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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 nm was shone on a single strand of hair, and the diffracted light was viewed on a screen 1.25 m away, the first dark fringes on either side of the central bright spot were 5.22 cm apart. How thick was this strand of hair?

Short Answer

Expert verified
The strand of hair is approximately 30.3 micrometers thick.

Step by step solution

01

Identify known variables

We'll start by listing the known values from the exercise. The wavelength of the light \( \lambda = 632.8 \) nm, which we convert to meters: \( 632.8 \times 10^{-9} \) m. The distance from the hair to the screen \( L = 1.25 \) m, and the distance between the first dark fringes \( y = 5.22 \) cm, which we convert to meters: \( 0.0522 \) m.
02

Understand the diffraction formula

For a straight object like a strand of hair acting as a single slit, we use the single-slit diffraction formula for the first dark fringe: \( a \sin \theta = m \lambda \). Here, \( m = 1 \) for the first dark fringe.
03

Compute angle \( \theta \) using small angle approximation

Since \( \theta \) is small, we use the approximation \( \sin \theta \approx \tan \theta \approx \frac{y}{2L} \). Here, \( y \) is the total fringe separation, so \( \theta \approx \frac{0.0522}{2 \times 1.25} \). Calculate \( \theta \approx 0.02088 \).
04

Solve for the thickness of the hair \( a \)

Substitute \( \sin \theta \approx 0.02088 \) into the diffraction formula: \( a \cdot 0.02088 = 632.8 \times 10^{-9} \). Solving for \( a \), we have \( a = \frac{632.8 \times 10^{-9}}{0.02088} \).
05

Calculate the final result

Upon calculating, \( a \approx 3.03 \times 10^{-5} \) meters or 30.3 micrometers.

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Key Concepts

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

light wavelength
The concept of light wavelength is essential when discussing diffraction. A light wave's wavelength is the distance between consecutive crests (or troughs) of the wave. In the context of diffraction, the wavelength affects how light spreads out after passing through a slit or around an object.
The wavelength is usually measured in nanometers (nm). Nanometers to meters conversion is crucial in physics problems. For example, a wavelength of 632.8 nm becomes 632.8 x 10^{-9} meters when converted.
Understanding the wavelength provides insights into the behavior of different colors of light. Generally, shorter wavelengths (like blue light) diffract less than longer ones (like red light). Thus, knowing the wavelength helps predict light behavior in diffraction scenarios.
diffraction formula
The diffraction formula is fundamental for solving problems involving diffraction patterns. This formula relates the physical characteristics of the diffracting object to the pattern it creates.
For single-slit diffraction, the formula is: \[ a \sin \theta = m \lambda \]
Here, \(a\) is the slit width (or the hair's thickness in the exercise), \( \sin \theta \) is the angle of diffraction, \(m\) is the order number of the fringe (1 for the first dark fringe), and \( \lambda \) is the light's wavelength.
This formula helps determine the width of an object by using known values of light wavelength and fringe separation. Understanding this formula is key to solving diffraction problems.
small angle approximation
The small angle approximation is a useful simplification in physics, particularly for diffraction problems. This approach is used when the angle \( \theta \) is small, making calculations easier and more manageable.
It allows us to approximate \( \sin \theta \approx \tan \theta \approx \frac{y}{L} \). Here, \(y\) is the fringe separation, and \(L\) is the distance from the object (or slit) to the screen.
In our exercise, the small angle approximation simplifies the relation between fringe separation and wavelength. This makes finding the angle straightforward without resorting to complex trigonometric functions. This technique is incredibly valuable in optics and other fields where small angles are common.
fringe separation
Fringe separation is a major output in diffraction experiments. It measures the distance between two adjacent dark or bright bands on a screen, observed after light passes through a slit or around an object.
The formula \( \frac{y}{L} = \tan \theta \) is used, where \(y\) is the fringe separation and \(L\) is the distance from the slit to the viewing screen.
  • In our exercise, the fringe separation was 5.22 cm, or 0.0522 meters.
  • This separation is critical for determining the thickness of the diffracting object using the diffraction formula.
Accurate measurement of fringe separation is essential for calculating other attributes in a diffraction setup, making it a key concept in our exercise.

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Most popular questions from this chapter

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

Laser light of wavelength \(500.0 \mathrm{nm}\) illuminates two identical slits, producing an interference pattern on a screen \(90.0 \mathrm{~cm}\) from the slits. The bright bands are \(1.00 \mathrm{~cm}\) apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. Calculate the wavelength of the light.

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (380-750 nm). The Arecibo radio telescope in Puerto Rico is 305 m (1000 ft) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength 75 cm. (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

Monochromatic light of wavelength \(\lambda\) = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity \(I_0\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) 1.00 mm; (b) 3.00 mm; (c) 5.00 mm?

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