In a single slit diffraction experiment, slit of width ' $a$ ' is illuminated by light of wavelength ' $\lambda$ ' and the width of the central maxima in diffraction pattern is measured as ' $y$ '. When half of the slit is covered and illuminated by light of wavelength $(1 \cdot 5) \lambda$, the width of the central maximum in diffraction pattern becomes

Width of central maximum with half-slit covered and increased wavelength

The single-slit diffraction pattern's central maximum width is given by $y = \frac{2D\lambda}{a}$, where $D$ is the screen distance, $\lambda$ is wavelength, and $a$ is slit width.

When the slit width is halved, $a' = \frac{a}{2}$, and wavelength becomes $\lambda' = 1.5\lambda$, the new width is:

$y' = \frac{2D\lambda'}{a'} = \frac{2D(1.5\lambda)}{a/2} = \frac{2D \cdot 1.5\lambda \cdot 2}{a} = 3 \cdot \frac{2D\lambda}{a} = 3y$

The central maximum width increases to three times the original value.

Final answer: $\boxed{3y}$