Q.A beam of light consisting of two wavelengths, $650\ \text{nm}$ and $520\ \text{nm}$, is used to obtain interference fringes in a Young's double-slit experiment.

$n$-th bright fringe: $y_n=\dfrac{n\lambda D}{d}$. Using the standard values for this NCERT problem, $D=1.2\ \text{m}$, $d=2\ \text{mm}$:

(a) Third bright fringe ($n=3$) for $\lambda_1=650\ \text{nm}$: $y_3=\dfrac{3\times650\text{nm}\times1.2\text{m}}{2\text{mm}}\approx1.17\ \text{mm}$. …

The governing relation

In Young's double-slit experiment, a bright fringe occurs at path difference $d\sin\theta=n\lambda$; for small angles ($\sin\theta\approx y/D$), the $n$-th bright fringe sits at

(This problem is the standard NCERT exercise, which supplies slit separation $d=2\ \text{mm}$ and screen distance $D=1.2\ \text{m}$; these values are used below.)

(a) Third bright fringe for $\lambda_1=650\ \text{nm}$

The central maximum is $n=0$, so the third bright fringe is $n=3$:

(b) Least distance where bright fringes of both wavelengths coincide

A bright fringe of $\lambda_1=650\ \text{nm}$ lands on a bright fringe of $\lambda_2=520\ \text{nm}$ when their positions match:

The smallest positive integers satisfying this are $n_1=4$, $n_2=5$ (i.e. the 4th bright fringe of the $650\ \text{nm}$ light coincides with the 5th bright fringe of the $520\ \text{nm}$ light - check: $4\times650=2600=5\times520$, confirmed). …

Method: Path Difference Approach for Young’s Double-Slit Interference

This method uses the condition for bright fringes based on path difference, then applies it to find positions on the screen.

Step 1: Recall the condition for bright fringes

For a bright fringe (constructive interference) in Young’s double-slit experiment:

where $n = 0, 1, 2, \dots$ gives the order of the bright fringe ($n=0$ is the central maximum).

The position of the $n^{\text{th}}$ bright fringe on the screen is:

where:

• $D$ = distance from slits to screen
• $d$ = separation between slits
• $\lambda$ = wavelength of light

Step 2: Solve part (a) — Third bright fringe for $\lambda = 650\ \text{nm}$

For the third bright fringe, $n = 3$.

Answer (a):

Note: Since $D$ and $d$ are not given in the problem, the answer is expressed in terms of these parameters. In an exam, if numerical values are provided, substitute them directly.

Step 3: Solve part (b) — Least distance where bright fringes coincide

Key idea: Two bright fringes coincide when their positions on the screen are equal.

Let $\lambda_1 = 650\ \text{nm}$ and $\lambda_2 = 520\ \text{nm}$.

For coincidence at some distance $y$ from the centre:

Cancelling $D/d$:

Substitute the wavelengths:

Step 4: Find the smallest integers $n_1, n_2$

Divide both sides by 10:

$65 n_1 = 52 n_2$ …

🧠 Common Mistakes & How to Avoid Them

✗ Mistake 1: Using the wrong formula for bright fringe position

• What students do: Use $y = \frac{n \lambda D}{d}$ but forget whether $n$ starts from 0 or 1.
• Why it’s wrong: For bright fringes, $n = 0$ is central maximum, $n = 1$ is first bright, etc. So third bright fringe means $n = 3$, not $n = 2$.
• ✓ Fix: Always write: $y_n = \frac{n \lambda D}{d}$, with $n = 0,1,2,\dots$

✗ Mistake 2: Not converting units (nm → m)

• What students do: Plug $650$ directly without converting to metres.
• Why it’s wrong: $d$ and $D$ are in metres — mismatch gives wrong answer.
• ✓ Fix: Convert: $650\ \text{nm} = 650 \times 10^{-9}\ \text{m}$.

✗ Mistake 3: Confusing “coincide” with “overlap of any fringe”

• What students do: Set $n_1 \lambda_1 = n_2 \lambda_2$ but forget both $n_1$ and $n_2$ must be integers.
• Why it’s wrong: Coincidence means bright fringe of one wavelength exactly at same position as bright fringe of the other.
• ✓ Fix: Solve $n_1 \lambda_1 = n_2 \lambda_2$ for smallest integers $n_1, n_2$.

✗ Mistake 4: Forgetting that $n$ starts from 0 for central maximum

• What students do: Take $n=1$ as the first coincidence.
• Why it’s wrong: At $n_1 = n_2 = 0$, both central maxima coincide — but question asks least distance from central maximum (excluding the centre itself).
• ✓ Fix: Find smallest non-zero integers satisfying $n_1 \lambda_1 = n_2 \lambda_2$.

✓ Correct Step-by-Step Solution

Given:

• $\lambda_1 = 650\ \text{nm} = 650 \times 10^{-9}\ \text{m}$
• $\lambda_2 = 520\ \text{nm} = 520 \times 10^{-9}\ \text{m}$
• Let slit separation $= d$, screen distance $= D$ (both in metres)

(a) Third bright fringe for $\lambda_1 = 650\ \text{nm}$

For bright fringe:

Third bright fringe $\Rightarrow n = 3$

Answer:

If $D$ and $d$ are given, substitute directly.

(b) Least distance where bright fringes coincide (excluding centre) …