# Young's double slit experiment

tanaygupta2000
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.

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Hence amplitude, a2 = 4a1
How much more energy is getting through the wider slit?

tanaygupta2000
How much more energy is getting through the wider slit?
Energy, E ∝ (amplitude)^2
So, E2 = 16E1

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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?

tanaygupta2000
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

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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?

tanaygupta2000
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 ?

tanaygupta2000
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 !

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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.

• tanaygupta2000