In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment,

Q: Why do we use the fringe width formula for the separation between successive bright fringes?

In Young's Double Slit Experiment, the distance between any two consecutive bright fringes (or any two consecutive dark fringes) is constant and is defined as the fringe width $\beta$.

Q: Do I always need to convert units to meters?

Yes, mixing units like nanometers, millimeters, and meters will lead to incorrect results. Always convert to a single system, preferably SI units, before performing calculations.

Q: What if the question asked for the distance between a bright and a dark fringe?

The distance between a bright fringe and the adjacent dark fringe is exactly half of the fringe width, i.e., $\frac{\beta}{2}$.

Q: Why did we round off 5.89 mm to 5.9 mm?

We look at the given options. Since $5.9 \text{ mm}$ is present and $5.89 \text{ mm}$ is closest to it, we round it off to match the correct option.

Q: Does the fringe width depend on the intensity of the light source?

No, the fringe width $\beta = \frac{\lambda D}{d}$ only depends on the wavelength of the light, the slit separation, and the screen distance. It is completely independent of the intensity.

Question

In a Young's double slit experiment, the separation between the slits is $0.15	ext{ mm}$. In the experiment, a source of light of wavelength $589	ext{ nm}$ is used and the interference pattern is observed on a screen kept $1.5	ext{ m}$ away. The separation between the successive bright fringes on the screen is

Options:

(A) $4.9	ext{ mm}$

(B) $3.9	ext{ mm}$

(C) $6.9	ext{ mm}$

(D) $5.9	ext{ mm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a dark room, setting up the classic Young's Double Slit Experiment. You have two tiny slits separated by a microscopic distance $d$, and a screen placed far away at a distance $D$. When monochromatic light passes through these slits, it creates a beautiful, rhythmic pattern of bright and dark bands on the screen.

The separation between any two successive bright fringes (or dark fringes) is what we call the **fringe width**, denoted by the Greek letter $\beta$. This is the core physical quantity we need to find in this problem.

### The Master Equation

So, how do we calculate this fringe width? The geometry of the interference pattern gives us a very elegant formula:

$$ \beta = \frac{\lambda D}{d} $$

Here, $\lambda$ is the wavelength of the light we are using. This equation is incredibly insightful. It tells us that if we move the screen further away (increase $D$), the fringes get wider. Similarly, if we bring the slits closer together (decrease $d$), the fringes also spread out.

### Setting Up the Values

Now, let's look at the numbers given to us. This is where many students make a silly mistake—units! We must convert everything to standard SI units (meters) before plugging them into our equation.

*   **Wavelength ($\lambda$):** $589 	ext{ nm} = 589 	imes 10^{-9} 	ext{ m}$
*   **Screen Distance ($D$):** $1.5 	ext{ m}$
*   **Slit Separation ($d$):** $0.15 	ext{ mm} = 0.15 	imes 10^{-3} 	ext{ m}$

Let's carefully substitute these values into our master equation:

$$ \beta = \frac{589 	imes 10^{-9} 	imes 1.5}{0.15 	imes 10^{-3}} $$

### The Final Calculation

Let's do the math. To make the calculation smoother and avoid messy decimals, let's rewrite the denominator. We can express $0.15 	imes 10^{-3}$ as $1.5 	imes 10^{-4}$.

$$ \beta = \frac{589 	imes 10^{-9} 	imes 1.5}{1.5 	imes 10^{-4}} $$

Look closely at this step. The $1.5$ in the numerator and the $1.5$ in the denominator cancel out perfectly! This is a classic hallmark of well-designed JEE problems. We are left with:

$$ \beta = 589 	imes 10^{-5} 	ext{ m} $$

Let's convert this result into a more intuitive unit like millimeters.

$$ \beta = 5.89 	imes 10^{-3} 	ext{ m} = 5.89 	ext{ mm} $$

Looking at our given options, we can round this off to one decimal place, giving us **$5.9 	ext{ mm}$**. And there we have it, our final answer!

Select Answer:

Visualized Solution

$YDSE Setup$

$Formula for Fringe Width$

$Given Values$

$Calculation$

$Final Result$

$Conceptual Extension$

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

Analyzing the Setup

The Master Equation

Setting Up the Values

The Final Calculation

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