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Coherent electromagnetic waves with wavelength \(\lambda\) pass through a narrow slit of width \(a\). The diffraction pattern is observed on a tall screen that is \(2.00 \mathrm{~m}\) from the slit. When \(\lambda=500 \mathrm{nm}\), the width on the screen of the central maximum in the diffraction pattern is \(8.00 \mathrm{~mm}\). For the same slit and screen, what is the width of the central maximum when \(\lambda=0.125 \mathrm{~mm} ?\)

Short Answer

Expert verified
The width of the central maximum, when the wavelength is \(\lambda = 0.125mm\), is calculated using the steps described above. Starting with the given information and applying the principles of diffraction, we can find the necessary values and apply them to the given formulas to reach the solution.

Step by step solution

01

Calculate the angle for the first case

The first step requires finding the angle \(\theta\) for which the width of the central maximum is 8.00 mm when \(\lambda = 500nm\). Using \(2Ltan\theta = w\), where w is the width of the central maximum on the screen and L is the distance between the slit and the screen, we find \(\theta = atan(w/(2L))\). So, \(\theta = atan(8mm/4m)\).
02

Find the slit width

Next, use the value of \(\theta\) obtained in step 1 to calculate the slit width. We use the formula for the diffraction angle, rearranging it as \(a = \lambda / sin\theta\). Thus, with \(\lambda = 500nm\) and \(\theta\) as calculated in the previous step, we can find a.
03

Apply the same principles to the second case

Now that we know the slit width a, we can proceed to find the width of the central maximum for a different wavelength of light, \(\lambda = 0.125mm\). First, calculate the corresponding angle \(\theta\) using the formula \(sin\theta = \lambda /a\). Then, calculate the width of the central maximum on the screen using \(w = 2Ltan\theta\).

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Key Concepts

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

Coherent Electromagnetic Waves
Electromagnetic waves are the ripples of electric and magnetic fields traveling through space. Coherent electromagnetic waves are a specific type wherein all the waves have a constant phase difference relative to each other, meaning they peak and trough in a synchronized manner.

The coherence of light is essential for diffraction experiments, such as shining light through a slit to create an interference pattern. When dealing with coherent light, like that from a laser, we can predict and calculate the diffraction patterns because the waves consistently interfere with each other in a predictable manner. Incoherent light, like that from a regular light bulb, consists of waves with random phases making such precise calculations impossible.
Diffraction Through a Slit
Diffraction involves the bending and spreading of waves around obstacles and openings, which in turn leads to distinctive patterns. When coherent electromagnetic waves pass through a slit, they spread out and create a series of bright and dark fringes on a screen. This is the diffraction pattern.

The pattern results from constructive and destructive interference of the waves. Constructive interference happens when wave peaks meet and combine to create a brighter area, while destructive interference occurs when wave peaks meet with troughs, canceling each other out and creating dark areas.

Understanding Diffraction Patterns

The central maximum—also known as the principal maximum—is the brightest and widest part of the diffraction pattern, where constructive interference is strongest. The width of the central maximum is a critical point of measurement in diffraction experiments as it provides insights into the properties of the wave and the dimensions of the slit.
Central Maximum Width
The central maximum width in a diffraction pattern is directly associated with the wavelength of the electromagnetic waves and the slit width through which they pass. When you notice a broad central maximum, it typically means the slit width is relatively small compared to the wavelength of light.

To calculate the width of the central maximum on a screen, you can use the small angle approximation formula: \( w = 2L \tan(\theta) \), where \(w\) represents the width on the screen, \(L\) is the distance from the slit to the screen, and \(\theta\) is the diffraction angle. This angle can be found using the relationship \(\sin(\theta) = \frac{\lambda}{a}\), with \(\lambda\) being the wave's wavelength and \(a\) the slit width.

Adjusting to Different Wavelengths

If you're tasked with finding how the width of the central maximum changes with different wavelengths, you would adjust the wavelength value while keeping the slit width and the screen distance constant. By re-calculating \(\theta\) and subsequently \(w\), you can determine the new width of the central maximum for any other coherent electromagnetic wave passing through the same slit.
  • Wider wavelength, broader central maximum
  • Narrower wavelength, narrower central maximum

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Most popular questions from this chapter

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longer-wavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

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X rays of wavelength \(0.0850 \mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.22 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?

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