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Chapter 36: Problem 46

Coherent monochromatic light of wavelength l passes through a narrow slit of width \(a\), and a diffraction pattern is observed on a screen that is a distance \(x\) from the slit. On the screen, the width \(w\) of the central diffraction maximum is twice the distance \(x\). What is the ratio \(a/ \lambda\) of the width of the slit to the wavelength of the light?

Short Answer

Expert verified
The ratio \( a/\lambda \) is 1.

Step by step solution

01

Identify the Condition for Central Maximum

The condition for the central maximum in a single-slit diffraction pattern is that the first minima occur where \( a \sin \theta = m \lambda \), with \( m \) as an integer. For the first minima adjacent to the central maximum, \( m = \pm 1 \).
02

Relate Width of Maximum to Angle

The width \( w \) of the central maximum is the angular spread between the first minima on either side. The angle \( \theta \) for the first minima can be given as \( \sin \theta = \lambda / a \).
03

Calculate the Width of the Central Maximum

Since the width \( w \) on the screen is given as twice the distance \( x \), we have \( w = 2x \). This translates to path lengths where \( \theta \) angles sum up as \( 2 \sin \theta \approx \sin(\theta_1) + \sin(\theta_2) \approx \theta_1 + \theta_2 = 2\theta \approx 2\lambda/a \).
04

Substitute and Rearrange

Given \( w = 2x \), substitute the expression \( 2x = 2x \lambda / a \). Simplifying, we get \( 2 \lambda x / a = 2x \), thus \( a = \lambda \), hence \( a/\lambda = 1 \).
05

Conclude the Ratio

Through simplifying the above steps, the ratio of the slit width \( a \) to the wavelength \( \lambda \) results in \( a / \lambda = 1 \).

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Key Concepts

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

Diffraction Pattern
When light passes through a narrow slit, it does not just go straight through in parallel lines. Instead, it spreads out in a pattern that we refer to as a diffraction pattern. This occurs due to the bending of waves around the edges of the slit. The result is a series of dark and light bands on a screen. The light bands, or maxima, are where light waves constructively interfere, and the dark bands, or minima, are where the waves destructively interfere. For a single-slit diffraction, the pattern forms because light waves emanating from different parts of the slit reach the screen at slightly different angles and interfere with each other.
  • The central maximum is the brightest and widest part of the pattern.
  • First-order minima flank the central maximum and are found where the path difference is \(m \lambda\) for \(m = \pm 1\).
Monochromatic Light
Monochromatic light refers to light consisting of a single wavelength. Think of it like a single note played on an instrument rather than a chord of several notes. When this type of light passes through a slit, it forms a distinct diffraction pattern. This is because light of only one wavelength interacts in a consistent way, leading to clear and predictable patterns on a screen.
  • Having a single wavelength allows precise calculations and predictions of the pattern's features.
  • Common sources of monochromatic light in experiments include lasers.
Utilizing monochromatic light simplifies the analysis because there are no overlapping patterns from different wavelengths.
Wavelength
Wavelength is the distance between successive peaks of a wave. In terms of light, it's a measure of the color; shorter wavelengths are towards the violet end of the spectrum, and longer wavelengths are toward the red. In the context of diffraction, the wavelength of the light determines how much the light waves will spread out after passing through a slit.
  • Shorter wavelengths bend less, causing less spreading in the diffraction pattern.
  • Longer wavelengths bend more, resulting in wider patterns.
The relationship between the slit width and the wavelength is crucial: if the slit width is comparable to the wavelength, a noticeable diffraction pattern will emerge.
Central Maximum
The central maximum is the principal bright band in a diffraction pattern. It is the most intense part of the pattern, sitting directly opposite the slit. This intensity is due to constructive interference from light waves that have similar path lengths as they travel through the slit and strike the screen. The width of the central maximum, denoted as \(w\), is determined by the geometry of the setup and how the light diffracts.
  • In a single-slit experiment, the width \(w\) of the central maximum can be observed to be related to the distance to the screen \(x\).
  • The angular spread of this maximum depends inversely on the slit width and directly on the light wavelength \(\frac{\lambda}{a}\).
Understanding the central maximum's formation helps in deducing properties like the wavelength when other parameters are known.

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Most popular questions from this chapter

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (380-750 nm). The Arecibo radio telescope in Puerto Rico is 305 m (1000 ft) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength 75 cm. (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

(a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time \(t\), the electric field at a distant point \(P\) is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where \(E_0\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi = (2\pi d\) sin \(\theta)/\lambda\), \(\theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrangement), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\). (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship \(e^{iz} = cos z + i \space sin \space z\), where \(i = \sqrt{-1}\). In this expression, cos \(z\) is the \(real\) \(part\) of the complex number \(e^{iz}\), and sin \(z\) is its \(imaginary\) \(part\). Show that the electric field \(E_P(t)\) is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that \(e^Ae^B = e^{(A+B)}\) and \((e^A)^n = e^{nA}\), show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship \(e^{iz}\) = cos \(z\) + \(i\) sin \(z\), show that the (real) electric field at point \(P\) is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at \(P\). (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where \(I_0\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N\) = 2. It will help to recall that sin 2\(A\) = 2 sin \(A\) cos \(A\). Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (\(Hint\): Is I0 defined in the same way in both expressions?)

A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm\)38.2\(^\circ\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm\)21.6\(^\circ\). Find the refractive index of the liquid.

On December 26, 2004, a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than 150,000 people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 h and the speed of the wave to be 800 km/h. Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about 4500 km, while the distance between the southern end of Australia and Antarctica is about 3700 km. As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many \(totally\) dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? 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 that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?
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