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Chapter 26: Problem 57

\(\bullet$$\bullet\) Thickness of human hair. 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?

Short Answer

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

Step by step solution

01

Understand the Context

The problem involves single-slit diffraction where the thin object (hair) causes the light to diffract. The formula for diffraction minima is similar to that for a single slit: \( a \cdot \sin \theta = m \cdot \lambda \), where \(a\) is the width (thickness) of the strand, \(m\) is the order of the minima, \(\lambda\) is the wavelength, and \(\theta\) is the diffraction angle.
02

Identify Given Values

We are given the light wavelength \( \lambda = 632.8 \text{ nm} = 632.8 \times 10^{-9} \text{ m} \), the distance to the screen \( D = 1.25 \, \text{m} \), and the separation of the first dark fringes \( x = 5.22 \, \text{cm} = 0.0522 \, \text{m} \). The order \( m = 1 \) for the first minimum.
03

Calculate the Angle for Minima

Use the approximation \( \tan \theta \approx \sin \theta \approx \theta \) for small angles, where \( \theta \approx \frac{x}{2D} \). Calculate \( \theta = \frac{0.0522}{2 imes 1.25} \approx 0.02088 \, \text{rad}\).
04

Apply Diffraction Formula

Substitute the known values into the diffraction formula \( a \cdot \sin \theta = m \cdot \lambda \). Since \( \sin \theta \approx \theta \), we have \( a \cdot 0.02088 = 1 \cdot 632.8 \times 10^{-9} \).
05

Solve for Thickness

Rearrange to find the thickness \( a = \frac{632.8 \times 10^{-9}}{0.02088} \approx 3.03 \times 10^{-5} \text{ m} \). Convert to micrometers: \( a \approx 30.3 \text{ micrometers} \).

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Key Concepts

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

Diffraction Formula
Single-slit diffraction occurs when light waves encounter an obstacle, like a thin strand of hair, and bend around it. This process creates a pattern of light and dark fringes due to interference. The diffraction formula helps us predict these patterns.
  • The equation used is: \( a \cdot \sin \theta = m \cdot \lambda \)
  • Here, \( a \) represents the width of the obstacle, \( m \) is the order of the minimum (fringe), and \( \lambda \) is the wavelength of light.
  • The angle \( \theta \) is the diffraction angle, formed between the original light path and the direction of the diffracted light.
This formula is crucial for calculating various physical dimensions, such as the thickness of a human hair, from light patterns.
Wavelength of Light
The wavelength of light, represented by \( \lambda \), is the distance between two consecutive peaks of a wave. It is a key factor in determining the behavior of light, especially in diffraction.
  • In our exercise, the wavelength given is \( 632.8 \) nanometers, which is a common wavelength for red light in experiments.
  • Knowing the wavelength helps us relate the physical properties of waves, like color and energy, to the patterns they produce.
A precise understanding of light's wavelength allows us to calculate dimensions, such as the hair thickness, when used in conjunction with the diffraction formula.
Calculation of Hair Thickness
To calculate the thickness of a strand of hair, we apply the diffraction formula mentioned earlier and solve for \( a \), the width of the strand.
  • Start by using the known values: wavelength \( \lambda = 632.8 \times 10^{-9} \) meters, and the angle \( \theta = 0.02088 \) radians (calculated from the fringe separation).
  • Substitute into the formula: \( a \cdot 0.02088 = 632.8 \times 10^{-9} \).
  • Rearrange to find \( a \): \( a = \frac{632.8 \times 10^{-9}}{0.02088} \approx 3.03 \times 10^{-5} \) meters, or approximately \( 30.3 \) micrometers.
This step-by-step calculation enables us to find an otherwise difficult-to-measure physical characteristic, like hair thickness.
Small Angle Approximation
The small angle approximation is a mathematical simplification used in diffraction calculations to make complex trigonometric equations more manageable.
  • The assumption is that for small angles, \( \sin \theta \approx \tan \theta \approx \theta \). This means the angle in radians is close to the sine of the angle.
  • In our example, \( \theta \) is calculated using this approximation: \( \theta \approx \frac{x}{2D} \), where \( x \) is the fringe separation and \( D \) the distance to the screen.
  • This simplification is especially useful when \( \theta \) is small, which commonly occurs in light diffraction problems.
Using the small angle approximation allows us to solve for the angle quickly, aiding in further calculations such as determining the thickness of thin objects like hair.

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Most popular questions from this chapter

\(\bullet\) Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge, so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546 nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

\(\bullet$$\bullet\) Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first inter- ference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(19.46^{\circ}\) instead. What is the index of refrac- tion of this liquid?

\(\bullet$$\bullet\) Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single fre- quency that can be varied between 300 and 600 Hz. The speed of sound is 340 \(\mathrm{m} / \mathrm{s} .\) You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 \(\mathrm{cm}\) toward you, the sound you hear has maximum inten- sity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

\(\bullet\) Light of wavelength 633 nm from a distant source is inci- dent on a slit 0.750 mm wide, and the resulting diffraction pat- tern is observed on a screen 3.50 \(\mathrm{m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

\(\bullet$$\bullet\) Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?
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