## Modified double-slit experiment

1. The problem statement, all variables and given/known data

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 $\theta$.

2. Relevant equations

Superposition of waves
Euler's theorem

3. 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.
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