Question 182 marksVery Short AnswerSection B

(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 n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2 at the smallest integer orders.


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

  • Minima position: y=nλD/ay = n\lambda D/a.
  • Coincidence: n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2 → n1(400)=n2(600)n_1(400) = n_2(600) → n1/n2=3/2n_1/n_2 = 3/2, so n1=3,n2=2n_1 = 3, n_2 = 2.
  • y=(3×400×10−9×1.5)/(1×10−3)y = (3 \times 400 \times 10^{-9} \times 1.5) / (1 \times 10^{-3})
  • y=1.8×10−3 m=1.8 mmy = 1.8 \times 10^{-3} \text{ m} = 1.8 \text{ mm}.

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

  • Maxima position: y=nλD/dy = n\lambda D/d.
  • Coincidence: n1(440)=n2(660)n_1(440) = n_2(660) → n1/n2=3/2n_1/n_2 = 3/2, so n1=3,n2=2n_1 = 3, n_2 = 2.
  • y=(3×440×10−9×1.5)/(0.6×10−3)y = (3 \times 440 \times 10^{-9} \times 1.5) / (0.6 \times 10^{-3})
  • y=3.3×10−3 m=3.3 mmy = 3.3 \times 10^{-3} \text{ m} = 3.3 \text{ mm}.
Answer keywords:coincidence of fringessingle slit diffraction minimaYoung's double slit maximan1 lambda1 = n2 lambda2least distancey = n lambda D / a1.8 mm3.3 mm

Marking scheme

  1. 1.Part (a): 1 mark for n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2 giving n1=3,n2=2n_1=3, n_2=2; 1 mark for y=3λ1D/a=1.8 mmy = 3\lambda_1 D/a = 1.8 \text{ mm}.
  2. 2.Part (b): 1 mark for n1=3,n2=2n_1=3, n_2=2 from 440/660440/660; 1 mark for y=3λ1D/d=3.3 mmy = 3\lambda_1 D/d = 3.3 \text{ mm}.
  3. 3.Either OR-part earns full 2 marks; correct SI substitution required.

Hint

Set n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2 for smallest integers; use y=nλD/ay = n\lambda D/a (single slit) or nλD/dn\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.

Common mistakes

  1. 1.Using slit separation d in part (a) where the slit width a is required (and vice versa in part b).
  2. 2.Not choosing the smallest integer order pair (n1=3,n2=2n_1=3, n_2=2), so the 'least' distance condition is missed.
  3. 3.Arithmetic slips converting nm and mm — keep all lengths in metres before dividing.

Analysis and 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 n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2.


Part (a) — diffraction minima:

  • Single-slit dark fringes occur at asin⁡θ=nλa \sin\theta = n\lambda, i.e. y=nλD/ay = n\lambda D/a.
  • Since λ1:λ2=400:600=2:3\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=nλD/dy = n\lambda D/d.
  • With 440:660=2:3440:660 = 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.

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⋅sin⁡θ=nλa \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.

Question details

Q18 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Wave Optics
Topic
Coincidence of fringes — diffraction and interference
Bloom's level
apply
Syllabus topic
Unit VI: Optics — Single-slit diffraction minima and Young's double-slit maxima; coincidence of fringes
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

When do fringes of two wavelengths coincide?

When n1λ1=n2λ2n_1\lambda_1 = n_2\lambda_2 for integer orders. The least distance uses the smallest integer pair, which here is n1=3n_1 = 3 and n2=2n_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 (asin⁡θ=nλa \sin\theta = n\lambda); part (b) is double-slit interference where maxima depend on slit separation d (y=nλD/dy = 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.