Animated Solution for Physics - Optics: In a Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda }{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity, then $I/I_0$ is equal to

The phenomenon of interference in Young's Double Slit Experiment (YDSE) is a beautiful demonstration of the wave nature of light. When light waves from two coherent sources meet, they superimpose, creating a pattern of bright and dark fringes. But what happens at a random point on the screen? Let's dive into the math and physics of finding the exact intensity at any given point!

Path Difference

Imagine you are standing at a point $P$ on the screen. Light waves from the two slits, $S_1$ and $S_2$, are traveling towards you. Because the slits are at slightly different distances from point $P$, one wave has to travel a bit further than the other. This extra 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-sixth of the wavelength of the light used. Mathematically, we write this as:

This path difference is the root cause of the interference pattern. It determines whether the waves will arrive in sync (constructive interference) or out of sync (destructive interference).

From Path to Phase

To calculate the intensity, we need to know how "out of sync" the waves are in terms of their phase. This is where the golden relation between path difference and phase difference comes into play. The phase difference, $\varphi$, is directly proportional to the path difference:

Let's substitute our known path difference into this equation. By replacing $\Delta x$ with $\frac{\lambda }{6}$, we get:

Notice how elegantly the wavelength $\lambda$ cancels out! This leaves us with:

So, the waves arriving at point $P$ have a phase difference of $\frac{\pi }{3}$ radians, which is equivalent to $60^{\circ }$.

Intensity

Now that we have the phase difference, we can determine the intensity of light at point $P$. The resultant intensity $I$ in a YDSE setup, assuming both slits emit light of equal intensity, is governed by the master equation:

The problem tells us that the maximum intensity (which occurs at the central bright fringe where $\varphi =0$) is denoted by $I_0$. Therefore, we can replace $I_{max}$ with $I_0$. We also substitute our calculated phase difference, $\varphi =\frac{\pi }{3}$:

Let's rearrange the equation to find the ratio of the intensity at point $P$ to the maximum intensity. We bring $I_0$ to the denominator on the left side. Inside the cosine function, dividing $\frac{\pi }{3}$ by $2$ gives us $\frac{\pi }{6}$:

Now, we just need to evaluate the trigonometric function. We know from basic trigonometry that $\cos \left(\frac{\pi }{6}\right)$, or $\cos (30^{\circ })$, is equal to $\frac{\sqrt{3}}{2}$. Substituting this value in, we get:

Finally, squaring the numerator and the denominator yields our answer:

And there we have it! The intensity at the point where the path difference is $\frac{\lambda }{6}$ is exactly three-quarters of the maximum intensity.