# Young's double slit experiment

• tanaygupta2000
In summary: 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 9:1I am not entirely unconvinced, though.
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.

tanaygupta2000 said:
Hence amplitude, a2 = 4a1
How much more energy is getting through the wider slit?

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

tanaygupta2000 said:
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?

haruspex said:
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

tanaygupta2000 said:
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?

haruspex said:
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 ?

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 !

tanaygupta2000 said:
Yes I think the energy will be halved.
Right, so back to the given problem. What energy goes
tanaygupta2000 said:
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

## 1. What is Young's double slit experiment?

Young's double slit experiment is a classic experiment in physics that demonstrates the wave-like nature of light. It involves shining a single beam of light through two parallel slits and observing the resulting interference pattern on a screen.

## 2. What is the significance of Young's double slit experiment?

The significance of Young's double slit experiment is that it provided evidence for the wave theory of light and helped to establish the field of wave optics. It also demonstrated the wave-particle duality of light, as the interference pattern can only be explained by light behaving as both a wave and a particle.

## 3. How does the distance between the slits affect the interference pattern in Young's double slit experiment?

The distance between the slits affects the interference pattern by changing the spacing between the bright and dark fringes. As the distance between the slits increases, the fringes become closer together and the pattern becomes more spread out.

## 4. What happens to the interference pattern if the light source is changed in Young's double slit experiment?

If the light source is changed in Young's double slit experiment, the interference pattern will also change. For example, if the light source is changed from a single wavelength to a broad spectrum of wavelengths, the interference pattern will become less defined and more blurred.

## 5. How is Young's double slit experiment related to the concept of superposition?

Young's double slit experiment is related to the concept of superposition because it demonstrates how two waves can interact and create a new wave pattern through constructive and destructive interference. This is similar to how multiple waves can combine to create a new wave in the principle of superposition.

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