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

Imagine you are standing in front of a classic Young's Double Slit Experiment setup. The geometry of this setup is the key to unlocking the entire problem. We are given the separation between the two slits, $d=0.3\text{ mm}$, which we must immediately convert to standard SI units: $0.3\times 10^{-3}\text{ m}$. The screen where the magic of interference happens is placed at a distance $D=1.5\text{ m}$ away.

Now, the problem presents a very specific geometric constraint: the distance between the fourth bright fringes on both sides of the central maximum is $2.4\text{ cm}$.

Think about what this means. The interference pattern is perfectly symmetric around the central bright fringe. If the total distance from the top 4th fringe to the bottom 4th fringe is $2.4\text{ cm}$, then the distance from the central maximum to just one of these 4th fringes (let's call it $Y_4$) is exactly half of that total distance.

Again, we must be disciplined with our units. Let's convert this to meters: $Y_4=1.2\times 10^{-2}\text{ m}$.

With our geometric parameters locked in, we turn to the master equation that governs the position of bright fringes in a YDSE setup. For the $n$-th bright fringe, the distance from the central maximum is given by:

Here, $n$ represents the order of the fringe (which is $4$ in our case), $\lambda$ is the unknown wavelength of the light, $D$ is the distance to the screen, and $d$ is the slit separation.

Let's substitute our known values into the master equation. This is where we must be careful not to make any silly algebraic mistakes.

Now, we isolate $\lambda$:

We have successfully found the wavelength of the light! But the journey isn't over yet. The question didn't ask for the wavelength; it asked for the frequency.

To bridge the gap between wavelength and frequency, we use the fundamental wave equation that connects them via the speed of light, $c$:

Rearranging this to solve for frequency, $f$, we get:

We know the speed of light in a vacuum is approximately $3\times 10^8\text{ m/s}$. Substituting our calculated wavelength:

To match the format requested by the question ($\times 10^{14}\text{ Hz}$), we adjust the decimal:

Comparing this with the given expression $........\times 10^{14}\text{ Hz}$, we can clearly see that the missing integer is $5$.