Finding the minimas from EM radiation.

In summary: Supposing we have constructive interference on the x-axis at a point x. What is the path difference between the rays from the two sources when they reach x? The path difference between the rays from the two sources when they reach x would be nt=kλ.
  • #1
sg001
134
0

Homework Statement



6. S1 and S2 in the diagram are effective
point sources of radiation, excited by
the same oscillator. They are coherent
and in phase with each other. Placed 4.0 y m apart, they emit equal amounts of
power in the form of 1.0 m wavelength electromagnetic waves.
(a) Find the positions of the first (that is, the nearest), the second, and the third maxima of the received signal, as the detector is moved out along axis OX.
(b) Is the intensity at the nearest minimum equal to zero? Justify your answer.
[Ans: (a) 7/6m, 3m, 7.5m; (b) No]


the diagram looks like this


^y
|
|
|
|*S2
|
|
|
|
|
O*S1______________________>X
where d is the distance between S1 and S2


Homework Equations





The Attempt at a Solution



(a)
let dx = the distance along the x-axis to the respective minima.

so sin θ = d/√(dx2+d2)

for maxima's

dsinθ=mλ

for m=0 (the first maxima)

dsinθ = 0

so x =1 -d2

x =√15 m

but this is wrong but I'm not sure where I have gone wrong.. it should be 7/6 m

for m=0.
 
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  • #2
This is not the same situation as with a slit, interference of the diffracted waves.

You have two different waves from different sources.
What is the condition of constructive interference when two waves meet? ehild
 
  • #3
ehild said:
This is not the same situation as with a slit, interference of the diffracted waves.

You have two different waves from different sources.
What is the condition of constructive interference when two waves meet?


ehild

ok so use 2nt = mλ
 
  • #4
that they are in phase...

so x=mλ
 
  • #5
What is your reasoning behind your answer for (b) No?
 
Last edited:
  • #6
sg001 said:
that they are in phase...

so x=mλ

Yes, a single wave has minima and maxima, but we speak about constructive interference of two waves. You said that the waves must be in phase. That is not quite true. The phase difference between them has to be integer multiple of 2pi. That corresponds an optical path difference nt=kλ, where n is the refractive index of the medium, t is the path difference, and k is an integer.
Supposing we have constructive interference on the x-axis at a point x. What is the path difference between the rays from the two sources when they reach x?

ehild
 

FAQ: Finding the minimas from EM radiation.

How do you find the minimas from EM radiation?

To find the minimas from EM radiation, you can use a process called spectroscopy. This involves analyzing the wavelengths of the radiation and identifying the points where the intensity is at its lowest. These points correspond to the minimas.

Why is it important to find the minimas from EM radiation?

Identifying the minimas from EM radiation can provide valuable information about the properties of the radiation, such as its frequency and energy. This can help us better understand the behavior and interactions of EM radiation in various environments.

What factors can affect the minimas in EM radiation?

The minimas in EM radiation can be affected by several factors, including the type of radiation, the medium through which it travels, and any obstacles or barriers it encounters. Additionally, changes in temperature and pressure can also impact the minimas.

Can the minimas in EM radiation be manipulated or controlled?

Yes, it is possible to manipulate or control the minimas in EM radiation. This can be achieved through the use of various techniques, such as adjusting the properties of the medium or using specialized materials that can alter the behavior of the radiation.

What is the significance of the minimas in EM radiation for practical applications?

The minimas in EM radiation have practical applications in fields such as telecommunications, medical imaging, and astronomy. By understanding and manipulating the minimas, we can develop technologies that use EM radiation for various purposes, such as transmitting signals or creating detailed images of the human body or distant objects in space.

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