The width of one of the two slits in a Young's double slit experiment is three times the other slit.

Q: Why is amplitude proportional to slit width?

In a single slit, the total electric field reaching a point is the sum of contributions from all secondary Huygens sources across the slit. A wider slit has more sources, so the total amplitude of the forward-propagating wave is directly proportional to the width.

Q: Why doesn't the minimum intensity drop to zero?

For destructive interference to yield zero intensity, the two waves must have exactly equal amplitudes so they can perfectly cancel each other out. Since one slit is wider, its wave is stronger, leaving leftover light even at the minima.

Q: Can I just use the ratio of widths directly in the intensity formula?

No! You must be very careful. Width gives you the ratio of amplitudes. You must square that ratio to get the ratio of intensities before plugging it into the $I_{\text{min}} / I_{\text{max}}$ formula.

Q: What happens to the fringe width if the slits have different widths?

The fringe width ($\beta = \lambda D / d$) depends only on the wavelength, the distance to the screen, and the separation *between* the slits. Changing the individual slit widths affects the intensity and contrast, but not the spacing of the fringes.

Q: Is the total energy conserved in this pattern?

Yes, absolutely. Interference doesn't create or destroy energy; it merely redistributes it. The bright fringes get brighter and the dark fringes get less dark, but the average intensity across the screen remains equal to the sum of the individual intensities ($I_1 + I_2$).

Question

The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is $x:4$ where $x$ is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a classic Young's Double Slit Experiment. Usually, we assume both slits are perfectly identical, allowing equal amounts of light to pass through. But what happens when we introduce a twist?

In this problem, the width of the first slit is three times the width of the second slit.

$$b_1 = 3b_2$$

This simple geometric change has a profound impact on the physics of the light waves emerging from the slits. The amplitude of the light wave coming from a slit is directly proportional to the width of that slit.

$$A \propto b$$

### The Master Equation

Because the first slit is three times wider, the amplitude of the light wave it emits will be exactly three times the amplitude of the wave from the second slit.

$$\frac{A_1}{A_2} = \frac{b_1}{b_2} = 3$$

Now, we must transition from amplitude to intensity, because our eyes and detectors measure the intensity of light, not its raw amplitude. The fundamental rule of wave optics tells us that intensity is proportional to the square of the amplitude.

$$I \propto A^2$$

By squaring the ratio of the amplitudes, we find the ratio of their intensities.

$$\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}ight)^2 = 3^2 = 9$$

This means the first slit is blasting nine times more light energy onto the screen than the second slit!

### Final Calculation

When these two unequal waves meet on the screen, they interfere. However, because their intensities are so mismatched, they can never perfectly cancel each other out. The dark fringes will not be completely black. To find the exact contrast, we use the standard formula for the ratio of minimum to maximum intensity in an interference pattern.

$$\frac{I_{	ext{min}}}{I_{	ext{max}}} = \left( \frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}} ight)^2$$

Let's substitute $I_1 = 9I_2$ into this master equation.

$$\frac{I_{	ext{min}}}{I_{	ext{max}}} = \left( \frac{\sqrt{9I_2} - \sqrt{I_2}}{\sqrt{9I_2} + \sqrt{I_2}} ight)^2$$

The square root of $9$ is $3$, allowing us to simplify the expression beautifully.

$$\frac{I_{	ext{min}}}{I_{	ext{max}}} = \left( \frac{3\sqrt{I_2} - 1\sqrt{I_2}}{3\sqrt{I_2} + 1\sqrt{I_2}} ight)^2$$

Notice how $\sqrt{I_2}$ is a common factor in every term. We can factor it out and cancel it completely, leaving us with pure numbers.

$$\frac{I_{	ext{min}}}{I_{	ext{max}}} = \left( \frac{3 - 1}{3 + 1} ight)^2 = \left( \frac{2}{4} ight)^2$$

Simplifying the fraction inside the parentheses gives us $1/2$. Squaring that yields our final ratio.

$$\frac{I_{	ext{min}}}{I_{	ext{max}}} = \frac{1}{4}$$

The problem states that this ratio is equal to $x:4$. By comparing our result, it is crystal clear that **$x = 1$**. The unequal slits have reduced the contrast, but the math reveals the exact pattern!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Slit Widths}

\text{Amplitude and Slit Width}

\text{Ratio of Amplitudes}

\text{Intensity and Amplitude}

\text{Ratio of Intensities}

\text{Interference Pattern}

\text{Formula for Max/Min Intensity}

\text{Substituting Intensities}

\text{Simplifying the Expression}

\text{Canceling Common Terms}

\text{Final Calculation}

\text{Comparing with Given Ratio}

\text{Final Answer}

\text{The Way Forward}

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

Analyzing the Setup

The Master Equation

Final Calculation

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