# Can Small Errors in Path Difference Affect Double Slit Interference Patterns?

• wcjy
In summary, a laser is needed to get the double slit experiment to work because ordinary light does not give the correct results. The phase of the light at the top and bottom slits is the same when it arrives at the detector.

#### wcjy

Homework Statement
(c) A double slit experiment is conducted using a laser source with a wavelength of 560 nm. As shown in Figure 3(a) on page 4, the two slits are separated by a distance of 2 mm, and a screen is located at 10 m away from the double slit. The laser light is directed normal to the double slit and forms an interference pattern on the screen. The refractive index of air is 1. Figure 3(a) is not drawn to scale.

(i) Calculate the distance between the third and the fifth dark lines of the interference pattern on the screen.

(ii) Calculate the phase difference between the two waves arriving at point Q which is 15 mm away from the centre point 0 on the screen, as shown in Figure 3(a).

(only ii is needed)
Relevant Equations
$$Δφ = 2π \frac{L_2 - L_1}{λ}$$
$$Δφ = 2π \frac{L_2 - L_1}{λ}$$
$$Δφ = 2π \frac{\sqrt{0.016^2+10^2}-\sqrt{0.014^2+10^2}}{560*10^{-9}}$$
$$Δφ = 33.659 rad$$Answer = 2.24 rad (unsure if the answer provided is correct/wrong)

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I didn't see your calculation, but, you know ... $$33.659 = 2.24 + 5*2\pi$$

wcjy
I didn't check your calculation because I cannot remember if the formula you give is right or wrong.

You don't need to apply a formula blindly to solve (ii) if you think about the physics of what is actually happening.

The slits are very wide apart compared with a normal double slit experiment and so, to get the experiment to work, it needs a laser and not an ordinary light.

So why is a laser needed? What does a laser give that an ordinary light does not give?

As it's a laser, what do you know about the phase of the light at the top slit and the phase of the light at the bottom slit?

So what is the phase of "the light from the top slit" when it arrives at Q? How far has it traveled from the slit? How many wavelengths is this?

So what is the phase of "the light from the bottom slit" when it arrives at Q? How far has it traveled from the slit? How many wavelengths is this?

Subtract and you have solved it.

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wcjy
Frodo said:
So why is a laser needed? What does a laser give that an ordinary light does not give?

Is it there will be no diffraction?
Frodo said:
As it's a laser, what do you know about the phase of the light at the top slit and the phase of the light at the bottom slit?

Frodo said:
So what is the phase of "the light from the top slit" when it arrives at Q? How far has it traveled from the slit? How many wavelengths is this?

So what is the phase of "the light from the bottom slit" when it arrives at Q? How far has it traveled from the slit? How many wavelengths is this?

Using top slit, the distance travelled, L1 can be found using pythagoras theorem. Number of wavelength can be found by taking the total L1 length and dividing by the wavelength of the laser source. Then to know the phase, just multiply by 2π.

Likewise, I did the same thing for the bottom slit. Then i subtract them.

However, I get the same thing though. I think I could be missing something out, or my understanding could be wrong. Need to figure out which.

The point about the laser is that laser light is coherent - all the light is in step or in phase. So, the phase of the light arriving at the top slit is the same as the phase of the light arriving at the bottom slit. Hence, the phase of the light leaving the top slit and the phase of the light leaving the bottom slit are both the same.

Both beams start their journeys to Q at the same phase - for simplicity, assume it is zero.

Calculate the difference in the path lengths and express it in wavelengths - it will be a number like 5.36 wavelengths. So, the upper slit light travels 5.36 wavelengths further than the lower slit light so it is 0.36 wavelengths "out of step" or the phase difference is 0.36 wavelengths. As one wavelength is equal to 2 pi you have the phase difference.

I think you will now understand much more about the experiment because you have thought about what is happening. Applying a formula without knowing what is happening is a hopeless mistake to make. If you understand the physics you don't need to remember a lot of formulae! You just work out the problem from first principles. I couldn't remember the formula but I did not need it.

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wcjy
Frodo said:
The point about the laser is that laser light is coherent - all the light is in step or in phase. So, the phase of the light arriving at the top slit is the same as the phase of the light arriving at the bottom slit. Hence, the phase of the light leaving the top slit and the phase of the light leaving the bottom slit are both the same. Both beams start their journeys to Q at the same phase - for simplicity, assume it is zero.

Calculate the difference in the path lengths and express it in wavelengths - it will be a number like 5.36 wavelengths. So, the upper slit light travels 5.36 wavelengths further than the lower slit light so it is 0.36 wavelengths "out of step" or the phase difference is 0.36 wavelengths. As one wavelength is equal to 2 pi you have the phase difference.
ohhh ok i finally get it! Thanks so much!

You have had that Aha! moment when you have understood something.

You can now solve any problem like this no matter how much the question is dressed up to make it look more difficult.

Remember what Rutherford said: " All of physics is either impossible or trivial. It is impossible until you understand it, and then it becomes trivial".

How much would you have learned if I had just given you the formula? Nothing!

Note also you can, by inspection of the diagram, work out which beam is ahead of the other so you know the sign of the phase. Always check any answer you calculate with the physics - does it make sense? is the sign correct?

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Incidentally, you can see how impractical the question actually is.

The path difference is ~5.3 wavelengths of light and the answer is the ~0.3 wavelengths of light.

So, to be accurate, the screen has to be perpendicular to a phenomenal accuracy. If it is just the tiniest bit non-perpendicular, it has a huge effect on the distances from the slits to Q, and the 2.24 radians will change enormously.

My bad. A quick calculation shows it isn't very sensitive even to a 0.1 degree error (4 sig fig) nor to a 1mm error in the 10m distance (3 sig fig).

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