Young's double slit experiment

  • #1
tanaygupta2000
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Homework Statement:
The two slits in a Young’s double-slit experiment are of unequal width, one being four
time wider than the other. If i(max) and i(min) denote the intensities at a neighbouring
maximum and a minimum, then the ratio i(min)/i(max) is: (a)1/9 (b)1/4 (c)3/5 (d)0 ?
Relevant Equations:
Intensity, i ∝ (amplitude)^2
Maximum i = i1 + i2
Minimum i = i1 -iI2
Since slit-2 = 4 × (slit-1)
Hence amplitude, a2 = 4a1
which gives i2 = 16×i1

So i(max) = i2 + i1 = 16i1 + i1 = 17i1
& i(min) = i2 - i1 = 16i1 - i1 = 15i1

=> i(min) /i(max) = 15/17
but there's no such option.
Kindly help me to figure out where am I doing error.
 

Answers and Replies

  • #3
tanaygupta2000
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How much more energy is getting through the wider slit?
Energy, E ∝ (amplitude)^2
So, E2 = 16E1
 
  • #4
haruspex
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Energy, E ∝ (amplitude)^2
So, E2 = 16E1
Does it seem reasonable that 16 times the energy gets through just because the slit is four times the width?
 
  • #5
tanaygupta2000
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Does it seem reasonable that 16 times the energy gets through just because the slit is four times the width?
I think, because if a2 = 4a1 (which I doubt) and since Energy ∝ Intensity ∝ (amplitude)^2
 
  • #6
haruspex
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I think, because if a2 = 4a1 (which I doubt) and since Energy ∝ Intensity ∝ (amplitude)^2
I see no reason why the amplitude should be proportional to the width of the slit.
Suppose I have slit width W, then split it into two slits width W/2. How much of the original energy would go through each?
 
  • #7
tanaygupta2000
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I see no reason why the amplitude should be proportional to the width of the slit.
Suppose I have slit width W, then split it into two slits width W/2. How much of the original energy would go through each?
Yes I think the energy will be halved.
So, I guess the amplitude becomes a/√2 from each ?
 
  • #8
tanaygupta2000
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Okay, I think I got the answer.
i(max) = i2 + i1 = 4i + i = 5i
i(min) = i2 - i1 = 4i - i = 3i
Hence i(min)/i(max) = 3i/5i = 3/5
which is option-(c)

Thank You so much for your help !
 
  • #9
haruspex
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Yes I think the energy will be halved.
Right, so back to the given problem. What energy goes
Okay, I think I got the answer.
i(max) = i2 + i1 = 4i + i = 5i
i(min) = i2 - i1 = 4i - i = 3i
Hence i(min)/i(max) = 3i/5i = 3/5
which is option-(c)

Thank You so much for your help !
No, it's not that simple.
Where the beams from the two slits overlap, the amplitudes add. That is why dark bands appear, positive amplitude plus negative amplitude.

I think the question intends you to argue like this:
- energies from the slits are in the ratio 4:1
- so amplitudes are in the ratio 2:1
- so where they fully reinforce the amplitude reaches 3, and where they maximally cancel the sum is only 1
- this gives an intensity ratio 9:1

I am not entirely unconvinced, though.
 
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