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 µm, and the separation between the slits is 4.83 µm.

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—an 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 (μm), 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

An instructor uses light of wavelength \(633 \mathrm{nm}\) to create a diffraction pattern with a slit of width \(0.135 \mathrm{~mm} .\) How far away from the slit must the instructor place the screen in order for the full width of the central maximum to be \(5.00 \mathrm{~cm} ?\)

A two-slit apparatus is covered with a red \((670 \mathrm{nm})\) filter. When white light is shone on the filter, on the screen beyond the two-slit apparatus, there are nine interference maxima within the 4.50 -cm-wide central diffraction maximum. When a blue \((450 \mathrm{nm})\) filter replaces the red, how many interference maxima will there be in the central diffraction maximum, and how wide will that diffraction maximum be?

You are making a diffraction grating that is required to separate the two spectral lines in the sodium \(D\) doublet, at wavelengths 588.9950 and \(589.5924 \mathrm{nm}\), by at least \(2.00 \mathrm{~mm}\) on a screen that is \(80.0 \mathrm{~cm}\) from the grating. The lines are to be ruled over a distance of \(1.50 \mathrm{~cm}\) on the grating. What is the minimum number of lines you should have on the grating?

Suppose the distance between the slits in a double-slit experiment is \(2.00 \cdot 10^{-5} \mathrm{~m} .\) A beam of light with a wavelength of \(750 \mathrm{nm}\) is shone on the slits. What is the angular separation between the central maximum and the adjacent maximum? a) \(5.00 \cdot 10^{-2} \mathrm{rad}\) b) \(4.50 \cdot 10^{-2} \mathrm{rad}\) c) \(3.75 \cdot 10^{-2} \mathrm{rad}\) d) \(2.50 \cdot 10^{-2} \mathrm{rad}\)

White light shines on a sheet of mica that has a uniform thickness of \(1.30 \mu \mathrm{m} .\) When the reflected light is viewed using a spectrometer, it is noted that light with wavelengths of \(433.3 \mathrm{nm}, 487.5 \mathrm{nm}, 557.1 \mathrm{nm}, 650.0 \mathrm{nm}\), and \(780.0 \mathrm{nm}\) is not present in the reflected light. What is the index of refraction of the mica?

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