The maximum number of possible interference maxima forming on screen for slit separation equal to 2.5 times the wavelength in YDSE is:

The path difference, $\mathrm{dsin}\mathrm{\theta }=\mathrm{n\lambda }$
$$\begin{gathered} \sin \mathrm{\theta }=\frac{\mathrm{n\lambda }}{\mathrm{d}}=\frac{\mathrm{n\lambda }}{2.5\mathrm{\lambda }}=\frac{\mathrm{n}}{2.5} \\ \text{ For, }\mathrm{n}=0,\sin \mathrm{\theta }=0,\mathrm{\theta }=0^{\circ } \\ \begin{array}{l}\mathrm{n}=1,\sin \mathrm{\theta }=\frac{1}{2.5}\\ \mathrm{n}=2,\sin \mathrm{\theta }=\frac{2}{2.5}\\ \mathrm{n}=3, \mathrm{sin\theta } =\frac{3}{2.5}>1 \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{possible}\\ \mathrm{so} \mathrm{there} \mathrm{will} \mathrm{be} \mathrm{one} \mathrm{central} \mathrm{maxima} \mathrm{for} \mathrm{\theta }=0 \\ \mathrm{and} \mathrm{two} \mathrm{maxima} \mathrm{will} \mathrm{form} \mathrm{on} \mathrm{each} \mathrm{side} \mathrm{of} \mathrm{central} \mathrm{maxima}.\\ \mathrm{hence} \mathrm{total} \mathrm{no} \mathrm{of} \mathrm{maxima} \mathrm{will} \mathrm{be} 5.\end{array} \end{gathered}$$