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Chapter 26: Problem 25

Parallel rays of green mercury light with a wavelength of \(546 \mathrm{nm}\) pass through a slit with a width of \(0.437 \mathrm{~mm}\). What is the distance from the central maximum to the first minimum on a screen \(1.75 \mathrm{~m}\) away from the slit?

Short Answer

Expert verified
The distance is approximately 2.19 mm.

Step by step solution

01

Understand the Problem

We need to find the distance of the first minimum from the central maximum on a screen using the given parameters: wavelength \( \lambda = 546 \text{ nm} = 546 \times 10^{-9} \text{ meters} \), slit width \( a = 0.437 \text{ mm} = 0.437 \times 10^{-3} \text{ meters} \), and screen distance \( L = 1.75 \text{ meters} \).
02

Apply Formula for Single-Slit Diffraction

For a single-slit diffraction pattern, the position of the first minimum is given by the formula \( a \sin \theta = m\lambda \), where \( m = 1 \) for the first minimum. Solving for \( \sin \theta \), we have \( \sin \theta = \frac{\lambda}{a} \).
03

Calculate the Angle \(\theta\)

Substitute the known values: \( \sin \theta = \frac{546 \times 10^{-9}}{0.437 \times 10^{-3}} \). Calculate \( \sin \theta \) to find \( \theta \). The calculation gives \( \sin \theta \approx 1.25 \times 10^{-3} \).
04

Approximate \( \theta \) using Small Angle Approximation

For small angles, \( \sin \theta \approx \tan \theta \approx \theta \) in radians. Thus, \( \theta \approx 1.25 \times 10^{-3} \text{ radians} \).
05

Calculate the Distance \( y \) to the First Minimum

The distance \( y \) from the central maximum to the first minimum on the screen is given by \( y = L \tan \theta \). Using \( \tan \theta \approx \theta \), substitute the known values: \( y = 1.75 \times 1.25 \times 10^{-3} \). The calculation gives \( y \approx 2.1875 \times 10^{-3} \text{ meters} \) or approximately \( 2.19 \text{ mm} \).

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Key Concepts

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

Wavelength
The wavelength of light is a fundamental concept in understanding diffraction patterns. It refers to the distance between consecutive peaks of a light wave. In the context of diffraction through a single slit, the wavelength determines how light interferes with itself to create patterns on a screen.
  • The wavelength for green mercury light is given as 546 nanometers (nm), which is a common measurement used in physics for visible light.
  • This value needs to be converted to meters when performing calculations that involve physical distances such as diffraction. In this case: \(546 \, \text{nm} = 546 \times 10^{-9} \, \text{m}\).
  • The smaller the wavelength, the closer the diffraction patterns appear on the screen for a given slit width.
Screen Distance
The screen distance is a crucial measurement in single-slit diffraction experiments. It refers to the distance between the slit through which light passes and the screen where the diffraction pattern is observed.
  • In this problem, the screen distance is given as 1.75 meters. This distance can affect the size and clarity of the diffraction pattern observed.
  • A larger distance will spread the pattern over a larger area, potentially making the bands wider and more distinct.
  • The screen distance (\( L \)) is used in calculating the position of the minima in the pattern through the equation \( y = L \tan \theta \), which connects it with the angle \( \theta \) formed at the central maximum.
Central Maximum
The central maximum is the brightest part of a diffraction pattern. It appears directly in line with the light source and the slit.
  • This feature occurs because light waves traveling straight through the slit interfere constructively at the center.
  • The width of the central maximum is typically wider than other maxima due to the nature of single-slit diffraction.
  • In calculations involving diffraction, such as finding the first minimum, the central maximum acts as a reference point. The distance from this central spot to the first minimum is calculated using the angle \( \theta \) determined by the light's wavelength and the slit width.
First Minimum
The first minimum in a diffraction pattern is the first point on either side of the central maximum where the light intensity falls to a minimum.
  • It results from destructive interference between light waves originating from different parts of the slit.
  • Mathematically, the position of the first minimum is found using the formula for single-slit diffraction: \( a \sin \theta = \lambda \), where \(a\) is the slit width, \(\lambda\) is the wavelength of light, and \(\theta\) represents the angle to the minimum from the center.
  • The position on the screen where this minimum occurs is calculated using: \( y = L \tan \theta \). For small angles, the approximation \( \tan \theta \approx \theta \) simplifies this calculation, allowing for easy conversion between angle and screen distance.

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Most popular questions from this chapter

What is the thinnest soap film (excluding the case of zero thickness) that appears black when viewed by reflected light with a wavelength of \(480 \mathrm{nm} ?\) The index of refraction of the film is 1.33 , and there is air on both sides of the film.

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index \(1.432,\) (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength \(550 \mathrm{nm}\) reflected toward the eye at normal incidence, and (b) will any other wavelengths of visible light be canceled or enhanced in the reflected light?

Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of \(8.94^{\circ} .\) What is the angular position of the fourth-order maximum?

The lenses of a particular set of binoculars have a coating with index of refraction \(n=1.38,\) and the glass itself has \(n=1.52 .\) If the lenses reflect a wavelength of \(525 \mathrm{nm}\) the most strongly, what is the minimum thickness of the coating?

Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength \(632.8 \mathrm{nm}\) was shone on a single strand of hair, and the diffracted light was viewed on a screen \(1.25 \mathrm{~m}\) away, the first dark fringes on either side of the central bright spot were \(5.22 \mathrm{~cm}\) apart. How thick was this strand of hair?
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