In the Young's double slit experiment, the interference pattern is found to have an intensity ratio

Q: Why is $I_{\text{max}}$ not just $I_1 + I_2$?

Because of constructive interference, the amplitudes add up ($A_1 + A_2$), and since intensity is proportional to the square of amplitude, $I_{\text{max}} \propto (A_1 + A_2)^2 = I_1 + I_2 + 2\sqrt{I_1 I_2}$.

Q: What does an intensity ratio of $9$ physically mean?

It means the brightest spots on the screen are $9$ times brighter than the darkest spots. The dark spots are not completely dark unless $I_1 = I_2$.

Q: Can we take the negative root when solving for $\sqrt{I_1}/\sqrt{I_2}$?

Since intensities are positive quantities, their square roots are positive, so the ratio $\sqrt{I_1}/\sqrt{I_2}$ must be positive.

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

In any wave phenomenon, the energy transported per unit area per unit time (intensity) is proportional to the square of the wave's amplitude.

Q: What if the slit widths are different?

Slit width is directly proportional to the intensity of light passing through it. So, a ratio of intensities $I_1/I_2 = 4/1$ also implies the ratio of slit widths is $4:1$.

Question

In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. This implies that

Options:

(A) the intensities at the screen due to the two slits are 5 units and 4 units respectively

(B) the intensities at the screen due to the two slits are 4 units and 1 unit respectively

(C) the amplitude ratio is 3

(D) the amplitude ratio is 2

Sigma Percentile · Accepted Answer

The beauty of Young's Double Slit Experiment (YDSE) lies in its vivid demonstration of the wave nature of light. When coherent light from two slits overlaps on a screen, it creates a mesmerizing pattern of bright and dark bands known as interference fringes. But what dictates how bright the bright fringes are, and how dark the dark fringes get? It all comes down to the individual intensities of the two slits!

### Analyzing the Setup

In our problem, we are given a fascinating piece of information: the ratio of the maximum intensity (the brightest bright fringe) to the minimum intensity (the darkest dark fringe) is exactly $9$.

Mathematically, this is written as:
$$ \frac{I_{	ext{max}}}{I_{	ext{min}}} = 9 $$

We know from the principles of wave interference that the maximum and minimum intensities are related to the individual intensities of the two slits, $I_1$ and $I_2$, by the following formulas:
$$ I_{	ext{max}} = (\sqrt{I_1} + \sqrt{I_2})^2 $$
$$ I_{	ext{min}} = (\sqrt{I_1} - \sqrt{I_2})^2 $$

### The Master Equation

Let's substitute these theoretical expressions into our given ratio. This gives us our master equation for the problem:
$$ \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = 9 $$

This equation might look a bit intimidating with all those squares and square roots, but it simplifies beautifully. By taking the square root of both sides, we strip away the outer squares:
$$ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = 3 $$

Now, we have a simple linear equation in terms of the square roots of the intensities. Let's cross-multiply to solve for their relationship:
$$ \sqrt{I_1} + \sqrt{I_2} = 3(\sqrt{I_1} - \sqrt{I_2}) $$
$$ \sqrt{I_1} + \sqrt{I_2} = 3\sqrt{I_1} - 3\sqrt{I_2} $$

### Final Calculation

To isolate the terms, we bring all the $\sqrt{I_2}$ terms to the left side and the $\sqrt{I_1}$ terms to the right side:
$$ \sqrt{I_2} + 3\sqrt{I_2} = 3\sqrt{I_1} - \sqrt{I_1} $$
$$ 4\sqrt{I_2} = 2\sqrt{I_1} $$

Dividing both sides by $2$, we get:
$$ 2\sqrt{I_2} = \sqrt{I_1} $$

Finally, to find the ratio of the actual intensities, we square both sides of this simplified equation:
$$ 4I_2 = I_1 $$
$$ \frac{I_1}{I_2} = \frac{4}{1} $$

This tells us that the intensity of one slit is four times the intensity of the other. Looking at our options, **option (b)** correctly states that the intensities are $4$ units and $1$ unit respectively!

But wait, we must also check the amplitude ratio. A fundamental law of wave physics states that the intensity of a wave is directly proportional to the square of its amplitude ($I \propto A^2$). Therefore, the ratio of the intensities is equal to the square of the ratio of the amplitudes:
$$ \frac{I_1}{I_2} = \left(\frac{A_1}{A_2}ight)^2 $$

Substituting our known intensity ratio:
$$ \left(\frac{A_1}{A_2}ight)^2 = 4 $$

Taking the square root gives us the amplitude ratio:
$$ \frac{A_1}{A_2} = 2 $$

This perfectly matches **option (d)**. Thus, both options (b) and (d) are the correct answers to this elegant problem!

Animated Solution for Physics - Optics: In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. This implies that

Select Answer:

Visualized Solution

$I_{max}$ and $I_{min}$ Ratio

Intensity Formulas

Substituting the Formulas

Taking the Square Root

Cross Multiplication

Rearranging Terms

Intensity Ratio

Intensity and Amplitude Relation

Amplitude Ratio

Fringe Visibility

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

Analyzing the Setup

The Master Equation

Final Calculation

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