Animated Solution for Physics - Optics: In a Young's double slit experiment, the path difference at a certain point on the screen between two interfering waves is $\frac{1}{8}$th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to

Interference is one of the most beautiful phenomena in physics, vividly demonstrating the wave nature of light. In Young's Double Slit Experiment (YDSE), light from a single source is split into two coherent sources, which then overlap on a screen to form a pattern of alternating bright and dark fringes. But the intensity of light doesn't just jump from maximum to zero; it varies continuously. Let's dive into a classic problem that asks us to find the exact intensity at a specific point on the screen based on its path difference.

Path Difference

Imagine you are standing on the screen at a point $P$. Light waves from the two slits, $S_1$ and $S_2$, have to travel different distances to reach you. This difference in distance is called the path difference, denoted by $\Delta x$.

In our specific problem, we are given that the path difference at point $P$ is exactly one-eighth of the wavelength of the light used. Mathematically, we write this as:

This path difference is the physical reality of the setup. However, to understand how the waves interfere (whether they add up constructively or cancel out destructively), we need to translate this physical distance into a phase angle.

Phase Difference

The relationship between path difference and phase difference is a fundamental bridge in wave optics. A full wavelength $\lambda$ corresponds to a full cycle of $2\pi$ radians. Therefore, the phase difference $\varphi$ is given by the formula:

Let's substitute our known path difference into this equation. We replace $\Delta x$ with $\frac{\lambda }{8}$:

Notice how beautifully the wavelength $\lambda$ cancels out. This leaves us with a pure angular measure:

So, the two waves arriving at point $P$ are out of phase by $\frac{\pi }{4}$ radians (or $45^{\circ }$).

Now that we have the phase difference, we can determine the resultant intensity. When two coherent waves with intensities $I_1$ and $I_2$ interfere, the resultant intensity $I^{\prime }$ is not simply their sum. It is governed by the master equation of interference:

In a standard YDSE setup, the two slits are identical, meaning they emit light of the same intensity. Let's call this individual intensity $I_0$. So, $I_1=I_0$ and $I_2=I_0$. Substituting these into our equation, along with our calculated phase difference $\varphi =\frac{\pi }{4}$, we get:

We know that $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$. Let's plug that in:

Using the standard approximation $\sqrt{2}\approx 1.414$, the intensity at point $P$ becomes:

The Peak of Brightness

To find the ratio requested by the problem, we need to know the intensity at the center of a bright fringe (the central maxima). At the exact center of the screen, the waves from both slits travel the exact same distance.

This means the path difference $\Delta x=0$, which in turn means the phase difference $\varphi =0$. Since $\cos (0)=1$, the intensity equation simplifies to:

This is a crucial result to remember: the maximum intensity in a standard YDSE is four times the intensity of a single slit.

We are finally ready to find the ratio of the intensity at point $P$ to the maximum intensity at the central bright fringe.

Substituting the values we derived:

The $I_0$ terms cancel out, leaving us with a simple numerical division:

Looking at our options, this value is closest to 0.85. This tells us that at a point where the path difference is $\frac{\lambda }{8}$, the screen is still quite bright, shining at $85\%$ of its maximum possible intensity. This continuous variation is the hallmark of wave interference!