1. Given data

• Slit separation: $d = 0.3\,\text{mm} = 3\times 10^{-4}\,\text{m}$
• Screen distance: $D = 1.5\,\text{m}$
• Distance between the 4th bright fringes on both sides of central bright: $2.4\,\text{cm} = 2.4\times 10^{-2}\,\text{m}$

2. Interpret the fringe positions

In Young's double slit experiment, position of the $n$th bright fringe from the central bright is $y_n = n\beta$ where $\beta$ is the fringe width.

So, the distance between the 4th bright fringes on both sides is $y_{+4} - y_{-4} = 4\beta - (-4\beta) = 8\beta$

Given: $8\beta = 2.4\times 10^{-2}$ $\beta = \frac{2.4\times 10^{-2}}{8} = 3\times 10^{-3}\,\text{m}$

3. Use fringe width formula

For YDSE, $\beta = \frac{\lambda D}{d}$ So, $\lambda = \frac{\beta d}{D}$

Substitute values: $\lambda = \frac{(3\times 10^{-3})(3\times 10^{-4})}{1.5}$ $\lambda = \frac{9\times 10^{-7}}{1.5} = 6\times 10^{-7}\,\text{m}$

4. Find frequency

Using $f = \frac{c}{\lambda}$ with $c = 3\times 10^8\,\text{m/s}$,

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

5. Final answer

The frequency is $5\times 10^{14}\,\text{Hz}$ So the required integer is 5.