Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. Th

Q: Why do we use the cosine function in the interference formula?

The interference term arises from the dot product of the electric field vectors of the two waves. Since the waves have a phase difference $\phi$, their superposition involves adding their amplitudes vectorially, which introduces the $\cos \phi$ term from the law of cosines.

Q: What does a phase difference of $\pi/2$ physically mean?

A phase difference of $\pi/2$ (or 90 degrees) means the two waves are out of phase by a quarter of a wavelength. At this point, the interference term $2\sqrt{I_1 I_2} \cos(\pi/2)$ becomes zero, so the resultant intensity is simply the sum of the individual intensities.

Q: Why is the intensity at point B not zero if it's destructive interference?

Destructive interference only results in zero intensity if the two interfering waves have exactly the same amplitude (and thus the same intensity). Here, the intensities are $I$ and $4I$, so their amplitudes are different. The minimum intensity is $(\sqrt{4I} - \sqrt{I})^2 = I$, not zero.

Q: Can the resultant intensity ever be greater than the sum of the individual intensities?

Yes, absolutely! During constructive interference (when $\phi = 0, 2\pi, \dots$), the $\cos \phi$ term is positive, making the resultant intensity greater than $I_1 + I_2$. In this problem, the maximum intensity is $9I$, which is greater than $I + 4I = 5I$. This is due to the redistribution of energy.

Q: How is phase difference related to path difference?

Phase difference $\phi$ and path difference $\Delta x$ are directly proportional. The relationship is given by $\phi = \frac{2\pi}{\lambda} \Delta x$, where $\lambda$ is the wavelength of the light. A phase difference of $\pi$ corresponds to a path difference of $\lambda/2$.

Question

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $\pi$ at point $B$. Then the difference between resultant intensities at $A$ and $B$ is

Options:

(A) $2 I$

(B) $4 I$

(C) $5 I$

(D) $7 I$

Sigma Percentile · Accepted Answer

The phenomenon of interference is one of the most beautiful and fundamental concepts in wave optics. When two light waves superimpose, their energies don't just add up linearly everywhere; instead, they redistribute, creating regions of maximum intensity (bright fringes) and minimum intensity (dark fringes).

In this problem, we are given two beams of light with intensities $I$ and $4I$. We need to find the difference in their resultant intensities at two specific points on the screen, $A$ and $B$, where the phase differences are $\frac{\pi}{2}$ and $\pi$, respectively.

### The Master Equation

To find the resultant intensity of two interfering waves, we rely on the standard interference formula. The resultant intensity $I_{res}$ is given by the sum of the individual intensities plus an interference term that depends on the phase difference $\phi$ between the waves.

$$I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$$

This equation is the heart of wave optics. The term $2\sqrt{I_1 I_2} \cos \phi$ is what makes interference so fascinating—it allows the total intensity to be greater or less than the simple sum of the two intensities, depending on the value of $\phi$.

### Substituting the Given Values

We are given the intensities of the two beams as $I_1 = I$ and $I_2 = 4I$. Let's substitute these values into our master equation to see how it simplifies.

$$I_{res} = I + 4I + 2\sqrt{I \cdot 4I} \cos \phi$$

Inside the square root, $I$ multiplied by $4I$ gives $4I^2$. The square root of $4I^2$ is $2I$. Multiplying this by the $2$ outside the square root gives us $4I$.

$$I_{res} = 5I + 4I \cos \phi$$

This simplified equation tells us exactly how the intensity varies across the screen as the phase difference $\phi$ changes.

### Analyzing Point A

Now, let's focus on point $A$. The problem states that the phase difference at point $A$ is $\frac{\pi}{2}$ (or 90 degrees).

$$I_A = 5I + 4I \cos\left(\frac{\pi}{2}\right)$$

We know from basic trigonometry that $\cos\left(\frac{\pi}{2}\right) = 0$. This means that at point $A$, the interference term completely vanishes! The waves are neither perfectly constructive nor perfectly destructive.

$$I_A = 5I + 0 = 5I$$

So, the resultant intensity at point $A$ is simply the sum of the individual intensities, which is $5I$.

### Analyzing Point B

Next, let's evaluate the intensity at point $B$. Here, the phase difference is given as $\pi$ (or 180 degrees).

$$I_B = 5I + 4I \cos(\pi)$$

The value of $\cos(\pi)$ is $-1$. This negative sign is crucial—it indicates that the waves are undergoing destructive interference at this point. The crest of one wave is aligning with the trough of the other.

$$I_B = 5I + 4I(-1) = 5I - 4I = I$$

Thus, the intensity at point $B$ drops to a minimum value of $I$. Notice that it doesn't drop to zero because the two interfering waves have different amplitudes.

### The Final Calculation

Finally, the question asks for the difference between the resultant intensities at point $A$ and point $B$. We simply subtract the intensity we found for $B$ from the intensity we found for $A$.

$$\Delta I = I_A - I_B$$

$$\Delta I = 5I - I = 4I$$

The difference in intensity between these two points is **$4I$**. This elegant result showcases how the phase difference dictates the distribution of light energy on the screen.

As an extra insight, consider the maximum possible intensity on this screen. It would occur when $\cos \phi = 1$, giving $I_{max} = 5I + 4I = 9I$. The minimum intensity, as we saw at point $B$, is $I$. The energy is conserved overall, but it is beautifully redistributed to form the interference pattern we observe!

Select Answer:

Visualized Solution

$I_{r es} = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ$

$I_{1} = I, I_{2} = 4 I$

$ϕ_{A} = \frac{π}{2}$

$ϕ_{B} = π$

$I_{A} - I_{B}$

$I_{ma x}$ and $I_{min}$

The Sigma Insight: Interference of Waves

The Master Equation

Substituting the Given Values

Analyzing Point A

Analyzing Point B

The Final Calculation

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