Animated Solution for Physics - Optics: In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $I_m$ is the maximum intensity, the resultant intensity $I$ when they interfere at phase difference $\varphi$, is given by

Imagine you are setting up a Young's Double Slit Experiment (YDSE), but with a twist. Instead of identical slits, one slit is intentionally made wider than the other. The problem tells us that this difference in width causes the amplitude of the light from the first slit to be exactly double the amplitude from the second slit.

Mathematically, we can write this as:

Now, a fundamental principle in wave optics is that the intensity of a wave is directly proportional to the square of its amplitude ($I\propto a^2$). This means if the amplitude is doubled, the intensity doesn't just double; it quadruples!

Let's assign a base intensity $I_0$ to the second slit. Therefore, the intensity of the first slit becomes:

Before we find the intensity at any random point, we need to determine the maximum possible intensity, denoted as $I_m$. Maximum intensity occurs during perfect constructive interference, where the phase difference is zero. The formula for maximum intensity is the square of the sum of the square roots of the individual intensities:

Let's substitute our values of $I_1$ and $I_2$ into this equation:

This is a crucial relationship. It tells us that the base intensity $I_0$ is simply $\frac{I_m}{9}$. We will need this later to match our final answer with the given options.

Now, let's look at a general point on the screen where the waves from the two slits arrive with a phase difference of $\varphi$. The general formula for the resultant intensity $I$ is:

Substituting $I_1=4I_0$ and $I_2=I_0$:

We have a neat expression for $I$, but the options are given in terms of $I_m$, not $I_0$. Let's substitute $I_0=\frac{I_m}{9}$ back into our equation:

We are very close! Option (a) looks tempting, but notice it has $(4+5\cos \varphi )$, not $(5+4\cos \varphi )$. Don't fall for the trap! We need to manipulate our expression further using trigonometry.

Let's split the number $5$ into $1+4$:

Now, factor out the $4$ from the last two terms:

Here comes the magic of half-angle identities. We know that $1+\cos \varphi =2{\cos }^2\left(\frac{\varphi }{2}\right)$. Substituting this into our equation gives:

And there we have it! This perfectly matches option (d). The journey from physical setup to algebraic substitution and finally trigonometric manipulation is what makes this a classic, beautiful JEE problem.