For the setup illustrated in the figure, y = 1.0 mm , d = 0.24 mm ,...

From the given figure, the distance from the central bright fringe to the first dark fringe in the interference pattern is $y$.

The position of the first dark fringe is given by:

$y = \dfrac{\lambda D}{2d}$

Rearranging for the wavelength $\lambda$ of the light source:

$\lambda = \dfrac{2yd}{D}$

Substituting the given values $y = 1.0 \text{ mm} = 10^{-3} \text{ m}$, $d = 0.24 \text{ mm} = 0.24 \times 10^{-3} \text{ m}$, and $D = 1.2 \text{ m}$:

$\lambda = \dfrac{2 \times 10^{-3} \times 0.24 \times 10^{-3}}{1.2} = 0.4 \times 10^{-6} \text{ m} = 400 \text{ nm}$

The energy of the incident photons is:

$E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{400 \text{ nm}} = 3.1 \text{ eV}$

Using Einstein's photoelectric equation, the stopping potential $V$ is given by:

$eV = E - \phi$

Substituting $E = 3.1 \text{ eV}$ and the work function $\phi = 2.2 \text{ eV}$:

$eV = 3.1 \text{ eV} - 2.2 \text{ eV} = 0.9 \text{ eV}$

Therefore, the stopping potential is $V = 0.9 \text{ V}$.