In a Young's double slit experiment, the width of the one of the slit is three times the other slit.

Q: Wait, isn't intensity proportional to slit width, not amplitude?

Excellent observation! In standard YDSE problems, $I \propto W$, which means $A \propto \sqrt{W}$. However, this specific question explicitly defines a custom rule: $A \propto W$. Always read the problem statement carefully and follow its specific constraints.

Q: Why do we square the sum and difference of amplitudes to get intensity?

Intensity is defined as the energy crossing per unit area per unit time, which is directly proportional to the square of the wave's amplitude ($I \propto A^2$). For constructive interference, the resultant amplitude is $A_1 + A_2$, so $I_{max} \propto (A_1 + A_2)^2$.

Q: What happens to the negative sign in the denominator?

The denominator represents destructive interference, where the waves are out of phase. The resultant amplitude is the magnitude of their difference, $|A_1 - A_2|$. Even if we write it as $A_1 - A_2$ and get a negative value, squaring it for intensity makes it positive.

Q: Would the ratio change if the sources were incoherent?

Yes, drastically! If the sources were incoherent, there would be no stable phase relationship, meaning no interference pattern would form. The intensity everywhere on the screen would simply be the sum of individual intensities: $I = I_1 + I_2$.

Q: Can the minimum intensity ever be zero?

Yes, but only if the two interfering waves have exactly the same amplitude ($A_1 = A_2$). In that case, the destructive interference is perfect, and $I_{min} = (A_1 - A_1)^2 = 0$. In our problem, since the amplitudes are different, the minimum intensity is non-zero.

Question

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Options:

(A) 4 : 1

(B) 2 : 1

(C) 1 : 4

(D) 3 : 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a Young's Double Slit Experiment setup. But there is a twist! The two slits are not identical. One slit is significantly wider than the other—three times wider, to be exact. Let us denote the width of the first slit as $W_1$ and the second as $W_2$. According to the problem, we have $W_2 = 3W_1$.

Now, here is where many students fall into a trap. In standard physics problems, the intensity of light passing through a slit is directly proportional to its width ($I \propto W$), which implies that the amplitude is proportional to the square root of the width ($A \propto \sqrt{W}$).

However, this specific question explicitly states a custom rule: **the amplitude of the light is proportional to the slit-width**. We must strictly follow the universe's rules as defined by the question! Therefore, we can confidently write $A \propto W$, which directly leads us to $A_2 = 3A_1$.

### The Master Equation

We are tasked with finding the ratio of the maximum intensity to the minimum intensity in the resulting interference pattern. To do this, we need our master equation that connects intensity to amplitude.

We know that the maximum intensity occurs during constructive interference, where amplitudes add up, and minimum intensity occurs during destructive interference, where they subtract. Since intensity is always proportional to the square of the amplitude ($I \propto A^2$), the ratio is given by:

$$ \frac{I_{max}}{I_{min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 $$

This elegant formula is the key to unlocking our answer.

### Final Calculation

Let us substitute our known relationship, $A_2 = 3A_1$, into the master equation.

$$ \frac{I_{max}}{I_{min}} = \left( \frac{A_1 + 3A_1}{A_1 - 3A_1} \right)^2 $$

Simplifying the numerator and the denominator, we get:

$$ \frac{I_{max}}{I_{min}} = \left( \frac{4A_1}{-2A_1} \right)^2 $$

The negative sign in the denominator might look intimidating, but remember, we are squaring the entire fraction, so it will vanish. The $A_1$ terms cancel out beautifully, leaving us with:

$$ \frac{I_{max}}{I_{min}} = \left( -2 \right)^2 = \frac{4}{1} $$

And there we have it! The ratio of the maximum to the minimum intensity is **$4:1$**.

Always remember to read the problem statement carefully. The custom condition $A \propto W$ was the secret to solving this without making a silly mistake!

Select Answer:

Visualized Solution

$W_{2} = 3 W_{1}$

$A \propto W$

$\frac{I _{ma x}}{I _{min}}$ Formula

Substitution

Final Ratio

Incoherent Sources?

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

Analyzing the Setup

The Master Equation

Final Calculation

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