Modified double-slit experiment

[Saketh]

Homework Statement

Suppose that one of the slits of a double-slit interference experiment is wider than the other, so the amplitude of the light reaching the central part of the screen from one slit, acting alone, is twice that from the other slit, acting alone. Derive an expression for the light intensity I at the screen as a function of $\theta$.

Homework Equations

Superposition of waves
Euler's theorem

The Attempt at a Solution

By the superposition of waves, $$\psi = \psi_1+\psi_2$$.

Because one wave has twice the amplitude of the other:
$$\psi_1 = 2Ae^{i(kr_1-\omega t)}$$

$$\psi_2 = Ae^{i(kr_2-\omega t)}$$.

Therefore,
$$\psi = 2Ae^{i(kr_1-\omega t)} + Ae^{i(kr_2-\omega t)}$$

I know how to manipulate this equation into the final form for intensity if the amplitudes of both waves are the same, but with different amplitudes I get stuck at the following step:

$$\psi = 2Ae^{i(kr_1-\omega t)}e^{ik\Delta r /2} \left ( e^{-ik\Delta r /2} + \frac{1}{2}e^{ik\Delta r /2} \right )$$

Normally, I wouldn't have the 1/2 in the second parenthesized term and I could proceed with Euler's theorem, but here I am stumped.

How can I solve this problem?

Thanks for the assistance.

 

[anathema]

Thank you for your question. In this scenario, we can use the principle of superposition to solve for the intensity at the screen. The intensity at any given point on the screen is proportional to the square of the amplitude of the electric field at that point. Therefore, we can rewrite the equation for \psi as:

\psi = 2A^2e^{i(kr_1-\omega t)} + A^2e^{i(kr_2-\omega t)}

Using Euler's theorem, we can rewrite this as:

\psi = A^2e^{i(kr_1-\omega t)} \left (2e^{ik\Delta r /2} + e^{ik\Delta r}\right)

Simplifying further, we get:

\psi = A^2e^{i(kr_1-\omega t)} \left (2e^{ik\Delta r /2} + e^{ik\Delta r /2}e^{ik\Delta r /2}\right)

Since the amplitude of one wave is twice that of the other, we can rewrite this as:

\psi = A^2e^{i(kr_1-\omega t)} \left (2e^{ik\Delta r /2} + \frac{1}{2}e^{ik\Delta r /2}e^{ik\Delta r}\right)

Using Euler's theorem again, we can expand the exponential term to get:

\psi = A^2e^{i(kr_1-\omega t)} \left (2e^{ik\Delta r /2} + \frac{1}{2}e^{ik\Delta r /2}(\cos(k\Delta r) + i\sin(k\Delta r))\right)

Now, we can take the square of the amplitude to get the intensity at any point on the screen:

I = A^4 \left (4 + \frac{1}{4}(2\cos(k\Delta r)+2i\sin(k\Delta r)+\cos^2(k\Delta r)+i\sin^2(k\Delta r))\right)

Simplifying further, we get:

I = A^4 \left (4 + \cos^2(k\Delta r) + \sin^2(k\Delta r)\right)

Using the identity \cos^2(x) + \sin^2

 

[m2003]

Thank you for sharing your work and question. It is important to consider the implications of a modified double-slit experiment, as it can provide valuable insights into the nature of light and its behavior. In this case, we can use the principles of superposition and Euler's theorem to derive an expression for the light intensity at the screen as a function of theta.

Firstly, we can rewrite the equation for the superposition of waves as:

\psi = 2Ae^{i(kr_1-\omega t)} + Ae^{i(kr_2-\omega t)} = Ae^{i(kr_1-\omega t)}(2e^{ik\Delta r/2} + e^{ik\Delta r/2})

Where \Delta r is the distance between the two slits. We can then use Euler's theorem to rewrite the exponential terms as cosine functions:

\psi = Ae^{i(kr_1-\omega t)}(2\cos(k\Delta r/2) + \cos(k\Delta r/2))

Next, we can use the following trigonometric identity to simplify the expression:

2\cos(x) + \cos(x) = 3\cos(x)

Therefore,

\psi = Ae^{i(kr_1-\omega t)}(3\cos(k\Delta r/2))

We can then use the equation for intensity, I = \psi^2, to obtain the final expression for the light intensity at the screen:

I = 9A^2\cos^2(k\Delta r/2)

This shows that the intensity at the screen will vary with theta, as the cosine function is dependent on the angle \theta. This result is consistent with the concept of interference in the double-slit experiment, where the intensity at a given point on the screen is determined by the superposition of waves from the two slits. In this case, the wider slit will contribute more to the overall intensity at the center of the screen, while the narrower slit will contribute less. This is an interesting modification to the traditional double-slit experiment and can provide valuable insights into the principles of light and wave behavior.