In the Young's double slit experiment the intensity produced by each one of the individual slits is $$I_{o}.$$ The distance between two slits is 2 mm . The distance of
screen from slits is 10 m. The wavelength of light is $$6000_A^\circ$$. The intensity of light on the screen in front of one of the slits is __________.

We need to find the intensity on the screen in front of one of the slits in Young's double slit experiment. The intensity from each slit is $$I_0$$, the slit separation is d = 2 mm = 2 × 10⁻³ m, the screen distance is D = 10 m, and the wavelength is $$\lambda = 6000 \text{ Å} = 6 \times 10^{-7}$$ m.

The fringe width is given by $$\beta = \frac{\lambda D}{d} = \frac{6 \times 10^{-7} \times 10}{2 \times 10^{-3}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm}$$.

The point in front of one slit lies at y = d/2 = 1 mm from the central maximum, so the path difference is $$\Delta = \frac{yd}{D} = \frac{1 \times 10^{-3} \times 2 \times 10^{-3}}{10} = 2 \times 10^{-7}$$ m.

The corresponding phase difference is $$\phi = \frac{2\pi}{\lambda} \times \Delta = \frac{2\pi}{6 \times 10^{-7}} \times 2 \times 10^{-7} = \frac{2\pi}{3}$$.

The resultant intensity is $$I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\cos\phi = 2I_0 + 2I_0\cos\frac{2\pi}{3},$$ which simplifies to $$= 2I_0 + 2I_0\left(-\frac{1}{2}\right) = 2I_0 - I_0 = I_0.$$

Therefore, the intensity is Option 3: $$I_0$$.