(a) A beam of light consisting of two wavelengths 400 nm and 600 nm is used to illuminate a single slit of width 1 mm. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed 1.5 m from the slit.

OR

(b) In a Young's double-slit experimental set-up with slit separation 0.6 mm a beam of light consisting of two wavelengths 440 nm and 660 nm is used to obtain interference pattern on a screen kept 1.5 m in front of the slits. Find the least distance of the point from the central maximum where the bright fringes due to both the wavelengths coincide.

Wave Optics

Coincidence of fringes — diffraction and interference

Official Answer

Lead: Fringes coincide when the two wavelengths satisfy $n_1\lambda_1 = n_2\lambda_2$ at the smallest integer orders.

Part (a) — single-slit dark fringes (minima):

Minima position: $y = n\lambda D/a$ .
Coincidence: $n_1\lambda_1 = n_2\lambda_2$ → $n_1(400) = n_2(600)$ → $n_1/n_2 = 3/2$ , so $n_1 = 3, n_2 = 2$ .
$y = (3 \times 400 \times 10^{-9} \times 1.5) / (1 \times 10^{-3})$
$y = 1.8 \times 10^{-3} \text{ m} = 1.8 \text{ mm}$ .

Part (b) — Young's double-slit bright fringes (maxima):

Maxima position: $y = n\lambda D/d$ .
Coincidence: $n_1(440) = n_2(660)$ → $n_1/n_2 = 3/2$ , so $n_1 = 3, n_2 = 2$ .
$y = (3 \times 440 \times 10^{-9} \times 1.5) / (0.6 \times 10^{-3})$
$y = 3.3 \times 10^{-3} \text{ m} = 3.3 \text{ mm}$ .

coincidence of fringessingle slit diffraction minimaYoung's double slit maximan1 lambda1 = n2 lambda2least distancey = n lambda D / a1.8 mm3.3 mm

Marking Scheme

1Part (a): 1 mark for $n_1\lambda_1 = n_2\lambda_2$ giving $n_1=3, n_2=2$ ; 1 mark for $y = 3\lambda_1 D/a = 1.8 \text{ mm}$ .
2Part (b): 1 mark for $n_1=3, n_2=2$ from $440/660$ ; 1 mark for $y = 3\lambda_1 D/d = 3.3 \text{ mm}$ .
3Either OR-part earns full 2 marks; correct SI substitution required.

Hint

Set $n_1\lambda_1 = n_2\lambda_2$ for smallest integers; use $y = n\lambda D/a$ (single slit) or $n\lambda D/d$ (double slit).

Quick Oral Answer

Fringes coincide when n1 lambda1 equals n2 lambda2 at the smallest whole numbers, giving orders 3 and 2; that yields 1.8 mm for the single-slit minima in part a and 3.3 mm for the double-slit maxima in part b.

Analysis & Explanation

Concept:

Both parts hinge on finding where a fringe of one colour lands exactly on a fringe of the other. This needs the lowest integers with $n_1\lambda_1 = n_2\lambda_2$ .

Part (a) — diffraction minima:

Single-slit dark fringes occur at $a \sin\theta = n\lambda$ , i.e. $y = n\lambda D/a$ .
Since $\lambda_1:\lambda_2 = 400:600 = 2:3$ , the third dark fringe of 400 nm meets the second dark fringe of 600 nm.

Part (b) — interference maxima:

Double-slit bright fringes occur at $y = n\lambda D/d$ .
With $440:660 = 2:3$ , the third bright fringe of 440 nm coincides with the second bright fringe of 660 nm.

Exam trap:

Use slit width a for single-slit diffraction, but slit separation d for the double slit — mixing them is the classic error.
Choose the smallest integer pair for the least distance.

Real-world:

Overlapping orders of different wavelengths is exactly what limits the resolving power of gratings and why spectrometers must separate overlapping spectral orders.

Common Mistakes

1Using slit separation d in part (a) where the slit width a is required (and vice versa in part b).
2Not choosing the smallest integer order pair ( $n_1=3, n_2=2$ ), so the 'least' distance condition is missed.
3Arithmetic slips converting nm and mm — keep all lengths in metres before dividing.

Interesting Facts

The ratio 2:3 of the wavelengths forces the first coincidence at the 3rd and 2nd fringes — a neat consequence of small integer ratios.

For single-slit diffraction it is the dark fringes (minima) that follow $a \cdot \sin\theta = n\lambda$ , unlike the bright-fringe formula of the double slit.

Young's 1801 double-slit experiment was the decisive evidence for the wave theory of light over Newton's corpuscular model.

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Frequently Asked Questions

When do fringes of two wavelengths coincide?

When $n_1\lambda_1 = n_2\lambda_2$ for integer orders. The least distance uses the smallest integer pair, which here is $n_1 = 3$ and $n_2 = 2$ since the wavelengths are in ratio 2:3.

Why is width a used in part (a) but separation d in part (b)?

Part (a) is single-slit diffraction where minima depend on slit width a ( $a \sin\theta = n\lambda$ ); part (b) is double-slit interference where maxima depend on slit separation d ( $y = n\lambda D/d$ ).

Do both OR options need to be solved in the exam?

No — a student attempts only one, either (a) giving 1.8 mm or (b) giving 3.3 mm. Both full solutions are shown here for completeness.