In Young's double slit experiment, the separation between the slits is halved and the distance betwe

Q: What happens if the wavelength $\lambda$ is also changed?

If the wavelength is changed, you simply substitute the new wavelength $\lambda'$ into the formula. The new fringe width will be directly proportional to the new wavelength.

Q: Why does decreasing the slit separation increase the fringe width?

Decreasing the slit separation causes the light waves to spread out more (diffract more) after passing through the slits, which leads to a wider interference pattern on the screen.

Q: Does the intensity of the fringes change when $D$ and $d$ are changed?

Changing $D$ affects the overall intensity on the screen (intensity decreases as $D$ increases due to the inverse square law), but the relative intensity of the bright and dark fringes remains the same.

Q: What if the entire setup is immersed in water?

If immersed in water, the wavelength of light decreases ($\lambda' = \lambda / \mu$), which would cause the fringe width to decrease proportionally.

Q: Is this formula valid for all values of $D$ and $d$?

The formula $\beta = \frac{\lambda D}{d}$ is an approximation valid when $D$ is much larger than $d$, and $d$ is much larger than $\lambda$.

Question

In Young's double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

Options:

(A) unchanged

(B) halved

(C) doubled

(D) quadrupled

Sigma Percentile · Accepted Answer

### The Magic of Interference

Imagine you are standing in a dark room, and light from a single source passes through two tiny slits. Instead of just two bright lines on the wall, you see a beautiful pattern of alternating bright and dark bands. This is the magic of interference, beautifully demonstrated by Young's Double Slit Experiment.

The distance between two consecutive bright or dark bands is what we call the **fringe width**. It tells us how spread out this beautiful pattern is.

### The Master Equation

To understand how the pattern changes, we need our master equation. The fringe width, denoted by $\beta$, is governed by a very elegant relationship:

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

Here, $\lambda$ is the wavelength of the light, $D$ is the distance from the slits to the screen, and $d$ is the tiny separation between the two slits. Notice how $D$ is in the numerator and $d$ is in the denominator. This means if we move the screen further away, the pattern expands. But if we move the slits further apart, the pattern shrinks!

### Tweaking the Setup

Now, let's look at what our problem asks us to do. We are playing with the physical setup of the experiment.

First, the separation between the slits is halved. This means our new slit separation is $d' = \frac{d}{2}$.

Second, the distance between the slits and the screen is doubled. So, our new screen distance is $D' = 2D$.

### The Grand Finale

Let's see what happens to our fringe width when we apply these tweaks. We substitute our new values into the master equation:

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

$\beta' = \frac{\lambda (2D)}{\frac{d}{2}}$

When we simplify this, the $2$ in the denominator of the denominator flips up to the numerator, multiplying with the other $2$:

$\beta' = 4 \left(\frac{\lambda D}{d}\right)$

Since the term in the parenthesis is our original fringe width $\beta$, we get:

$\beta' = 4\beta$

The new fringe width is exactly four times the original! The pattern has expanded massively. In mathematical terms, the fringe width is **quadrupled**.

Animated Solution for Physics - Optics: In Young's double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

Select Answer:

Visualized Solution

$Fringe Width Formula$

$Applying New Conditions$

$Calculating New Fringe Width$

$Final Result$

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

The Magic of Interference

The Master Equation

Tweaking the Setup

The Grand Finale

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