Q.In a Young's double-slit experiment, the slits are separated by $0.28\ \text{mm}$ and the screen is placed $1.4\ \text{m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\ \text{cm}$. Determine the wavelength of light used in the experiment.

Concept: Young’s Double-Slit Interference — the position of bright fringes is given by $y_n = n \frac{\lambda D}{d}$.

Reasoning:

1. For the $n$-th bright fringe (central is $n=0$), the distance from the centre is

   $y_n = n \frac{\lambda D}{d}$.

2. Here $n = 4$, $y_4 = 1.2\ \text{cm} = 1.2 \times 10^{-2}\ \text{m}$,

   $d = 0.28\ \text{mm} = 2.8 \times 10^{-4}\ \text{m}$,

   $D = 1.4\ \text{m}$.

3. Rearranging: …

The key to solving this problem is understanding that in Young’s double-slit experiment, bright fringes (constructive interference) occur at positions where the path difference from the two slits is an integer multiple of the wavelength. The formula $y_n = n \frac{\lambda D}{d}$ directly relates the fringe position to the wavelength, slit separation, and screen distance. Here, we know the distance to the fourth bright fringe ($n=4$), so we can solve for $\lambda$ directly.

Let’s work through it step by step.

1. Identify the known quantities.

   Slit separation: $d = 0.28\ \text{mm} = 0.28 \times 10^{-3}\ \text{m} = 2.8 \times 10^{-4}\ \text{m}$.

   Screen distance: $D = 1.4\ \text{m}$.

   Distance to the fourth bright fringe from the central maximum: $y_4 = 1.2\ \text{cm} = 1.2 \times 10^{-2}\ \text{m}$.

   Fringe order: $n = 4$.

2. Recall the formula for bright fringe positions.

   For constructive interference in Young’s double-slit, the $n$-th bright fringe (where $n = 0, 1, 2, \dots$) is located at a distance from the central maximum given by:

Here $n=0$ gives the central bright fringe, $n=1$ the first bright fringe, and so on. The problem states “the distance between the central bright fringe and the fourth bright fringe” — that is exactly $y_4$.

3. Substitute the known values into the formula.

4. Solve for $\lambda$. First, simplify the right-hand side:

So the equation becomes:

Divide both sides by $2 \times 10^{4}$:

$\lambda = \frac{1.2 \times 10^{-2}}{2 \times 10^{4}} = 0.6 \times 10^{-6}\ \text{m} = 6 \times 10^{-7}\ \text{m}$ …

Method: Fringe Width Method for Young’s Double-Slit Experiment

This method uses the relationship between fringe spacing, slit separation, screen distance, and wavelength.

Step-by-step solution

Step 1: Identify the given data

• Slit separation, $d = 0.28\ \text{mm} = 0.28 \times 10^{-3}\ \text{m}$
• Screen distance, $D = 1.4\ \text{m}$
• Distance from central bright to fourth bright fringe, $y_4 = 1.2\ \text{cm} = 1.2 \times 10^{-2}\ \text{m}$
• Order of fringe, $n = 4$

Step 2: Recall the formula for bright fringe position

For $n$th bright fringe from centre:

Step 3: Substitute for the fourth bright fringe

Step 4: Solve for wavelength $\lambda$

Step 5: Plug in the values

$\lambda = 6.0 \times 10^{-7}\ \text{m}$ …

Common Mistakes in Young's Double-Slit Problems

Students often lose marks on this exact type of question. Here are the most frequent errors and how to avoid each:

1. Confusing fringe order with fringe number

The Mistake:

Students take $n = 4$ for the "fourth bright fringe" but incorrectly use the formula for the first fringe.

Why it happens:

The central bright fringe is $n = 0$, so the fourth bright fringe corresponds to $n = 4$, not $n = 3$.

How to avoid:

Always remember:

• Central bright fringe → $n = 0$
• First bright fringe → $n = 1$
• Fourth bright fringe → $n = 4$

So here, $n = 4$ is correct.

2. Unit conversion errors

The Mistake:

Using $d = 0.28$ (in mm) or $D = 1.4$ (in m) without converting to consistent units.

Why it happens:

The slit separation is given in mm, the screen distance in m, and the fringe distance in cm — three different units.

How to avoid:

Convert everything to metres before plugging into the formula:

• $d = 0.28\ \text{mm} = 0.28 \times 10^{-3}\ \text{m} = 2.8 \times 10^{-4}\ \text{m}$
• $D = 1.4\ \text{m}$
• $y_4 = 1.2\ \text{cm} = 1.2 \times 10^{-2}\ \text{m}$

3. Using the wrong formula for fringe position

The Mistake:

Using $y_n = \frac{n\lambda D}{d}$ when the problem gives the distance from the central fringe.

Why it happens:

Students sometimes use the formula for fringe spacing ($\beta = \frac{\lambda D}{d}$) and multiply by $n$, which is actually correct — but they forget that $y_n$ is measured from the centre.

How to avoid:

For bright fringes, the distance from the central fringe to the $n$th bright fringe is:

Here, $y_4 = 1.2 \times 10^{-2}\ \text{m}$, $n = 4$, $D = 1.4\ \text{m}$, $d = 2.8 \times 10^{-4}\ \text{m}$.

4. Algebraic rearrangement errors

The Mistake:

Solving for $\lambda$ incorrectly — e.g., multiplying instead of dividing.

How to avoid:

Rearrange step by step:

Multiply both sides by $d$:

$y_n d = n\lambda D$ …