Please help with optics question

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The last term can be simplified using the identity cosx+cosy = 2cos((x+y)/2)cos((x-y)/2). |E|^2 = Eo^2 [cos^2αcos^2wt+sin^2αsin^2wt+sin^2α+sin^2α+4sinαsinwtcosαcoswtcos(δ/2)cos(δ/2+δ/2+δ/2)]Simplifying further, we get:|E|^2 = Eo^2 [cos^2αcos^2wt+sin^2αsin^2wt+sin^2α
  • #1
cuti_pie75
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Imagine that you have an opaque screen with three horizontal very narrow parallel slits in it. The second is a center-to-center distance a beneath the first, and the third is a distance 5a/2 beneath the first. (a)Write a complex exponential expression in terms of δ for the amplitude of the electric field at some point P at an elevation θ on a distant screen where δ=ka sinθ. Prove that I(θ) = I(0)/3 + 2I(0)/9 (cos δ + cos 3δ/2 + cos 5δ/2)
Verify that at θ=0, I(θ)=I(0)

I think i got the first part (a) right for the exponential expression.
E=Eo Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)]

But my problem is that i don't know how to prove this Intensity equation.

So any help would be greatly appreciated... :frown:
 
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  • #2


Hello,

I can help you with proving the intensity equation. Let's start by breaking down the exponential expression you have written for the electric field at point P on the distant screen.

E = Eo Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)]

First, we can simplify the exponent by using the trigonometric identity e^ix = cosx + isinx.

E = Eo Re[e^iα(coswt+isinwt)(1+e^-iδ +e^i3δ/2)]

Next, we can distribute the terms inside the parentheses to get:

E = Eo Re[cosα(coswt+isinwt)+isinα(coswt+isinwt)(1+e^-iδ +e^i3δ/2)]

Expanding further, we get:

E = Eo Re[cosαcoswt+isinαcoswt+icosαsinwt-sinαsinwt+isinα+isinαe^-iδ +isinαe^i3δ/2]

Using the property Re[iz] = Im[z], we can simplify this to:

E = Eo [cosαcoswt+sinαsinwt+sinαe^-iδ +sinαe^i3δ/2]

Now, let's consider the amplitude of the electric field at point P, which is given by |E|^2.

|E|^2 = Eo^2 [cosαcoswt+sinαsinwt+sinαe^-iδ +sinαe^i3δ/2]^2

Using the property |z|^2 = z*z*, we can simplify this to:

|E|^2 = Eo^2 [cosαcoswt+sinαsinwt+sinαe^-iδ +sinαe^i3δ/2][cosαcoswt+sinαsinwt+sinαe^-iδ +sinαe^i3δ/2]*

Expanding and simplifying, we get:

|E|^2 = Eo^2 [cos^2αcos^2wt+sin^2αsin^2wt+sin^2α+sin^2α+2sinαsinwtcosαcoswt+2sinαsinwtcosαcoswtcosδ+2sinαsin
 
  • #3


To prove the intensity equation, we first need to understand what it represents. Intensity, denoted by I, is the power per unit area of an electromagnetic wave. In this case, it represents the power per unit area of the electric field at point P on the distant screen.

We can use the expression for the electric field, E, that you have derived in part (a) to find the intensity, I. The intensity is given by the square of the magnitude of the electric field, so we have:

I = |E|^2 = E*E

Substituting the expression for E that you have derived, we get:

I = E*E = (Eo Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)]) * (Eo Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)])

Expanding the expression and using the fact that |e^ix| = 1, we get:

I = E*E = Eo^2 Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)] * Re[e^-i(α-wt) (1+e^iδ +e^-i3δ/2)]

Using the trigonometric identity cos(A+B) = cosAcosB - sinAsinB, we can simplify the expression to:

I = E*E = Eo^2 Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)] * (1 + cos δ + cos 3δ/2)

Now, we need to use the fact that the intensity is proportional to the square of the amplitude of the electric field. This means that I is directly proportional to Eo^2, so we can write:

I = E*E = Eo^2 Re[e^i(α-wt) (1+e^-iδ +e^i3δ/2)] * (1 + cos δ + cos 3δ/2)
= Eo^2 [Re[e^i(α-wt)]^2 (1+e^-iδ +e^i3δ/2)^2] * (1 + cos δ + cos 3δ/2)
= Eo^2 [1
 

1. What is optics?

Optics is a branch of physics that deals with the behavior and properties of light, including its interactions with matter and the instruments used to detect and manipulate it.

2. What is the question trying to solve?

The question is trying to solve a problem related to optics, such as understanding a concept or calculating a measurement.

3. Can you explain the difference between reflection and refraction?

Reflection is when light bounces off a surface, while refraction is when light passes through a medium and changes direction due to a change in its speed.

4. How do you calculate the speed of light in a specific medium?

The speed of light in a specific medium can be calculated by dividing the speed of light in a vacuum by the refractive index of the medium.

5. What are some real-life applications of optics?

Optics has many real-life applications, such as in the design of lenses for glasses and cameras, in the development of fiber optic communication systems, and in medical imaging techniques like MRI and ultrasound.

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