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

Finding Intensity in YDSE

Imagine you are standing in front of the classic Young's Double Slit Experiment (YDSE) setup. Two coherent light waves emerge from their respective slits, $S_1$ and $S_2$, and travel towards a distant screen. At the exact center of the screen, point $O$, both waves travel the exact same distance. They arrive perfectly in sync, crest to crest and trough to trough, creating a brilliant central maximum. We call this maximum intensity $I_{max}$.

But what happens when we move away from the center to an arbitrary point $P$? The symmetry is broken. The wave from the lower slit, $S_2$, has to travel a slightly longer distance than the wave from the upper slit, $S_1$. This extra distance is the path difference, denoted by $\Delta x$.

In our specific problem, we are given a fascinating constraint: the path difference at point $P$ is exactly one-eighth of a wavelength, or $\Delta x=\frac{\lambda }{8}$.

This path difference is the physical reason why the waves are no longer perfectly in sync. It creates a phase difference ($\varphi$) between the two waves when they superimpose. The relationship between path difference and phase difference is one of the most fundamental bridges in wave optics:

Let's substitute our given path difference into this elegant equation:

Notice how the wavelengths ($\lambda$) cancel out beautifully, leaving us with pure geometry:

So, the waves arrive at point $P$ with a phase difference of $45^{\circ }$.

Now, how does this phase difference dictate the brightness at point $P$? When two coherent waves interfere, the resultant intensity $I$ is given by the general formula:

Assuming our slits are identical, they both emit light of the same intensity, let's call it $I_0$. Substituting $I_1=I_2=I_0$, the formula simplifies dramatically:

At the central maximum, where the phase difference $\varphi$ is zero, $\cos (0)=1$. Therefore, the maximum intensity is:

The question asks for the ratio of the intensity at point $P$ to the maximum intensity at the center. Let's set up this ratio by dividing our expression for $I$ by $I_{max}$:

The $I_0$ terms cancel out, and the fraction simplifies to a very clean expression:

Now, we simply bring back the phase difference we calculated earlier, $\varphi =\frac{\pi }{4}$:

We know from basic trigonometry that $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$, which is approximately $0.707$. Let's crunch the numbers:

Rounding it off, we get $0.853$. This means the intensity at our point $P$ is roughly $85.3\%$ of the maximum possible intensity at the center. Looking at our given choices, this perfectly matches option (b). A beautiful, logical progression from physical geometry to mathematical intensity!