= Solution
Take the quoted separation to mean adjacent slit centre spacing $d=50\,\mu\mathrm m$. For three equal very narrow slits at $-d,0,d$, the <Fraunhofer diffraction> <complex amplitude> in the transverse direction is
$$
\psi(\theta)=\psi_1\left(e^{-2i\beta}+1+e^{2i\beta}\right)
=\psi_1(1+2\cos2\beta),\qquad
\beta=\frac{\pi d\sin\theta}{\lambda}.
$$
With $I_1$ denoting one narrow slit's contribution, the <light intensity> is
$$
\boxed{\frac{I}{I_1}=\left(1+2\cos2\beta\right)^2
=\left(\frac{\sin3\beta}{\sin\beta}\right)^2.}
$$
The quotient is interpreted by its limit where both numerator and denominator vanish. At small angles on a screen at $R=1\,\mathrm m$, $\sin\theta\simeq X/R$ and $\beta=\pi X/(10\,\mathrm{mm})$. Principal <interference fringes> occur at
$$
\boxed{X_m\simeq m\frac{\lambda R}{d}=m(10.0\,\mathrm{mm}),\qquad I(X_m)=9I_1.}
$$
Between successive principal maxima there are two zeros, at $X=(m+1/3)(10\,\mathrm{mm})$ and $X=(m+2/3)(10\,\mathrm{mm})$. Halfway between them is a smaller maximum $I=I_1$, one ninth of a principal peak. The pattern is symmetric about $X=0$. The positions quoted in millimetres use the <paraxial approximation>; exactly $d\sin\theta_m=m\lambda$ and $X_m=R\tan\theta_m$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2-b7-three-slits.png]
{title=Original calculated <Fraunhofer diffraction> patterns for three narrow slits and for three finite-width slits}
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