Path differenece of light ray to produce first minima

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SUMMARY

The discussion centers on the conditions necessary for producing the first minima in single-slit diffraction, specifically addressing the path difference of light rays. It is established that the path difference must equal λ/2 (half a wavelength) to achieve destructive interference, resulting in the first minima. The confusion arises from the assertion that a path difference of λ leads to in-phase light rays, which contradicts the conditions for minima. The relationship between slit width (a), angle (θ), and wavelength (λ) is clarified through the equation a sinθ = λ.

PREREQUISITES
  • Understanding of single-slit diffraction
  • Familiarity with the concept of path difference in wave interference
  • Knowledge of the wavelength (λ) and its role in interference patterns
  • Basic grasp of Fraunhofer diffraction principles
NEXT STEPS
  • Study the principles of Fraunhofer diffraction in detail
  • Learn how to derive the conditions for minima in single-slit diffraction
  • Explore the mathematical relationships involving slit width, angle, and wavelength
  • Investigate the effects of varying slit widths on diffraction patterns
USEFUL FOR

Students studying wave optics, physics educators explaining diffraction concepts, and anyone interested in the mathematical foundations of light interference patterns.

desmond iking
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Homework Statement


for part a , in order to produce first minima from the central bright ray , the path differenece between 2 ray should be equal to λ/2 or 180 degree am i right? why the notes give the path
differenece = λ
...If path differenece = λ , the two light ray are in phase right? how can destructive intefrence occur?

Homework Equations





The Attempt at a Solution

 

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In case of a single slit, the waves emerging from the upper half of the slit interfere with those, emerging from the bottom half.

So the ray starting from the upper edge of the slit interferes with the ray from the centre and their path difference should be half wavelength to produce the first minimum on the screen. The ray from the lower edge interferes also with a central ray.

That means (0.5a)sinθ=0.5λ --->a sinθ =λ

The other minima occur where (0.5a)sinθ=(0.5+k)λ that is when asinθ=(2k+1)λ

ehiild
 
Ha, here's Des again, violating copyright and PF rules simultaneously :smile:. Why would I want to help this guy ? Well, because he comes with interesting topics. And he saves his energy for when it really becomes complicated, I suppose.
This time I can be lazy too: just refer him (/her ?) to Fraunhofer diffraction !
 

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