Intensity and electric field amplitude

In summary, the author is discussing how to find the electric field amplitude at a screen with two slits of different widths. The author states that his approximation is incorrect, and that the solution provides a different answer.
  • #1
LCSphysicist
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Homework Statement
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Relevant Equations
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Hello. I am having some trouble to understand the resolution of this question.

We could easily try to calculate the electric field relative resultant at the screen. The problem i am having is about the amplitude of the electric field:

Generally, we have that the intensity part dependent of the width of the slits is $$I = C(\frac{sin \beta}{\beta})^2$$, where ##\beta = \pi w_i \theta / 2##, so that ##|E| = C' \sqrt{I} = C'' / w_i##, that is, the ampltitude of E is inverselly proportional to w. This would imply that the ##w_2## slit produce a electric field with amplitude weaker than the other one, in fact, $$|E_1|/|E_2|= 2$$.

BUT, the solution says that ##|E_2|/|E_1| = 2##, so where am i wrong?

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  • #2
Herculi said:
##|E| = C' \sqrt{I} = C'' / w_i##, that is, the ampltitude of E is inverselly proportional to w.
Are you sure? You could therefore use a zero-width slit to obtain infinite intensity – the world’s energy problems solved at a stroke!
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html
 
  • #3
Steve4Physics said:
Are you sure? You could therefore use a zero-width slit to obtain infinite intensity – the world’s energy problems solved at a stroke!
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html
Hello. As you can see, i am using a approximation to simplify my problem. Your quoted link says the same thing as i am saying, the intensity is still ##I = \frac{C}{w^2}## Note that, when i say is the amplitude of E is ##E = \frac{C}{w}##, i am ignoring the oscillatory factor that comes with it.

Of course, my approach is ad hoc for this unique problem. If the problem was, as you quoted, something that does not talk about the amplitude of the field that diffract, but the energy itself, i wouldn't disconsider the factor the cos factor. we can note that as x tends to zero, $sinx/x$ will tends to one.
 
  • #4
Herculi said:
... we can note that as x tends to zero, $sinx/x$ will tends to one.
But, as far as I can tell from what you have written, you seem to have applied the (incorrect) appproximation:
as x tends to zero, ##\frac {sinx}{x}## tends to ##\frac 1 x##
which is wrong!

Also, note that the original question is to find δx, the spacing between adjacent maxima. But the solution presented has nothing to do with the original question!
 
  • #5
Steve4Physics said:
But, as far as I can tell from what you have written, you seem to have applied the (incorrect) appproximation:
as x tends to zero, ##\frac {sinx}{x}## tends to ##\frac 1 x##
which is wrong!

Also, note that the original question is to find δx, the spacing between adjacent maxima. But the solution presented has nothing to do with the original question!
Actually, it has something with the question yes. After all, to calculate the relativies intesity at the screen, you will need to calculate the resultand electric field at each point of the screen, and so it is of great importante to know what will be the amplitude of the eletric field of the diffracted wave at each of the slits. SInce after find the eletric field amplitude the solution becomes trivial, the main problem here is to find the amplitude itself.

But, for while i can't see an error on my idea. Even so, the solution provides a diferent answer, and so probably mine is wrong. Since you are knowing better than me why my idea is wrong, i think you could give me a tip of how to deduce the relativity electric field amplitudes after diffract in each slits? Maybe if i understand your argument of how to find the amplitudes, i can see where is the error on mine approach.
 
  • #6
Q1 , ##I = \frac {C}{\omega^2}##, in Post #3, is wrong. Do you understand why?

Q2. Do you undertsand how to find δx (the spacing between maxima) for 2 slits of equal widths? (No electric fields or intensities are needed - it's a geometry problem.)
 
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  • #7
LCSphysicist said:
Homework Statement: .
Relevant Equations: .

View attachment 291222
Hello. I am having some trouble to understand the resolution of this question.

We could easily try to calculate the electric field relative resultant at the screen. The problem i am having is about the amplitude of the electric field:

Generally, we have that the intensity part dependent of the width of the slits is $$I = C(\frac{sin \beta}{\beta})^2$$, where ##\beta = \pi w_i \theta / 2##, so that ##|E| = C' \sqrt{I} = C'' / w_i##, that is, the ampltitude of E is inverselly proportional to w. This would imply that the ##w_2## slit produce a electric field with amplitude weaker than the other one, in fact, $$|E_1|/|E_2|= 2$$.

BUT, the solution says that ##|E_2|/|E_1| = 2##, so where am i wrong?

View attachment 291223
The question given here talks about interference phenomenon.
According to me, the Intensity formula that you have stated is regarding the diffraction phenomenon.
For interference, the wider the slit is, the individual electric field produced will be more. Qualitatively, looking at the values, since the width has doubled, and therefore the amplitude of electric field doubles, and as Intensity is proportional to the square of the electric field, you will get the answer. Finding the individual and resultant fields at a point can be done through geometry using second order approximations.
 
  • #8
omkar260206 said:
The question given here talks about interference phenomenon.
According to me, the Intensity formula that you have stated is regarding the diffraction phenomenon.
For interference, the wider the slit is, the individual electric field produced will be more. Qualitatively, looking at the values, since the width has doubled, and therefore the amplitude of electric field doubles, and as Intensity is proportional to the square of the electric field, you will get the answer. Finding the individual and resultant fields at a point can be done through geometry using second order approximations.
Hi @omkar260206 and welcome to PF.

It might be worth noting that this thread is over 2 years old, so the OP may have moved on.
 

Related to Intensity and electric field amplitude

1. What is the relationship between intensity and electric field amplitude?

The intensity of an electromagnetic wave is directly proportional to the square of the electric field amplitude. This means that as the electric field amplitude increases, the intensity of the wave will increase by a greater amount.

2. How is intensity related to the energy carried by an electromagnetic wave?

The intensity of an electromagnetic wave is a measure of the energy carried by the wave per unit area. A higher intensity means that more energy is being carried by the wave, while a lower intensity indicates less energy.

3. Can the electric field amplitude of a wave be negative?

Yes, the electric field amplitude of an electromagnetic wave can be negative. This simply indicates that the electric field is oscillating in the opposite direction compared to when it is positive.

4. How does the distance from the source affect the intensity and electric field amplitude of a wave?

As an electromagnetic wave travels further away from its source, both the intensity and electric field amplitude will decrease. This is due to the spreading out of the wave over a larger area, resulting in a lower concentration of energy.

5. What units are used to measure intensity and electric field amplitude?

The intensity of an electromagnetic wave is typically measured in watts per square meter (W/m^2), while the electric field amplitude is measured in volts per meter (V/m). These units help quantify the strength and energy of the wave at a given point in space.

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