 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 isWith $I_1$ denoting one narrow slit's contribution, the light intensity isThe 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 atBetween 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 $X_m=R\tan {\theta }_m$.