Finding the width of a diffraction slit

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SUMMARY

The discussion focuses on calculating the width of a diffraction slit that is seven times larger than the wavelength of various electromagnetic waves. The wavelength is determined using the formula λ = c/f, where c is the speed of light (approximately 3 x 10^8 m/s). For the given frequencies: 100 MHz (FM radio), 2.0 GHz (cell phone), 5e14 Hz (red light), and 4e20 Hz (X-rays), the slit widths are calculated as follows: 2.1 meters for FM radio, 0.105 meters for cell phone signals, 0.00042 meters for red light, and 0.0000000525 meters for X-rays. Understanding these calculations is essential for applications in optics and telecommunications.

PREREQUISITES
  • Understanding of electromagnetic wave properties
  • Familiarity with the relationship between frequency and wavelength
  • Basic knowledge of the speed of light (c = 3 x 10^8 m/s)
  • Concept of diffraction in wave physics
NEXT STEPS
  • Study the principles of single-slit diffraction and its mathematical derivations
  • Learn how to calculate wavelength from frequency using λ = c/f
  • Explore applications of diffraction in telecommunications and optics
  • Investigate the effects of slit width on diffraction patterns
USEFUL FOR

Students in physics, optical engineers, telecommunications professionals, and anyone interested in the principles of wave behavior and diffraction.

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Homework Statement


Single-slit diffraction can be observed with any type of electromagnetic wave (not just light). Suppose you want to make a diffraction slit whose width is seven times larger than the wavelength for the following cases. How wide would the slit be?

(a) A radio wave for your favorite FM station (f = 100 MHz)


(b) The waves that carry your cell phone signals (f = 2.0 GHz)


(c) Red light (f = 5e14 Hz)


(d) The X-rays used in your dentist's office (f ≈ 4e20Hz)



Homework Equations





The Attempt at a Solution


I honestly have no idea where to begin with this.
I do know that the answer to a) is 7*3.. but I don't know why?
If anyone can point me in the right direction it would be appreciated.
 
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The relationship between frequency and wavelength is

\lambda=\frac{c}{f}

where c is the speed of light. The problem has nothing to do with diffraction.
 

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