known values, used alongside other variables to find the angles and positions of maxima and minima.
Single-Slit Diffraction
Single-slit diffraction occurs when a wave passes through a narrow opening and spreads out. In a single-slit setup, the diffraction pattern consists of a prominent central maximum with various smaller side maxima and minima.

The mathematical analysis of single-slit diffraction uses the formula for angular positions of minima, given by: \[ \theta = \arcsin\left( \frac{m\lambda}{w} \right) \] Where:
  • \( m \) is the order of the minimum, with \( m = 1, 2, 3, \ldots \)
  • \( \lambda \) is the wavelength of the wave
  • \( w \) is the width of the slit
These calculations help us predict the positions of dark bands in the pattern.

Single-slit diffraction is essential in fields demanding precision measurement and imaging, where understanding and controlling wave spreading are crucial.

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Most popular questions from this chapter

\(\mathrm{A}\) pair of narrow, parallel slits separated by \(0.250 \mathrm{~mm}\) is illuminated by the green component from a mercury vapor lamp \((\lambda=546.1 \mathrm{~nm})\). The interference pattern is observed on a screen \(1.20 \mathrm{~m}\) from the plane of the parallel slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands in the interference pattern.

A helium-neon laser \((\lambda=632.8 \mathrm{~nm})\) is used to calibrate a diffraction grating. If the first-order maximum occurs at \(20.5^{\circ}\), what is the spacing between adjacent grooves in the grating?

A diffraction pattern is produced on a screen \(140 \mathrm{~cm}\) from a single slit, using monochromatic light of wavelength \(500 \mathrm{~nm} .\) The distance from the center of the central maximum to the first-order maximum is \(3.00 \mathrm{~mm}\). Calculate the slit width. Hint: Assume that the first- order maximum is halfway between the first-and second-order minima.

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 with a wavelength in vacuum of \(546.1 \mathrm{~nm}\) falls perpendicularly on a biological specimen that is \(1.000 \mu \mathrm{m}\) thick. The light splits into two beams polarized at right angles, for which the indices of refraction are \(1.320\) and \(1.333\), respectively. (a) Calculate the wavelength of each component of the light while it is traversing the specimen. (b) Calculate the phase difference between the two beams when they emerge from the specimen.
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