At two points P and Q on screen in Young's double slit experiment, waves from slits S1 and S2 have a

Q: Why is the maximum intensity $4I_0$ and not just $2I_0$? Shouldn't two slits of intensity $I_0$ just add up to $2I_0$?

That's a very common intuition, but it applies to *incoherent* light (like two separate light bulbs). In YDSE, the light is *coherent*. We don't add intensities directly; we add the wave amplitudes. If each slit has amplitude $A$, constructive interference gives a total amplitude of $A + A = 2A$. Since intensity is proportional to the square of the amplitude, the maximum intensity is proportional to $(2A)^2 = 4A^2$, which corresponds to $4I_0$.

Q: What does a path difference of $0$ physically mean for point $P$?

A path difference of $0$ means that the distance from slit $S_1$ to point $P$ is exactly the same as the distance from slit $S_2$ to point $P$. The waves travel the exact same distance, so they arrive perfectly in sync (in phase), creating the brightest possible spot, known as the central maximum.

Q: Why do we use $\cos^2(\phi/2)$ in the formula instead of just $\cos(\phi)$?

The general formula for interference of two waves with intensities $I_1$ and $I_2$ is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$. When the slits are identical ($I_1 = I_2 = I_0$), this becomes $I = 2I_0 + 2I_0 \cos\phi = 2I_0(1 + \cos\phi)$. Using the trigonometric half-angle identity $1 + \cos\phi = 2\cos^2(\phi/2)$, we get the simplified and elegant form $I = 4I_0 \cos^2(\phi/2)$.

Q: Can the intensity ever be negative if the cosine term is negative?

No, intensity can never be negative. Notice that the formula uses $\cos^2(\phi/2)$. Because the cosine term is squared, the result is always positive or zero. Physically, intensity represents energy per unit area per unit time, which must be a non-negative quantity.

Q: What would happen if the two slits had different widths, and therefore different intensities?

If the slits had different intensities, say $I_1$ and $I_2$, we could no longer use the simplified $4I_0 \cos^2(\phi/2)$ formula. We would have to revert to the general equation: $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$. The maximum intensity would be $(\sqrt{I_1} + \sqrt{I_2})^2$ and the minimum would be $(\sqrt{I_1} - \sqrt{I_2})^2$.

Question

At two points $P$ and $Q$ on screen in Young's double slit experiment, waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$, respectively. The ratio of intensities at $P$ and $Q$ will be

Options:

(A) 3:2

(B) 2:1

(C) $\sqrt{2}:1$

(D) 4:1

Sigma Percentile · Accepted Answer

### The Setup: Two Slits, One Screen

Imagine you are standing in a dark room, observing the classic Young's Double Slit Experiment. Light from a single source passes through two narrow slits, $S_1$ and $S_2$, creating two coherent waves. These waves travel towards a screen, overlapping and interfering with each other.

When they meet, they don't just add up simply. Depending on how their peaks and troughs align, they can create bright spots (constructive interference) or dark spots (destructive interference). If we assume both slits emit light of the same intensity, say $I_0$, the resultant intensity $I$ at any point on the screen is governed by a beautiful mathematical relationship.

### The Master Equation: Phase and Intensity

The intensity at any point is determined by the phase difference $\phi$ between the two arriving waves. The master equation for this is:

$$I = 4I_0 \cos^2\left(\frac{\phi}{2}ight)$$

But how do we find this phase difference? It all comes down to the extra distance one wave travels compared to the other, known as the **path difference** ($\Delta x$). The phase difference is directly proportional to the path difference:

$$\phi = \frac{2\pi}{\lambda} \Delta x$$

This equation is our bridge. It connects the physical geometry of the setup (path difference) to the abstract wave mathematics (phase difference).

### Point P: The Brightest Spot

Let's analyze the first point, $P$. The problem states that the path difference at $P$ is exactly $0$. Physically, this means point $P$ is exactly equidistant from both slits. It lies right on the central axis of the setup.

Because $\Delta x_P = 0$, the phase difference $\phi_P$ is also $0$. Let's plug this into our intensity equation:

$$I_P = 4I_0 \cos^2\left(\frac{0}{2}ight)$$

Since $\cos(0) = 1$, the intensity maximizes:

$$I_P = 4I_0$$

This is the central maximum, the brightest spot on the screen. Notice how the intensity is four times the intensity of a single slit, not just double. This is the magic of constructive interference!

### Point Q: The Phase Shift

Now, let's move to point $Q$. Here, the waves arrive with a path difference of $\frac{\lambda}{4}$. One wave has traveled a quarter of a wavelength further than the other. Let's find the corresponding phase difference:

$$\phi_Q = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}ight) = \frac{\pi}{2}$$

A path difference of a quarter wavelength corresponds to a phase shift of $90^\circ$ or $\frac{\pi}{2}$ radians. Now, we substitute this into our intensity formula. Be careful not to make a silly mistake with the half-angle!

$$I_Q = 4I_0 \cos^2\left(\frac{\pi/2}{2}ight) = 4I_0 \cos^2\left(\frac{\pi}{4}ight)$$

We know that $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$. Squaring this gives $\frac{1}{2}$. Therefore:

$$I_Q = 4I_0 \left(\frac{1}{2}ight) = 2I_0$$

At point $Q$, the intensity is exactly half of the maximum possible intensity.

### The Final Showdown: Comparing Intensities

The question asks for the ratio of the intensities at these two points. We have all the pieces; we just need to put them together.

$$	ext{Ratio} = \frac{I_P}{I_Q} = \frac{4I_0}{2I_0}$$

The $I_0$ terms cancel out beautifully, leaving us with:

$$	ext{Ratio} = \frac{2}{1} = 2:1$$

And there we have it! By systematically linking path difference to phase difference, and then to intensity, we've decoded the interference pattern. The correct option is **(b)**.

Animated Solution for Physics - Optics: At two points $P$ and $Q$ on screen in Young's double slit experiment, waves from slits $S_{1}$ and $S_{2}$ have a path difference of $0$ and $\frac{λ}{4}$ , respectively. The ratio of intensities at $P$ and $Q$ will be

Select Answer:

Visualized Solution

Young's Double Slit Setup

Phase Difference Formula

Phase Difference at Point $P$

Calculating Intensity at $P$

Phase Difference at Point $Q$

Calculating Intensity at $Q$

Ratio of Intensities

The Way Forward

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Setup

The Master Equation

Point P

Point Q

The Final Showdown

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