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Chapter 35: Problem 6

Coherent light of wavelength \(500 \mathrm{nm}\) is incident on two very narrow and closely spaced slits. The interference pattern is observed on a very tall screen that is \(2.00 \mathrm{~m}\) from the slits. Near the center of the screen the separation between two adjacent interference maxima is \(3.53 \mathrm{~cm}\). What is the distance on the screen between the \(m=49\) and \(m=50\) maxima?

Short Answer

Expert verified
The distance between the m=49 and m=50 maxima is approximately 3.52 cm.

Step by step solution

01

Converting units

First, convert all units to match each other. Convert the wavelength from nm to m, \(500 \mathrm{nm}\) becomes \(500 \times 10^{-9} \mathrm{m}\) or \(5.0 \times 10^{-7} \mathrm{m}\). Convert the separation between maxima from cm to m, \(3.53 \mathrm{cm}\) becomes \(3.53 \times 10^{-2} \mathrm{m}\) or \(0.0353 \mathrm{m}\).
02

Find the width between the slits

Utilize the equation \(y_m = \frac{m\lambda L}{d}\) to find the slit separation \(d\). However, instead of using the equation once, it is used twice, once for the m1 and once for the m2, where m1 and m2 are the order numbers of adjacent maxima. Set the two equations equal to each other to solve for \(d\):\(\frac{L\lambda m_1}{y_{m1}} = \frac{L\lambda m_2}{y_{m2}}\).\nThis simplifies to \(d = \frac{(m_2 - m_1) \lambda L}{(y_{m2} - y_{m1})}\). Assuming m1 = 0 and m2 = 1, we then substitute the known values to find \(d\), which is approximately 0.0283 m.
03

Calculate the distance between the m=49 and m=50 maxima

To find the distance between the m=49 and m=50 maxima, refer back to the equation \(y_m = \frac{m\lambda L}{d}\). Instead of setting the equations equal this time, subtract the position of the m=49 maxima from the position of the m=50 maxima to find the difference, yielding \(y_{50} - y_{49} = \frac{(50-49)\lambda L}{d}\). Further calculation would yield the distance to be approximately 3.52 cm.

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Key Concepts

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

Wave Interference
Wave interference is a phenomenon that occurs when two or more waves overlap and combine to form a new wave pattern. This can result in areas where the waves either enhance each other (constructive interference) or cancel each other out (destructive interference).

Imagine throwing two pebbles into a pond and watching the ripples from each pebble spread out. When the ripples meet, they will either add up to make a bigger wave if their peaks align (constructive), or they will subtract from each other and flatten out when a peak meets a trough (destructive).

In physics, this principle is used to explain a variety of phenomena, from the patterns created by sound waves to the behavior of light. Interference patterns can be complex and beautiful, and they can tell scientists a lot about the nature of the waves causing them.
Young's Double-Slit Experiment
Young's double-slit experiment dramatically demonstrates wave interference using light. Performed first by Thomas Young in 1801, this experiment shines coherent light (light of a single wavelength) through two closely spaced slits. The light waves emerging from these slits overlap, creating an interference pattern of bright and dark fringes on a screen.

These bright spots, called maxima, are the result of constructive interference, where the light waves add together. The dark spots, known as minima, are where destructive interference occurs, and the waves cancel out. Young's experiment was groundbreaking because it showed light could behave as both a particle and a wave, leading to the development of quantum mechanics and revolutionizing optical physics.
Optical Physics
Optical physics, also known as optics, is a branch of physics that studies the behavior of light and its interactions with matter. It encompasses a variety of phenomena, including reflection, refraction, diffraction, dispersion, and polarization.

In the context of Young's experiment and wave interference, optics helps us understand how light waves can constructively and destructively interfere with each other to produce patterns of light and dark spots. By studying these patterns, optical physicists can infer properties of light, such as its wavelength and frequency, and apply these in technologies ranging from microscopes and telescopes to fiber optic communication and laser surgery.
Wavelength Calculation
Wavelength calculation is often an essential task in optical physics, particularly when analyzing interference patterns. The formula to determine the distance between interference fringes is given by
\[ y_m = \frac{m\lambda L}{d} \]
where \( y_m \) is the distance from the central maximum to the m-th maximum, \( m \) is the order number of the fringe, \( \lambda \) is the wavelength of the light, \( L \) is the distance from the slits to the screen, and \( d \) is the separation between the slits.

Using this formula, if we know the order of the fringes, the wavelength of the light, and the distance to the screen, we can calculate the slit separation or vice versa. Such calculations are vital in experiments and applications involving lasers, diffraction gratings, and even in understanding the limits of optical resolution in microscopes and telescopes.

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Most popular questions from this chapter

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index \(1.432,\) (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength \(550 \mathrm{nm}\) reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

The index of refraction of a glass rod is 1.48 at \(T=20.0^{\circ} \mathrm{C}\) and varies linearly with temperature, with a coefficient of \(2.50 \times 10^{-5} / \mathrm{C}^{\circ} .\) The coefficient of linear expansion of the glass is \(5.00 \times 10^{-6} / \mathrm{C}^{\circ} .\) At \(20.0^{\circ} \mathrm{C}\) the length of the rod is \(3.00 \mathrm{~cm}\) A Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of \(5.00 \mathrm{C}^{\circ} / \mathrm{min} .\) The light source has wavelength \(\lambda=589 \mathrm{nm},\) and the rod initially is at \(T=20.0^{\circ} \mathrm{C}\). How many fringes cross the field of view each minute?

Two thin parallel slits that are \(0.0116 \mathrm{~mm}\) apart are illuminated by a laser beam of wavelength \(585 \mathrm{nm}\). (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{~Hz}\) passes through two thin slits and falls on a screen \(85.0 \mathrm{~cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{~cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

Laser light of wavelength \(510 \mathrm{nm}\) is traveling in air and shines the at normal incidence onto the flat end of a transparent plastic rod that has \(n=1.30 .\) The end of the rod has a thin coating of a transparent material that has refractive index \(1.65 .\) What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?
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