In a Young's double slit experiment, the ratio of the slit's width is 4 : 1. The ratio of the intens

Q: Why do we take the square root of the width ratio?

Intensity is directly proportional to the slit width ($I \propto W$). However, the interference formula uses amplitudes. Since intensity is proportional to the square of the amplitude ($I \propto a^2$), the amplitude is proportional to the square root of the intensity, and thus the square root of the width ($a \propto \sqrt{W}$).

Q: Can I just use the intensities directly in the formula?

Yes! The formula $\frac{I_{max}}{I_{min}} = \left(\frac{a_1 + a_2}{a_1 - a_2}\right)^2$ can be rewritten as $\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2$. Since $I \propto W$, you can even plug in the widths directly: $\left(\frac{\sqrt{W_1} + \sqrt{W_2}}{\sqrt{W_1} - \sqrt{W_2}}\right)^2$.

Q: What does 'close to the central fringe' mean in the question?

It implies that we are looking at the interference pattern near the center of the screen where the path difference is small, and the intensities of the individual waves haven't dropped off significantly due to the single-slit diffraction envelope. It ensures we can treat $I_1$ and $I_2$ as constants.

Q: What happens if the slit widths are equal?

If $W_1 = W_2$, then $a_1 = a_2$. The minimum intensity $I_{min} \propto (a_1 - a_2)^2$ becomes exactly zero. This results in perfectly dark minima and maximum contrast, which is the ideal setup for observing interference.

Q: Does the distance between the slits ($d$) or the screen ($D$) affect this ratio?

No. The distances $d$ and $D$, along with the wavelength $\lambda$, determine the *fringe width* (how far apart the fringes are). The *ratio* of maximum to minimum intensity depends solely on the relative amplitudes (or widths) of the two interfering sources.

Question

In a Young's double slit experiment, the ratio of the slit's width is $4 : 1$. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

Options:

(A) $4 : 1$

(B) $25 : 9$

(C) $9 : 1$

(D) $(\sqrt{3} + 1)^4 : 16$

Sigma Percentile · Accepted Answer

Imagine you are standing in a dark room, observing the beautiful, rhythmic bands of light and dark on a screen. The Young's Double Slit Experiment (YDSE) is not just a laboratory setup; it is a profound demonstration of the wave nature of light.

In the standard textbook version of this experiment, we almost always assume a perfect symmetry: the two slits are identical. This symmetry leads to perfectly dark minima, creating a sharp, beautiful contrast. But what happens when we break this symmetry? What if one slit is wider than the other? This is exactly the twist presented in this problem, and it forces us to deeply understand the relationship between physical geometry and wave properties.

### Analyzing the Setup: Slit Width and Intensity

The problem states that the ratio of the slit widths is $4 : 1$. Let's call the width of the first slit $W_1$ and the second slit $W_2$. Therefore, we have:

$$W_1 : W_2 = 4 : 1$$

Now, we must ask ourselves: what physical quantity does the slit width directly control? The answer is the **intensity** of the light passing through it. Think of the slit as a window. A wider window allows more light energy to stream into the room per second. Mathematically, the intensity ($I$) of the light emerging from a slit is directly proportional to the area of the slit. Assuming the slits have the same height, the area is directly proportional to the width ($W$).

$$I \propto W$$

Consequently, the ratio of the intensities of the light emerging from the two slits is equal to the ratio of their widths:

$$\frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{4}{1}$$

### The Master Equation: Amplitude and Intensity

While intensity tells us about the energy, interference is fundamentally a phenomenon of wave superposition, which deals with **amplitudes**. We need to bridge the gap between intensity and amplitude. From the core principles of wave optics, we know that the intensity of a wave is directly proportional to the square of its amplitude ($a$).

$$I \propto a^2$$

By taking the square root of both sides, we can see that the amplitude is proportional to the square root of the intensity, and therefore, proportional to the square root of the slit width.

$$a \propto \sqrt{I} \propto \sqrt{W}$$

### Calculating the Amplitude Ratio

Armed with this relationship, we can easily find the ratio of the amplitudes of the two interfering light waves.

$$\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{W_1}{W_2}}$$

Substituting the given width ratio into our equation:

$$\frac{a_1}{a_2} = \sqrt{\frac{4}{1}} = \frac{2}{1}$$

This tells us that the light wave emerging from the wider slit has twice the amplitude of the wave emerging from the narrower slit.

### The Interference Pattern: Maxima and Minima

Now, let's shift our focus to the screen where the two waves meet and interfere. The maximum intensity ($I_{max}$) occurs where the waves interfere constructively (crest meets crest), meaning their amplitudes add up.

$$a_{max} = a_1 + a_2$$

Conversely, the minimum intensity ($I_{min}$) occurs where they interfere destructively (crest meets trough), meaning their amplitudes subtract.

$$a_{min} = a_1 - a_2$$

Since intensity is the square of the amplitude, the ratio of maximum to minimum intensity is given by the master formula:

$$\frac{I_{max}}{I_{min}} = \left(\frac{a_1 + a_2}{a_1 - a_2}\right)^2$$

### Final Calculation

To make the calculation effortless, we can divide the numerator and the denominator inside the parenthesis by $a_2$. This allows us to directly plug in the amplitude ratio we found earlier.

$$\frac{I_{max}}{I_{min}} = \left(\frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1}\right)^2$$

Substitute $\frac{a_1}{a_2} = 2$ into the equation:

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

$$\frac{I_{max}}{I_{min}} = \left(\frac{3}{1}\right)^2 = \frac{9}{1}$$

And there we have it! The ratio of the maximum intensity to the minimum intensity in the interference pattern is **$9 : 1$**.

This problem beautifully illustrates how a simple geometric change in the experimental setup cascades through the physics of waves—from width to intensity, from intensity to amplitude, and finally back to the intensity of the resulting interference pattern. Always remember to trace the physical logic step-by-step!

Animated Solution for Physics - Optics: In a Young's double slit experiment, the ratio of the slit's width is $4 : 1$ . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

Select Answer:

Visualized Solution

$Visual Anchor: The Setup$

$I \propto W$

$I \propto a^{2}$

$Amplitude Ratio$

$Intensity Ratio Formula$

$Substitution$

$Final Calculation$

$The Way Forward$

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

Analyzing the Setup

The Master Equation

Calculating the Amplitude Ratio

The Interference Pattern

Final Calculation

Similar Questions

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.

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 ......... .

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

In a Young's double slit experiment, the intensity at a point where the path difference is $\frac{λ}{6}$ ( $λ$ being the wavelength of the light used) is $I$ . If $I_{0}$ denotes the maximum intensity, then $I / I_{0}$ is equal to

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $I_{m}$ is the maximum intensity, the resultant intensity $I$ when they interfere at phase difference $ϕ$ , is given by

At two points $P$ and $Q$ on screen in Young's double slit experiment, waves from slits $S_{1}$ and $S_{2}$ have a path difference of $0$ and $\frac{λ}{4}$ , respectively. The ratio of intensities at $P$ and $Q$ will be

In a Young's double slit experiment, the path difference at a certain point on the screen between two interfering waves is $\frac{1}{8}$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to

In Young's double slit experiment, the intensity at a point is (1/4) of the maximum intensity. Angular position of this point is

In the Young's double slit experiment, the distance between the slits varies in time as $d (t) = d_{0} + a_{0} sin ω t$ , where $d_{0}, ω$ and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as