anonymous
  • anonymous
A double-slit pattern is observed on a screen 3m from the slits. Given that the light is incident normally and has a wavelength of 390nm, what is the minimum slit seperation for a point 3.8mm from the centre of the middle bright fringe to be A) A maximum b) a minimum
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What have you got so far?
anonymous
  • anonymous
I feel like I'm missing data. I can't use \[dsin \theta= x _{1}-x2\]
anonymous
  • anonymous
the question doesn't even specify if they are in phase or out of phase :/

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anonymous
  • anonymous
Who are 'they' ? And no, you are not missing data, you just need to think a little - I'm sure, you can do it..
anonymous
  • anonymous
the two light rays... if they are out of phase the equation will change.
anonymous
  • anonymous
Its okay I figured it out :) thanks thoug
anonymous
  • anonymous
Well, I'd claim the problem implies, there is only one light source behind the double-slit plate. Also, if different areas of the source are emitting with different phases - independent from one another - which is the case for most sources, the phases of the resulting waves in slit 1 and slit 2 will vary (statistically). However, this will _not_ mess with the intensity of the wave at a given point on the screen, given the variations are synchronous, cause the phase-difference of the outgoing waves (Huygens principle) from S1 and S2 is time-invariant. In this case the two slits are forming two coherent light-sources that will show a time-independent interference-pattern on the screen. (Sry, english is a 2nd language - hope that makes sense...)
anonymous
  • anonymous
You did? Congratulations ;)

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