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Chapter 34: Problem 46

A red laser pointer with a wavelength of \(635 \mathrm{nm}\) is shined on a double slit producing a diffraction pattern on a screen that is \(1.60 \mathrm{~m}\) behind the double slit. The central maximum of the diffraction pattern has a width of \(4.20 \mathrm{~cm}\) and the fourth bright spot is missing on both sides. What is the size of the individual slits, and what is the separation between them?

Short Answer

Expert verified
Based on the given problem, we found that the size of each individual slit is 3.623 碌m, and the separation between the slits is 4.83 碌m.

Step by step solution

01

Calculate the distance between the central maximum and the first bright fringe

Given that the width of the central maximum (which spans from the first dark fringe on one side to the first dark fringe on the other side) is 4.20 cm, we can find the distance from the central maximum to the first bright fringe as half of this width: $$ y_1 = \frac{1}{2} \times (4.20\,\text{cm}) = 2.10\,\text{cm} $$
02

Use the double-slit interference formula to find the separation between slits

The double-slit interference formula is: $$ y_m = \frac{m\lambda L}{d} $$ where \(y_m\) is the distance between the central maximum and the \(m\)th bright fringe, \(\lambda\) is the wavelength of the light, \(L\) is the distance from the double slits to the screen, \(m\) is the order of the bright fringe, and \(d\) is the separation between the slits. For \(m=1\) (the first bright fringe) and rearranging the formula for \(d\), we get: $$ d = \frac{1\lambda L}{y_1} $$ Substituting the given values, we find the separation \(d\): $$ d = \frac{1 \times (635\,\text{nm}) \times (1.60\,\text{m})}{2.10\,\text{cm}} = \frac{1 \times (635 \times 10^{-9}\,\text{m})\times (1.60\,\text{m})}{0.021\,\text{m}} = 4.83 \times 10^{-6}\,\text{m}\,\,(4.83\,\mathrm{\mu m}) $$
03

Use the single-slit diffraction formula to find the width of the individual slits

The single-slit diffraction formula is: $$ x_n = \frac{2n-1}{2} \frac{\lambda L}{a} $$ where \(x_n\) is the position of the missing bright spot on the screen, \(n\) is the dark fringe order, and \(a\) is the width of each slit. We are given that the fourth bright spot is missing, which corresponds to \(n=2\): $$ x_{2} = \frac{2(2)-1}{2} \frac{\lambda L}{a} $$ We have the double-slit interference formula: $$ y_m = \frac{m\lambda L}{d} $$ Comparing both formulas and knowing that at the position of the missing fourth bright spot, \(x_2 = y_4,\) we can write: $$ \frac{2(2)-1}{2} \frac{\lambda L}{a} = \frac{4\lambda L}{d} $$ Rearranging the formula for \(a\) and substituting the known values, we get: $$ a = \frac{d(2(2)-1)}{4} = \frac{(4.83 \times 10^{-6}\,\text{m})(3)}{4} = 3.623 \times 10^{-6}\,\text{m}\,\,(3.623\,\mathrm{\mu m}) $$
04

State the size and separation of the slits

The size of each individual slit is \(a = 3.623\,\mathrm{\mu m}\), and the separation between the slits is \(d = 4.83\,\mathrm{\mu m}\).

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Key Concepts

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

Interference and Diffraction
The phenomena of interference and diffraction are fundamental concepts in wave physics and form the basis for understanding various optical patterns created by light. Interference occurs when two or more waves overlap and combine to form a new wave pattern. Depending on the phase difference between the overlapping waves, the interference can be constructive or destructive, leading to an increase or decrease in the wave intensity respectively.

Diffraction, on the other hand, involves the bending of waves around obstacles or the spreading of waves as they pass through openings; in essence, it allows light to apparently 'turn corners'. The most intriguing effects of diffraction are observed when the size of the obstacle or opening is comparable to the wavelength of the light. In such cases, one can observe distinct patterns on a screen, characterized by a series of bright and dark fringes. An important aspect of diffraction is that it demonstrates the wave nature of light.
Double-Slit Experiment
The double-slit experiment is a crucial demonstration of the wave-like properties of light. In this experiment, a beam of light is directed towards two very closely spaced slits. As the light passes through these slits, it diffracts and the resulting wavefronts interfere with each other.

A screen placed behind the double slits captures the resulting pattern of light, known as an interference pattern. This pattern consists of a series of light and dark fringes鈥攁n artifact of constructive and destructive interference, respectively. The central maximum is the brightest fringe located at the center of the pattern, and bright fringes (maxima) and dark fringes (minima) are symmetrically arranged about it. By carefully analyzing the spacing and location of these fringes, one can deduce information about the wavelength of light and dimensions of the slits.
Single-Slit Diffraction Formula
The single-slit diffraction formula is key to understanding the pattern produced when light passes through a single, narrow aperture. The formula is given by:
\[ x_n = \frac{2n-1}{2} \frac{\lambda L}{a} \]
where \( x_n \) is the position of the dark fringe on a screen, \( n \) is the order of the fringe, \( \lambda \) is the wavelength of the light, \( L \) is the distance from the slit to the screen, and \( a \) is the width of the slit. This equation predicts the locations where the intensity of light is minimal or essentially zero due to destructive interference, known as dark fringes. The absence of a bright fringe in the double-slit pattern can be correlated with the position of a dark fringe in the single-slit pattern, thereby providing insights into the characteristics of the slits used in experiments.
Wavelength of Light
The concept of the wavelength of light plays a central role in understanding diffraction and interference patterns. It is defined as the distance between successive crests (or troughs) of a wave and is a critical parameter in the equations that describe light's behavior in wave terms. The wavelength determines the color of light within the visible spectrum and also influences how light diffracts and interferes when it encounters obstacles or openings.

In optical experiments, the wavelength can be used to calculate the dimensions of slits or spacings in gratings that produce particular interference and diffraction patterns, as observed in the double-slit experiment. Wavelength is typically measured in nanometers (nm) or micrometers (渭m), and the solution to the textbook problem uses the known wavelength of a red laser pointer to determine the slit dimensions, demonstrating the interplay between these concepts and their practical applications.

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Most popular questions from this chapter

Some mirrors for infrared lasers are constructed wit alternating layers of hafnia and silica. Suppose you want to produce constructive interference from a thin film of hafnia \((n=1.90)\) on BK-7 glass \((n=1.51)\) when infrared radiation of wavelength \(1.06 \mu \mathrm{m}\) is used. What is the smallest film thickness that would be appropriate, assuming the laser beam is oriented at right angles to the film?

The Michelson interferometer is used in a class of commercially available optical instruments called wavelength meters. In a wavelength meter, the interferometer is illuminated simultaneously with the parallel beam of a reference laser of known wavelength and that of an unknown laser. The movable mirror of the interferometer is then displaced by a distance \(\Delta d,\) and the number of fringes produced by each laser and passing by a reference point (a photo detector) is counted. In a given wavelength meter, a red He-Ne laser \(\left(\lambda_{\mathrm{Red}}=632.8 \mathrm{nm}\right)\) is used as a reference laser. When the movable mirror of the interferometer is displaced by a distance \(\Delta d\), a number \(\Delta N_{\text {Red }}=6.000 \cdot 10^{4}\) red fringes and \(\Delta N_{\text {unknown }}=7.780 \cdot 10^{4}\) fringes pass by the reference photodiode. a) Calculate the wavelength of the unknown laser. b) Calculate the displacement, \(\Delta d\), of the movable mirror.

Coherent, monochromatic light of wavelength \(450.0 \mathrm{nm}\) is emitted from two locations and detected at another location. The path difference between the two routes taken by the light is \(20.25 \mathrm{~cm}\). Will the two light waves interfere destructively or constructively at the detection point?

A Newton's ring apparatus consists of a convex lens with a large radius of curvature \(R\) placed on a flat glass disc. (a) Show that the distance \(x\) from the center to the air, thickness \(d,\) and the radius of curvature \(R\) are given by \(x^{2}=2 R d\) (b) Show that the radius of nth constructive interference is given by \(x_{\mathrm{n}}=\left[\left(n+\frac{1}{2}\right) \lambda R\right]^{1 / 2} .\) (c) How many bright fringes may be seen if it is viewed by red light of wavelength 700\. \(\mathrm{nm}\) for \(R=10.0 \mathrm{~m},\) and the plane glass disc diameter is \(5.00 \mathrm{~cm} ?\)

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?
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