# Min. Phase Diff.: Find x_p at Wavelength λ, 2 Sources S1 & S2

• noppawit
In summary: For the first dark point, the path difference is 1/2 wavelength. Using the formula above, you can solve for the distance x_p.In summary, the problem discusses two isotropic point sources separated by a distance of 6.00λ, emitting light in phase at the same amplitude and wavelength. The sources lie on the x-axis, with a viewing screen located at a distance of 2.00λ. The origin is located on the perpendicular bisector between the sources. The question asks for the minimum possible phase difference at point P on the screen, which can be calculated by solving a quadratic equation using the path lengths from each source to P. The minimum phase difference occurs when the path difference is 1/
noppawit
Two isotropic (radiating the same intensity of EM wave in all directions) point sources S1and S2 emit light in phase at wavelength λ and at the same amplitude. The sources are separated by distance 2d=6.00λ. They lie on the axis that parallel to the x-axis, which runs along a viewing screen at distance D=2.00λ. The origin lies on the perpendicular bisector between the sources. The figure below shows two rays reaching point P on the screen, at position xp

At what value of xp do the rays have the minimum possible phase difference?

http://www.wisheyebio.com/images/screen.png

From question, the minimum possible phase difference, does it mean that the first dark point, or not?

I tried like this...
$$\Delta L=d\sin \theta) = (m+0.5)\lambda$$
$$\frac{2dx_{p}}{D} = 0.5\lambda$$
$$\frac{6 \lambda x_{p}}{20\lambda}$$
$$x_{p} = 1.66\lambda$$

Am I correct or not?

Last edited by a moderator:
Your attachment hasn't been approved yet, so I can't see it.
Are the sources a distance 2 wavelengths above the screen?
If the sources are that close to the screen, I would forget the formulas and just work out the distance from each source to P - expressions involving Xp, d and D. Choose Xp so the difference between the distances is one wavelength, to get the first constructive interference spot.

The formulas are approximate and they work well when the distance is large, but not close up.

Delphi is correct, noppawit. The formula you use involves the approximation that d is much smaller than D. That's not true here. So you have got to write down the formulas for the path lengths direct : take the distances between each slit and the point x_p. Constructive interference occurs first when the two path lengths differ by a wavelength.

You have to solve a quadratic equation to get the answer, I reckon.

For the S1 to Xp path, draw a line straight up from S1 to make a right triangle. Use the pythagorean formula to find the length of the path. Do the same for the other path. Write that the longer one is equal to the shorter one plus one wavelength. Solve for the distance Xp.

Does one wavelength represent the minimum possible phase difference?

Sorry to jump in here, but it isn't quite clear to me what is being asked in the original problem statement.

The minimum phase difference is zero. It will occur when the path difference is 0, 1, 2, 3 ... wavelengths.

## 1. What is the purpose of finding xp at a specific wavelength and with two sources?

The purpose of finding xp at a specific wavelength and with two sources is to determine the phase difference between the two sources at that particular wavelength. This information can be used to analyze interference patterns and understand the behavior of waves in different mediums.

## 2. How is the phase difference between two sources calculated?

The phase difference is calculated by finding the difference in path length between the two sources and converting it to a phase angle. This can be done by dividing the path length by the wavelength and multiplying by 360 degrees (or 2π radians).

## 3. Can the phase difference change at different wavelengths?

Yes, the phase difference between two sources can change at different wavelengths. This is because the wavelength affects the path length and therefore the phase angle. In some cases, the phase difference may even be 0 or 180 degrees, resulting in constructive or destructive interference.

## 4. How does the distance between the two sources affect the phase difference?

The distance between the two sources affects the phase difference by changing the path length and therefore the phase angle. As the distance increases, the phase difference also increases. This is why interference patterns change when the distance between sources is altered.

## 5. How is the phase difference used in practical applications?

The phase difference is used in practical applications such as antenna design, signal processing, and holography. It is also important in understanding the behavior of light and sound waves, and can be used to analyze the properties of different materials and mediums.

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