Range of wavelength from white light through a diffraction grating

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SUMMARY

The discussion centers on calculating the range of visible wavelengths that pass through a 1 cm square hole in a screen positioned 30 cm from a transmission grating with 3150 lines per cm. The relevant equation used is Nλ = d sin θ, where d represents the grating spacing. The user seeks to determine the angle range corresponding to the hole's position, which is 5 cm from the zeroth order. The solution involves applying the derived formula for wavelength variation based on the angle.

PREREQUISITES
  • Understanding of diffraction gratings and their properties
  • Familiarity with the equation Nλ = d sin θ
  • Basic knowledge of trigonometry and angles
  • Concept of collimated light and its behavior through optical devices
NEXT STEPS
  • Study the principles of diffraction and interference patterns
  • Learn about the calculation of grating spacing and its impact on wavelength dispersion
  • Explore the application of the formula Δλ/θΔ = d/N in practical scenarios
  • Investigate the characteristics of visible light wavelengths and their measurement techniques
USEFUL FOR

Students in physics, optical engineers, and anyone involved in the study of light behavior through diffraction gratings and wavelength analysis.

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Homework Statement



A narrow beam of collimated white light falls at normal incidence upon a transmission grating with 3150 lines per cm. A spectrum is formed by the grating on a screen 30 cm away. If a 1 cm square hole is cut in the screen its inner edge being 5 cm from the zeroth order, what range of visible wavelengths passes through the hole?

Homework Equations



N\lambda = dsin\theta

The Attempt at a Solution



\frac{\Delta\lambda}{\theta\Delta} = \frac{d}{N}\sqrt{1 - sin^2\theta}
 
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What is the range of angle determined by the square hole ?

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