When diffraction is performed in water then why will the fringes will be smaller in number than when it is performed in air?

Maxima for a double slit occur when n lambda = d sin (theta)

so the angles between maxima (and hence the spacing) depends on sin(theta)

Reaaranging: sin (theta) = n lambda/d

When the apparatus is put in water, lambda changes. The velocity in water is less so mthe distance travelled durin on e period is less ie the wavelength (lambda) decreases. So sin(theta) decreases and hence the spacing between the maxima.