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Since this is one dimension problem and the crack is of width d, and I am given at first that the screen/monitor is placed at infinity (far away)

Therefore I can use Fraunhofer's diffraction equation

where a plane wave is:

$$E = E_0e^{i({kx-\omega t)}}$$

Then the diffraction pattern is given by

$$\int^\frac{d}{2}_\frac{d}{2}E_0e^{i({kx-\omega t)}}e^{-ikx}dx$$

My problem here is that we're left with something constant, which means that the pattern is proportional to the fourier transform of the shape of the crack, which is:

$$t(x)=rect\left(\frac{2x}{d}\right)$$

Still I'm unsure of my solution and would appreciate some feedback.

EDIT unless the integral is:

$$\int^\frac{d}{2}_\frac{d}{2}E_0e^{i({kR-\omega t)}}e^{-ikx}dx$$

since the screen is very far

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While your exact question is unclear to me, the $e^{-ikx}$ in your second equation is unnecessary. Saying the diffraction pattern far away is the Fourier transform of the shape of the crack, is equivalent to saying a plane wave is emitted from each point. So you are using the same logic twice, but with different convention of the sign of the complex phase, which is why they incorrectly cancel each other. You can also drop the $e^{-i\omega t}$ as it won't make any difference here.

$$\int_{-d/2}^{d/2} E_0 e^{ikx} \, dx$$ should give you the correct solution. Also note your lower limit of the integral should be $-d/2$.

You could also use $e^{-ikx}$ instead of $e^{ikx}$, as long as you are persistent in your sign convention.

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