= Solution
Normalize the central slit amplitude to one. The two outer amplitudes are $1/2$ each, not $1/\sqrt2$, because the specified attenuation concerns amplitude. For centre spacing $d=5\lambda$, adjacent slits have phase difference $\delta=(2\pi d/\lambda)\sin\theta=10\pi\sin\theta$. The far-field amplitude and intensity are therefore
$$
A=\tfrac12e^{-i\delta}+1+\tfrac12e^{i\delta}=1+\cos\delta,
\qquad
\boxed{\frac I{I_0}=(1+\cos(10\pi\sin\theta))^2=4\cos^4(5\pi\sin\theta),}
$$
where $I_0$ is the central slit's intensity alone in the same narrow-slit approximation. Principal maxima have $\sin\theta=m/5$ and height $4I_0$; between them, zeros occur at $\sin\theta=(2m+1)/10$. Only values in $[-1,1]$ are physical. The endpoints are grazing-direction limits of this ideal angular model.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-weighted-three-slit.png]
{title=Weighted three-slit interference with outer amplitudes one half of the central amplitude}
{description=Original narrow-slit far-field calculation. A finite slit width supplies an additional single-slit envelope, which is omitted here.}
The amplitude weights form a binomial pattern. They remove the secondary maxima that an equal-amplitude three-slit array would have between its principal peaks. No <phase shift> is needed: changing the coherent weights already changes the <interference> structure.
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