In a Young's double slit experiment, two slits are separated by 2 mm and the screen is placed one me

Q: Why do we call it fringe separation instead of fringe width?

They mean exactly the same thing! Fringe separation refers to the linear distance between the centers of two consecutive bright (or dark) fringes, which is the definition of fringe width.

Q: Why is it so important to convert everything to meters first?

Mixing units like millimeters and nanometers in the same equation will lead to massive power-of-ten errors. Converting everything to the standard SI unit (meters) acts as a safety net.

Q: Does the fringe width change if we look at higher-order fringes?

No, in a standard Young's Double Slit Experiment, the fringe width $\beta$ is constant across the entire interference pattern near the center.

Q: What if the experiment was performed in water instead of air?

The wavelength of light $\lambda$ would decrease because $\lambda_{water} = \frac{\lambda_{air}}{\mu}$. Since $\beta$ is directly proportional to $\lambda$, the fringe width would also decrease.

Q: Can I use this formula if the screen is very close to the slits?

The formula $\beta = \frac{\lambda D}{d}$ relies on the approximation that $D$ is much, much larger than $d$. If the screen is too close, this approximation fails, and the formula becomes inaccurate.

Question

In a Young's double slit experiment, two slits are separated by $2	ext{ mm}$ and the screen is placed one metre away. When a light of wavelength $500	ext{ nm}$ is used, the fringe separation will be

Options:

(A) $0.25	ext{ mm}$

(B) $0.50	ext{ mm}$

(C) $0.75	ext{ mm}$

(D) $1	ext{ mm}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine the classic Young's Double Slit Experiment. We have two tiny slits separated by a small distance $d$, and a screen placed far away at a distance $D$. Light waves pass through these slits and interfere to form a beautiful pattern of bright and dark fringes on the screen.

### The Master Equation

The question asks for the fringe separation, which is just another name for the fringe width. Do you remember the formula?

The fringe width, denoted by $\beta$, is given by the equation:

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

This elegant formula connects the macroscopic world (the fringe width we can measure) to the microscopic world (the wavelength of light).

### Setting Up the Values

Let's carefully write down what we know and ensure all our units are consistent. The slit separation $d$ is $2	ext{ mm}$, which we convert to $2 	imes 10^{-3}	ext{ m}$. The screen distance $D$ is exactly $1	ext{ m}$.

The wavelength $\lambda$ is $500	ext{ nm}$. Converting this to meters gives us $500 	imes 10^{-9}	ext{ m}$. Now, we substitute these into our formula:

$$ \beta = \frac{(500 	imes 10^{-9})(1)}{2 	imes 10^{-3}} $$

### Executing the Calculation

Now, let's simplify the numerator first. $500 	imes 10^{-9}$ can be written as $5 	imes 10^{-7}$. Multiplying by $1$ doesn't change anything.

Next, we divide $5 	imes 10^{-7}$ by $2 	imes 10^{-3}$. $5$ divided by $2$ is $2.5$. For the powers of ten, $-7 - (-3)$ gives us $-4$.

$$ \beta = 2.5 	imes 10^{-4}	ext{ m} $$

### The Final Answer

Finally, let's look at our options. They are all in millimeters. To convert our answer to millimeters, we multiply by $1000$.

$$ \beta = 0.25	ext{ mm} $$

And that matches option (a) perfectly!

### The Way Forward

What would happen to the fringe width if we immersed this entire experimental setup in water? The wavelength of light would decrease by a factor of the refractive index. Since fringe width is directly proportional to wavelength, the fringes would come closer together. Keep this concept in mind, it's a favorite for JEE!

Animated Solution for Physics - Optics: In a Young's double slit experiment, two slits are separated by $2 mm$ and the screen is placed one metre away. When a light of wavelength $500 nm$ is used, the fringe separation will be

Select Answer:

Visualized Solution

$Visualizing the Setup$

$The Master Formula$

$Substituting Values$

$Simplifying the Numerator$

$Executing the Division$

$Final Answer$

$The Way Forward$

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

Visualizing the Setup

The Master Equation

Setting Up the Values

Executing the Calculation

The Final Answer

The Way Forward

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