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Chapter 36: Problem 45

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 thickness (a) of the hair was found to be approximately \(1.78 \times 10^{-4}\) meters, or 178 micro-meters.

Step by step solution

01

Understand the basics of diffraction

When light waves encounter a barrier (in this case, it's a strand of human hair) with an opening on the order of the wavelength of the light, the waves bend around the barrier and interferes. This causes a pattern of bright and dark bands to be formed, known as a diffraction pattern. The formula used to calculate the dimensions involved in the diffraction pattern is derived from the geometry of the situation and is given by \(sin(\Theta) = \frac{m \lambda}{a}\) where m is the order of the fringe (in this case 1 for the first dark fringe), \( \lambda \) is the wavelength of the light, and a is the width of the slit (or in our case, the thickness of human hair).
02

Calculate the angle using given data

Given that the screen is 1.25m away and the first dark fringes on either side of the central bright spot were 5.22cm apart, which means the distance from the center to frist dark fringe is \(d=5.22/2 cm = 0.0261m\). \n\nThen, the angle theta \(\Theta\) that light has been diffracted can be calculated as \(\Theta = tan^{-1} \frac{d}{L} = tan^{-1}(\frac{0.0261}{1.25})\).
03

Finishing up

Solve for thickness of the hair using our initial formula \(sin(\Theta) = \frac{m \lambda}{a}\), we can get a by \( a= \frac{m \lambda}{sin(\Theta)} = \frac{1 \times 632.8e-9}{sin(tan^{-1}(\frac{0.0261}{1.25}))}\), which gives us the thickness of the hair.

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Key Concepts

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

Diffraction Patterns
When light encounters an obstacle or a slit that is similar in size to its wavelength, a fascinating phenomenon occurs known as diffraction.
As the light waves bend around the obstacle, they interfere with each other, producing patterns of alternating light and dark areas.
This is called a diffraction pattern. Imagine these patterns as ripples on a pond, where light waves add together (constructive interference) to make bright fringes, and cancel each other out (destructive interference) to create dark fringes.
  • Bright Fringes: Where the waves align in phase.
  • Dark Fringes: Where the waves are out of phase.
This alternating fringe pattern is what researchers measure to analyze various properties like the thickness of a hair or the width of a slit.
Light Wavelength
Wavelength is a crucial factor in understanding diffraction patterns.
It refers to the distance between two consecutive peaks (or troughs) of a wave. In this problem, the wavelength is given as 632.8 nm, which is typical for visible red light.
The wavelength determines how much the light will diffract when passing by an object.
  • Shorter Wavelengths: Less diffraction, sharper shadows.
  • Longer Wavelengths: More diffraction, broader patterns.
The wavelength's size compared to the obstacle defines how pronounced the diffraction pattern will be.
Thin Object Diffraction
This concept takes the idea of diffraction and applies it to thin objects like hairs or wires.
Just like slits, these objects can cause light to bend around them, forming a diffraction pattern on a screen.
The width of the object, rather than a conventional slit, plays a pivotal role here. The size of the object with respect to the wavelength of the light is critical.
  • Thin Object: Acts as a barrier causing diffraction.
  • Light Bending: The light bends around the object similar to a single slit.
In the given problem, the hair acts as such an obstacle, and its thickness can be measured using these diffraction principles.
Fringe Analysis
Analyzing the positions of the fringes in the diffraction pattern helps in calculating the dimensions related to the obstacle.
The formula for single-slit diffraction plays a key role here: \( \sin(\Theta) = \frac{m \lambda}{a} \). Here, \(m\) represents the order of the fringe, where 1 corresponds to the first dark fringe.
  • Central Bright Fringe: The main, intense bright spot.
  • Dark Fringes: Located on either side, are used to calculate angles and thicknesses.
Using the geometry of the situation (distance from center of the pattern to dark fringe and the distance to the screen), we can derive the angle \( \Theta \) and subsequently find the thickness of the object.
This method effectively allows researchers to measure even tiny dimensions with impressive accuracy.

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Most popular questions from this chapter

Parallel rays of monochromatic light with wavelength \(568 \mathrm{nm}\) illuminate two identical slits and produce an interference pattern on a screen that is \(75.0 \mathrm{~cm}\) from the slits. The centers of the slits are \(0.640 \mathrm{~mm}\) apart and the width of each slit is \(0.434 \mathrm{~mm}\). If the intensity at the center of the central maximum is \(5.00 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at a point on the screen that is \(0.900 \mathrm{~mm}\) from the center of the central maximum?

A series of parallel linear water wave fronts are traveling directly toward the shore at \(15.0 \mathrm{~cm} / \mathrm{s}\) on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of \(3.20 \mathrm{~m}\) away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm 61.3 \mathrm{~cm}\) from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

Two satellites at an altitude of \(1200 \mathrm{~km}\) are separated by \(28 \mathrm{~km} .\) If they broadcast \(3.6 \mathrm{~cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits \(/ \mathrm{cm} ?\) (b) What is the second-order deviation of this wavelength? Assume normal incidence.

Quasars, an abbreviation for quasi-stellar radio sources, are distant objects that look like stars through a telescope but that emit far more electromagnetic radiation than an entire normal galaxy of stars. An example is the bright object below and to the left of center in Fig. \(\mathrm{P} 36.58 ;\) the other elongated objects in this image are normal galaxies. The leading model for the structure of a quasar is a galaxy with a supermassive black hole at its center. In this model, the radiation is emitted by interstellar gas and dust within the galaxy as this material falls toward the black hole. The radiation is thought to emanate from a region just a few light-years in diameter. (The diffuse glow surrounding the bright quasar shown in Fig. \(\mathrm{P} 36.58\) is thought to be this quasar's host galaxy.) To investigate this model of quasars and to study other exotic astronomical objects, the Russian Space Agency has placed a radio telescope in a large orbit around the earth. When this telescope is \(77,000 \mathrm{~km}\) from earth and the signals it receives are combined with signals from the ground-based telescopes of the VLBA, the resolution is that of a single radio telescope \(77,000 \mathrm{~km}\) in diameter. What is the size of the smallest detail that this arrangement can resolve in quasar \(3 \mathrm{C} 405,\) which is \(7.2 \times 10^{8}\) light-years from earth, using radio waves at a frequency of \(1665 \mathrm{MHz}\) ? (Hint: Use Rayleigh's criterion.) Give your answer in lightyears and in kilometers.
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