In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright fringe is 2.4 cm. The frequency of light used is $$x \times 10^{14}$$ Hz.

We have a Young’s double-slit experiment in which the distance between the slits is given as $$d = 0.3\ \text{mm}$$. First we convert this separation into metres so that every quantity is in SI units: $$0.3\ \text{mm} = 0.3 \times 10^{-3}\ \text{m} = 3 \times 10^{-4}\ \text{m}$$.

The screen is placed at a distance $$D = 1.5\ \text{m}$$ from the slits.

The statement says that the distance between the fourth bright fringes on the two opposite sides of the central bright fringe is $$2.4\ \text{cm}$$. Converting into metres gives $$\Delta y = 2.4\ \text{cm} = 2.4 \times 10^{-2}\ \text{m}$$.

In Young’s experiment, the position of the $$m^{\text{th}}$$ bright fringe (measured from the central maximum) is described by the formula

$$y_m = \frac{m \lambda D}{d},$$

where $$\lambda$$ is the wavelength of the light. Thus, the fourth bright fringe on the right of the centre is at

$$y_{+4} = \frac{4 \lambda D}{d},$$

and the fourth bright fringe on the left of the centre is at

$$y_{-4} = -\,\frac{4 \lambda D}{d}.$$

The separation between these two fringes is therefore

$$\Delta y = y_{+4} - y_{-4} = \frac{4 \lambda D}{d} - \Bigl(-\frac{4 \lambda D}{d}\Bigr) = \frac{8 \lambda D}{d}.$$

We substitute the known values now. Rearranging the above relation gives

$$\lambda = \frac{\Delta y\, d}{8D}.$$

Putting in $$\Delta y = 2.4 \times 10^{-2}\ \text{m},\quad d = 3 \times 10^{-4}\ \text{m},\quad D = 1.5\ \text{m},$$ we obtain

$$\lambda = \frac{(2.4 \times 10^{-2})(3 \times 10^{-4})}{8 \times 1.5}.$$

First, multiply the numerators:

$$(2.4 \times 3) \times 10^{-2-4} = 7.2 \times 10^{-6}\ \text{m}.$$

Next, multiply the denominators:

$$8 \times 1.5 = 12.$$

Hence,

$$\lambda = \frac{7.2 \times 10^{-6}}{12} = 0.6 \times 10^{-6}\ \text{m} = 6 \times 10^{-7}\ \text{m}.$$

We now need the frequency $$\nu$$, and we use the fundamental relation between speed of light, wavelength, and frequency:

$$c = \lambda \nu.$$

So,

$$\nu = \frac{c}{\lambda}.$$

Taking $$c = 3 \times 10^{8}\ \text{m s}^{-1}$$ and $$\lambda = 6 \times 10^{-7}\ \text{m},$$ we get

$$\nu = \frac{3 \times 10^{8}}{6 \times 10^{-7}} = \frac{3}{6} \times 10^{8+7}\ \text{Hz} = 0.5 \times 10^{15}\ \text{Hz} = 5 \times 10^{14}\ \text{Hz}.$$

This matches the form $$x \times 10^{14}\ \text{Hz}$$ with $$x = 5$$.

So, the answer is $$5$$.