In an interference experiment, the ratio of amplitudes of coherent waves is a1/a2 = 1/3. The ratio o

Q: Why is intensity proportional to the square of the amplitude?

In wave mechanics, the energy transported by a wave is proportional to the square of its amplitude. Since intensity is defined as power (energy per unit time) per unit area, it naturally follows that intensity is also proportional to the square of the amplitude.

Q: What does it mean for the waves to be 'coherent'?

Coherent waves have a constant phase difference and the same frequency. This is absolutely necessary to produce a stable, observable interference pattern. If they were incoherent, the phase difference would fluctuate randomly, and we would just see a uniform average intensity everywhere.

Q: Why do we take the absolute value for the minimum amplitude?

Amplitude represents the maximum displacement from the equilibrium position, which is a physical magnitude (a positive quantity). When a crest meets a trough, we subtract their values, but the resulting amplitude of the combined wave is just the absolute difference between them.

Q: Could I have solved this by finding the individual intensities first?

Yes, absolutely! You could say $I_1 \propto a^2$ and $I_2 \propto 9a^2$. Then use the formula $\frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$. Substituting the values will give you the exact same result of 4.

Q: What if the ratio of amplitudes was 1:1?

If the amplitudes were equal ($a_1 = a_2$), then $a_{min} = a - a = 0$. The minimum intensity would be zero, meaning the dark fringes would be perfectly dark. The ratio $\frac{I_{max}}{I_{min}}$ would approach infinity.

Question

In an interference experiment, the ratio of amplitudes of coherent waves is $\frac{a_1}{a_2} = \frac{1}{3}$. The ratio of maximum and minimum intensities of fringes will be

Options:

(A) 2

(B) 18

(C) 4

(D) 9

Sigma Percentile · Accepted Answer

### The Dance of Waves: Unraveling Interference Intensities

Imagine standing by a calm pond and dropping two pebbles into the water simultaneously. As the ripples spread outward, they eventually meet and interact. This beautiful phenomenon is known as **interference**. In the realm of light, when two coherent waves superimpose, they create a pattern of bright and dark fringes. But how bright is the brightest spot, and how dark is the darkest? Let's dive into the mathematics of wave interference to find out.

### Analyzing the Setup

In our specific problem, we are dealing with two coherent waves. Coherence means they maintain a constant phase difference over time—a crucial requirement for a stable interference pattern. We are given that the ratio of their amplitudes is $1:3$.

Let's assign some variables to make this concrete. If the amplitude of the first wave is $a_1 = a$, then the amplitude of the second wave must be $a_2 = 3a$.

### Constructive and Destructive Interference

When these two waves meet, they obey the principle of superposition.

**Constructive interference** occurs when the crest of one wave aligns perfectly with the crest of the other. Their amplitudes add up, creating a maximum possible amplitude:
$$a_{max} = a_1 + a_2 = a + 3a = 4a$$

Conversely, **destructive interference** happens when the crest of one wave meets the trough of the other. They work against each other, resulting in a minimum possible amplitude. Since amplitude represents a physical magnitude, we take the absolute difference:
$$a_{min} = |a_1 - a_2| = |a - 3a| = 2a$$

### The Master Equation: Intensity and Amplitude

Now, what we actually observe on a screen (the brightness of the fringes) is not the amplitude, but the **intensity** of the light. A fundamental law of wave physics states that the intensity ($I$) of a wave is directly proportional to the square of its amplitude ($a$):
$$I \propto a^2$$

This means that to find the ratio of the maximum intensity to the minimum intensity, we simply need to square the ratio of their respective amplitudes:
$$\frac{I_{max}}{I_{min}} = \left(\frac{a_{max}}{a_{min}}\right)^2$$

### Final Calculation

We have all the pieces of the puzzle. Let's substitute our calculated maximum and minimum amplitudes into the intensity ratio formula:
$$\frac{I_{max}}{I_{min}} = \left(\frac{4a}{2a}\right)^2$$

The variable $a$ elegantly cancels out from the numerator and the denominator, leaving us with a simple fraction:
$$\frac{I_{max}}{I_{min}} = (2)^2 = 4$$

And there we have it! The ratio of the maximum intensity to the minimum intensity in this interference pattern is exactly **4**. This means the brightest fringes will be four times as intense as the darkest fringes.

Animated Solution for Physics - Optics: In an interference experiment, the ratio of amplitudes of coherent waves is $\frac{a _{1}}{a _{2}} = \frac{1}{3}$ . The ratio of maximum and minimum intensities of fringes will be

Select Answer:

Visualized Solution

$a_{1}$ and $a_{2}$

$a_{ma x}$ and $a_{min}$

Calculate Amplitudes

$I \propto a^{2}$

Substitute Values

Final Ratio

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

The Dance of Waves

Analyzing the Setup

Constructive and Destructive Interference

The Master Equation

Final Calculation

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