Coherent Sources
S₁ and S₂ have the same frequency and a constant phase difference because both are illuminated by one source.
Wave Optics · Chapter 03
Young Double Slit Experiment
Build YDSE from coherent sources and path difference to fringe width, fringe shift, missing orders and intensity distribution.
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01
YDSE Setup
Young's experiment demonstrates the wave nature of light by dividing one incident wavefront into two coherent secondary sources S₁ and S₂. The slits are narrow, have separation d, and illuminate a screen at distance D where D ≫ d.
Experimental Assumptions
Monochromatic light, narrow slits, D ≫ d, small observation angle and overlapping diffracted beams.
Central Fringe
At O, both optical paths are equal: Δ = 0. Hence the central fringe is bright.
Δ = path differenceφ = phase difference
02
Fringe Formation
For a point P at transverse displacement x, the rays make a small angle θ.
Path difference: Δ = S₂P − S₁P = d sinθ.
For small θ: sinθ ≈ tanθ = x/D.
Therefore: Δ = xd/D.
Phase difference: φ = (2π/λ)Δ = 2πxd/(λD).
Δ = xd/Dφ = (2π/λ)Δ
Notation rule: Δ is used only for path difference. φ is used only for phase difference. No other symbol is used for path difference.
03
Position of Bright Fringes
For constructive interference: Δ = nλ, where n = 0, 1, 2, 3, ...
Using Δ = dxₙ/D: dxₙ/D = nλ.
Hence xₙ = nλD/d = nβ.
For n = 0, x₀ = 0, giving the central bright fringe.
04
Position of Dark Fringes
For destructive interference: Δ = (2n−1)λ/2, where n = 1, 2, 3, ...
Using Δ = dxₙ/D: dxₙ/D = (2n−1)λ/2.
Hence xₙ = (2n−1)λD/(2d) = (2n−1)β/2.
The first dark fringe is at x = β/2 on either side of O.
05
Fringe Width
The positions of consecutive bright fringes are xₙ = nλD/d and xₙ₊₁ = (n+1)λD/d.
β = xₙ₊₁ − xₙ = λD/d.
The same separation is obtained between consecutive dark fringes.
| Change | Effect on β | Reason |
|---|---|---|
| λ increases | β increases | β ∝ λ |
| D increases | β increases | β ∝ D |
| d increases | β decreases | β ∝ 1/d |
06
Angular Fringe Width
Angular position θₙ ≈ xₙ/D.
For consecutive bright fringes, Δθ = β/D.
Using β = λD/d: Δθ = λ/d.
07
Effect of Changing λ, D and d
Increase λ
Bright and dark fringes move farther apart. Red light gives wider fringes than blue light.
Increase D
The pattern expands linearly on the screen, although angular fringe width remains unchanged.
Increase d
The pattern contracts because the path difference changes more rapidly with x.
08
Source Shift
Moving the illuminating source introduces a constant initial path difference between S₁ and S₂. The zero-path point therefore moves, shifting the entire fringe system.
Source shifted upward
The pattern shifts in the opposite sense required to restore equal total paths.
Source shifted downward
The shift reverses. Its magnitude depends on source geometry.
Source shift moves the complete pattern but does not change fringe width.
09
Screen Shift
Screen moved away
D increases, so β = λD/d increases and the fringes spread out.
Screen moved closer
D decreases, so β decreases and the fringes come closer.
10
Thin Film in One Path
A plate of thickness t and refractive index μ replaces the same thickness of air.
Additional optical path: Δp = μt − t = (μ−1)t.
A shift Δx produces compensating geometrical path dΔx/D.
Equating: dΔx/D = Δp.
Therefore Δx = DΔp/d = D(μ−1)t/d.
Since β = λD/d: Δx = β(μ−1)t/λ.
Δp = (μ−1)tΔx = D(μ−1)t/d = β(μ−1)t/λ
The entire fringe system shifts toward the slit containing the plate. Fringe width remains unchanged.
11
Immersion in Liquid
In a liquid of refractive index μ, wavelength becomes λ′ = λ/μ.
β′ = λ′D/d = λD/(μd).
Thus β′ = β/μ.
Frequency is unchanged on entering the liquid, but wave speed and wavelength decrease, so the fringes become narrower.
12
Missing Orders
Let each slit have width a and slit separation d.
Interference maximum: d sinθ = nλ.
Single-slit diffraction minimum: a sinθ = mλ.
If both coincide, n = m(d/a).
Therefore interference orders equal to integer multiples of d/a are missing when d/a is an integer.
Example: If d = 4a, the 4th, 8th, 12th, ... interference maxima are absent.
13
Intensity Distribution
General result: I = I₁ + I₂ + 2√(I₁I₂)cosφ.
For I₁ = I₂ = I₀: I = 2I₀(1+cosφ).
Using 1+cosφ = 2cos²(φ/2): I = 4I₀cos²(φ/2).
At φ = 2nπ, Imax = 4I₀. At φ = (2n−1)π, Imin = 0.
Visibility V = (Imax−Imin)/(Imax+Imin).
14
50 Solved Numericals
NEET
1. Fringe width
For λ = 500 nm, D = 1 m and d = 0.5 mm, find β.
Answer: 1.000 mm
Step-by-step solution: β = λD/d = (500×10⁻⁹×1)/(0.5×10⁻³) = 1.000e-3 m = 1.000 mm.
NEET
2. Bright fringe position
If β = 1.10 mm, find the position of the 2nd bright fringe.
Answer: 2.20 mm
Step-by-step solution: xₙ = nβ = 2 × 1.10 = 2.20 mm.
NEET
3. Dark fringe position
For β = 1.20 mm, find the 3rd dark fringe from the center.
Answer: 3.00 mm
Step-by-step solution: xₙ = (2n−1)β/2 = 5×1.20/2 = 3.00 mm.
NEET
4. Thin sheet shift
A sheet of μ = 1.55 and thickness 5 μm is inserted. D = 2.5 m, d = 0.5 mm. Find Δx.
Answer: 13.750 mm
Step-by-step solution: Δp = (μ−1)t. Therefore Δx = D(μ−1)t/d = 13.750 mm.
NEET
5. Liquid immersion
The apparatus is immersed in a liquid of refractive index 1.40. Find β′/β.
Answer: 0.714
Step-by-step solution: β′ = β/μ, so β′/β = 1/1.40 = 0.714.
NEET
6. Missing order
Interference slit separation is 3 times the slit width. Which interference orders are missing?
Answer: Multiples of 3
Step-by-step solution: Diffraction minimum a sinθ = mλ coincides with interference maximum d sinθ = nλ. Thus n = m(d/a) = 3m; orders 3, 6, 9, ... are missing.
NEET
7. Intensity ratio
Two slit intensities are 4 and 1 units. Find Imax and Imin.
Answer: Imax = 9, Imin = 1
Step-by-step solution: Imax = (√I₁+√I₂)² and Imin = (√I₁−√I₂)². Substitution gives the stated values.
NEET
8. Path and phase
At a point, path difference Δ = 0.75λ. Find phase difference φ.
Answer: 1.50π rad
Step-by-step solution: φ = (2π/λ)Δ = 2π(0.75) = 1.50π rad.
NEET
9. Fringe width
For λ = 650 nm, D = 1 m and d = 1 mm, find β.
Answer: 0.650 mm
Step-by-step solution: β = λD/d = (650×10⁻⁹×1)/(1×10⁻³) = 6.500e-4 m = 0.650 mm.
NEET
10. Bright fringe position
If β = 2.10 mm, find the position of the 5th bright fringe.
Answer: 10.50 mm
Step-by-step solution: xₙ = nβ = 5 × 2.10 = 10.50 mm.
NEET
11. Dark fringe position
For β = 1.33 mm, find the 3rd dark fringe from the center.
Answer: 3.33 mm
Step-by-step solution: xₙ = (2n−1)β/2 = 5×1.33/2 = 3.33 mm.
NEET
12. Thin sheet shift
A sheet of μ = 1.55 and thickness 3 μm is inserted. D = 2.5 m, d = 1 mm. Find Δx.
Answer: 4.125 mm
Step-by-step solution: Δp = (μ−1)t. Therefore Δx = D(μ−1)t/d = 4.125 mm.
NEET
13. Liquid immersion
The apparatus is immersed in a liquid of refractive index 1.40. Find β′/β.
Answer: 0.714
Step-by-step solution: β′ = β/μ, so β′/β = 1/1.40 = 0.714.
NEET
14. Missing order
Interference slit separation is 3 times the slit width. Which interference orders are missing?
Answer: Multiples of 3
Step-by-step solution: Diffraction minimum a sinθ = mλ coincides with interference maximum d sinθ = nλ. Thus n = m(d/a) = 3m; orders 3, 6, 9, ... are missing.
NEET
15. Intensity ratio
Two slit intensities are 25 and 9 units. Find Imax and Imin.
Answer: Imax = 64, Imin = 4
Step-by-step solution: Imax = (√I₁+√I₂)² and Imin = (√I₁−√I₂)². Substitution gives the stated values.
NEET
16. Path and phase
At a point, path difference Δ = 0.25λ. Find phase difference φ.
Answer: 0.50π rad
Step-by-step solution: φ = (2π/λ)Δ = 2π(0.25) = 0.50π rad.
NEET
17. Fringe width
For λ = 550 nm, D = 1 m and d = 0.75 mm, find β.
Answer: 0.733 mm
Step-by-step solution: β = λD/d = (550×10⁻⁹×1)/(0.75×10⁻³) = 7.333e-4 m = 0.733 mm.
NEET
18. Bright fringe position
If β = 0.90 mm, find the position of the 3rd bright fringe.
Answer: 2.70 mm
Step-by-step solution: xₙ = nβ = 3 × 0.90 = 2.70 mm.
JEE Main
19. Dark fringe position
For β = 2.60 mm, find the 3rd dark fringe from the center.
Answer: 6.50 mm
Step-by-step solution: xₙ = (2n−1)β/2 = 5×2.60/2 = 6.50 mm.
JEE Main
20. Thin sheet shift
A sheet of μ = 1.55 and thickness 6 μm is inserted. D = 2.5 m, d = 0.75 mm. Find Δx.
Answer: 11.000 mm
Step-by-step solution: Δp = (μ−1)t. Therefore Δx = D(μ−1)t/d = 11.000 mm.
JEE Main
21. Liquid immersion
The apparatus is immersed in a liquid of refractive index 1.40. Find β′/β.
Answer: 0.714
Step-by-step solution: β′ = β/μ, so β′/β = 1/1.40 = 0.714.
JEE Main
22. Missing order
Interference slit separation is 3 times the slit width. Which interference orders are missing?
Answer: Multiples of 3
Step-by-step solution: Diffraction minimum a sinθ = mλ coincides with interference maximum d sinθ = nλ. Thus n = m(d/a) = 3m; orders 3, 6, 9, ... are missing.
JEE Main
23. Intensity ratio
Two slit intensities are 9 and 4 units. Find Imax and Imin.
Answer: Imax = 25, Imin = 1
Step-by-step solution: Imax = (√I₁+√I₂)² and Imin = (√I₁−√I₂)². Substitution gives the stated values.
JEE Main
24. Path and phase
At a point, path difference Δ = 1λ. Find phase difference φ.
Answer: 2.00π rad
Step-by-step solution: φ = (2π/λ)Δ = 2π(1) = 2.00π rad.
JEE Main
25. Fringe width
For λ = 700 nm, D = 1 m and d = 0.5 mm, find β.
Answer: 1.400 mm
Step-by-step solution: β = λD/d = (700×10⁻⁹×1)/(0.5×10⁻³) = 1.400e-3 m = 1.400 mm.
JEE Main
26. Bright fringe position
If β = 1.00 mm, find the position of the 1st bright fringe.
Answer: 1.00 mm
Step-by-step solution: xₙ = nβ = 1 × 1.00 = 1.00 mm.
JEE Main
27. Dark fringe position
For β = 1.10 mm, find the 3rd dark fringe from the center.
Answer: 2.75 mm
Step-by-step solution: xₙ = (2n−1)β/2 = 5×1.10/2 = 2.75 mm.
JEE Main
28. Thin sheet shift
A sheet of μ = 1.55 and thickness 4 μm is inserted. D = 2.5 m, d = 0.5 mm. Find Δx.
Answer: 11.000 mm
Step-by-step solution: Δp = (μ−1)t. Therefore Δx = D(μ−1)t/d = 11.000 mm.
JEE Main
29. Liquid immersion
The apparatus is immersed in a liquid of refractive index 1.40. Find β′/β.
Answer: 0.714
Step-by-step solution: β′ = β/μ, so β′/β = 1/1.40 = 0.714.
JEE Main
30. Missing order
Interference slit separation is 3 times the slit width. Which interference orders are missing?
Answer: Multiples of 3
Step-by-step solution: Diffraction minimum a sinθ = mλ coincides with interference maximum d sinθ = nλ. Thus n = m(d/a) = 3m; orders 3, 6, 9, ... are missing.
JEE Main
31. Intensity ratio
Two slit intensities are 1 and 1 units. Find Imax and Imin.
Answer: Imax = 4, Imin = 0
Step-by-step solution: Imax = (√I₁+√I₂)² and Imin = (√I₁−√I₂)². Substitution gives the stated values.
JEE Main
32. Path and phase
At a point, path difference Δ = 0.5λ. Find phase difference φ.
Answer: 1.00π rad
Step-by-step solution: φ = (2π/λ)Δ = 2π(0.5) = 1.00π rad.
JEE Main
33. Fringe width
For λ = 600 nm, D = 1 m and d = 1 mm, find β.
Answer: 0.600 mm
Step-by-step solution: β = λD/d = (600×10⁻⁹×1)/(1×10⁻³) = 6.000e-4 m = 0.600 mm.
JEE Main
34. Bright fringe position
If β = 1.95 mm, find the position of the 4th bright fringe.
Answer: 7.80 mm
Step-by-step solution: xₙ = nβ = 4 × 1.95 = 7.80 mm.
JEE Main
35. Dark fringe position
For β = 1.87 mm, find the 3rd dark fringe from the center.
Answer: 4.67 mm
Step-by-step solution: xₙ = (2n−1)β/2 = 5×1.87/2 = 4.67 mm.
JEE Main
36. Thin sheet shift
A sheet of μ = 1.55 and thickness 2 μm is inserted. D = 2.5 m, d = 1 mm. Find Δx.
Answer: 2.750 mm
Step-by-step solution: Δp = (μ−1)t. Therefore Δx = D(μ−1)t/d = 2.750 mm.
JEE Advanced
37. Liquid immersion
The apparatus is immersed in a liquid of refractive index 1.40. Find β′/β.
Answer: 0.714
Step-by-step solution: β′ = β/μ, so β′/β = 1/1.40 = 0.714.
JEE Advanced
38. Missing order
Interference slit separation is 3 times the slit width. Which interference orders are missing?
Answer: Multiples of 3
Step-by-step solution: Diffraction minimum a sinθ = mλ coincides with interference maximum d sinθ = nλ. Thus n = m(d/a) = 3m; orders 3, 6, 9, ... are missing.
JEE Advanced
39. Intensity ratio
Two slit intensities are 16 and 9 units. Find Imax and Imin.
Answer: Imax = 49, Imin = 1
Step-by-step solution: Imax = (√I₁+√I₂)² and Imin = (√I₁−√I₂)². Substitution gives the stated values.
JEE Advanced
40. Path and phase
At a point, path difference Δ = 1.25λ. Find phase difference φ.
Answer: 2.50π rad
Step-by-step solution: φ = (2π/λ)Δ = 2π(1.25) = 2.50π rad.
JEE Advanced
41. Fringe width
For λ = 500 nm, D = 1 m and d = 0.75 mm, find β.
Answer: 0.667 mm
Step-by-step solution: β = λD/d = (500×10⁻⁹×1)/(0.75×10⁻³) = 6.667e-4 m = 0.667 mm.
JEE Advanced
42. Bright fringe position
If β = 0.82 mm, find the position of the 2nd bright fringe.
Answer: 1.65 mm
Step-by-step solution: xₙ = nβ = 2 × 0.82 = 1.65 mm.
JEE Advanced
43. Dark fringe position
For β = 2.40 mm, find the 3rd dark fringe from the center.
Answer: 6.00 mm
Step-by-step solution: xₙ = (2n−1)β/2 = 5×2.40/2 = 6.00 mm.
JEE Advanced
44. Thin sheet shift
A sheet of μ = 1.55 and thickness 5 μm is inserted. D = 2.5 m, d = 0.75 mm. Find Δx.
Answer: 9.167 mm
Step-by-step solution: Δp = (μ−1)t. Therefore Δx = D(μ−1)t/d = 9.167 mm.
JEE Advanced
45. Liquid immersion
The apparatus is immersed in a liquid of refractive index 1.40. Find β′/β.
Answer: 0.714
Step-by-step solution: β′ = β/μ, so β′/β = 1/1.40 = 0.714.
JEE Advanced
46. Missing order
Interference slit separation is 3 times the slit width. Which interference orders are missing?
Answer: Multiples of 3
Step-by-step solution: Diffraction minimum a sinθ = mλ coincides with interference maximum d sinθ = nλ. Thus n = m(d/a) = 3m; orders 3, 6, 9, ... are missing.
JEE Advanced
47. Intensity ratio
Two slit intensities are 4 and 4 units. Find Imax and Imin.
Answer: Imax = 16, Imin = 0
Step-by-step solution: Imax = (√I₁+√I₂)² and Imin = (√I₁−√I₂)². Substitution gives the stated values.
JEE Advanced
48. Path and phase
At a point, path difference Δ = 0.75λ. Find phase difference φ.
Answer: 1.50π rad
Step-by-step solution: φ = (2π/λ)Δ = 2π(0.75) = 1.50π rad.
JEE Advanced
49. Fringe width
For λ = 650 nm, D = 1 m and d = 0.5 mm, find β.
Answer: 1.300 mm
Step-by-step solution: β = λD/d = (650×10⁻⁹×1)/(0.5×10⁻³) = 1.300e-3 m = 1.300 mm.
JEE Advanced
50. Bright fringe position
If β = 1.40 mm, find the position of the 5th bright fringe.
Answer: 7.00 mm
Step-by-step solution: xₙ = nβ = 5 × 1.40 = 7.00 mm.
15
PYQ-Pattern and Exam-Style Practice
Academic note: These are original questions in the requested examination patterns, not verbatim reproductions of copyrighted papers.
CBSE PYQ-Pattern Practice · 30 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q21For bright position, select the correct YDSE result.
Answer: A. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q22For dark position, select the correct YDSE result.
Answer: D. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q23For fringe width, select the correct YDSE result.
Answer: C. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q24For angular fringe width, select the correct YDSE result.
Answer: B. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q25For thin-sheet optical path, select the correct YDSE result.
Answer: A. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q26For thin-sheet shift, select the correct YDSE result.
Answer: D. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q27For liquid fringe width, select the correct YDSE result.
Answer: C. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q28For equal-beam intensity, select the correct YDSE result.
Answer: B. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q29For central fringe, select the correct YDSE result.
Answer: A. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q30For source shift effect, select the correct YDSE result.
Answer: D. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
NEET PYQ-Pattern Practice · 30 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q21For bright position, select the correct YDSE result.
Answer: A. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q22For dark position, select the correct YDSE result.
Answer: D. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q23For fringe width, select the correct YDSE result.
Answer: C. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q24For angular fringe width, select the correct YDSE result.
Answer: B. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q25For thin-sheet optical path, select the correct YDSE result.
Answer: A. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q26For thin-sheet shift, select the correct YDSE result.
Answer: D. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q27For liquid fringe width, select the correct YDSE result.
Answer: C. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q28For equal-beam intensity, select the correct YDSE result.
Answer: B. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q29For central fringe, select the correct YDSE result.
Answer: A. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q30For source shift effect, select the correct YDSE result.
Answer: D. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
JEE Main PYQ-Pattern Practice · 30 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q21For bright position, select the correct YDSE result.
Answer: A. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q22For dark position, select the correct YDSE result.
Answer: D. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q23For fringe width, select the correct YDSE result.
Answer: C. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q24For angular fringe width, select the correct YDSE result.
Answer: B. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q25For thin-sheet optical path, select the correct YDSE result.
Answer: A. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q26For thin-sheet shift, select the correct YDSE result.
Answer: D. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q27For liquid fringe width, select the correct YDSE result.
Answer: C. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q28For equal-beam intensity, select the correct YDSE result.
Answer: B. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q29For central fringe, select the correct YDSE result.
Answer: A. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q30For source shift effect, select the correct YDSE result.
Answer: D. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
JEE Advanced PYQ-Pattern Practice · 30 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q21For bright position, select the correct YDSE result.
Answer: A. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q22For dark position, select the correct YDSE result.
Answer: D. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q23For fringe width, select the correct YDSE result.
Answer: C. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q24For angular fringe width, select the correct YDSE result.
Answer: B. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q25For thin-sheet optical path, select the correct YDSE result.
Answer: A. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q26For thin-sheet shift, select the correct YDSE result.
Answer: D. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q27For liquid fringe width, select the correct YDSE result.
Answer: C. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q28For equal-beam intensity, select the correct YDSE result.
Answer: B. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q29For central fringe, select the correct YDSE result.
Answer: A. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q30For source shift effect, select the correct YDSE result.
Answer: D. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
IB Physics Exam-Style Questions · 20 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
IGCSE Exam-Style Questions · 20 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
A-Level Exam-Style Questions · 20 Questions
Q1For path-difference symbol, select the correct YDSE result.
Answer: A. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q2For phase-difference symbol, select the correct YDSE result.
Answer: D. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q3For YDSE path difference, select the correct YDSE result.
Answer: C. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q4For bright condition, select the correct YDSE result.
Answer: B. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q5For dark condition, select the correct YDSE result.
Answer: A. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
Q6For bright position, select the correct YDSE result.
Answer: D. xₙ = nλD/d
Explanation: The correct result is xₙ = nλD/d. Here Δ denotes path difference and φ denotes phase difference.
Q7For dark position, select the correct YDSE result.
Answer: C. xₙ = (2n−1)λD/(2d)
Explanation: The correct result is xₙ = (2n−1)λD/(2d). Here Δ denotes path difference and φ denotes phase difference.
Q8For fringe width, select the correct YDSE result.
Answer: B. β = λD/d
Explanation: The correct result is β = λD/d. Here Δ denotes path difference and φ denotes phase difference.
Q9For angular fringe width, select the correct YDSE result.
Answer: A. λ/d
Explanation: The correct result is λ/d. Here Δ denotes path difference and φ denotes phase difference.
Q10For thin-sheet optical path, select the correct YDSE result.
Answer: D. Δp = (μ−1)t
Explanation: The correct result is Δp = (μ−1)t. Here Δ denotes path difference and φ denotes phase difference.
Q11For thin-sheet shift, select the correct YDSE result.
Answer: C. Δx = D(μ−1)t/d
Explanation: The correct result is Δx = D(μ−1)t/d. Here Δ denotes path difference and φ denotes phase difference.
Q12For liquid fringe width, select the correct YDSE result.
Answer: B. β′ = β/μ
Explanation: The correct result is β′ = β/μ. Here Δ denotes path difference and φ denotes phase difference.
Q13For equal-beam intensity, select the correct YDSE result.
Answer: A. I = 4I₀cos²(φ/2)
Explanation: The correct result is I = 4I₀cos²(φ/2). Here Δ denotes path difference and φ denotes phase difference.
Q14For central fringe, select the correct YDSE result.
Answer: D. bright
Explanation: The correct result is bright. Here Δ denotes path difference and φ denotes phase difference.
Q15For source shift effect, select the correct YDSE result.
Answer: C. pattern shifts but β stays unchanged
Explanation: The correct result is pattern shifts but β stays unchanged. Here Δ denotes path difference and φ denotes phase difference.
Q16For path-difference symbol, select the correct YDSE result.
Answer: B. Δ
Explanation: The correct result is Δ. Here Δ denotes path difference and φ denotes phase difference.
Q17For phase-difference symbol, select the correct YDSE result.
Answer: A. φ
Explanation: The correct result is φ. Here Δ denotes path difference and φ denotes phase difference.
Q18For YDSE path difference, select the correct YDSE result.
Answer: D. Δ = xd/D
Explanation: The correct result is Δ = xd/D. Here Δ denotes path difference and φ denotes phase difference.
Q19For bright condition, select the correct YDSE result.
Answer: C. Δ = nλ
Explanation: The correct result is Δ = nλ. Here Δ denotes path difference and φ denotes phase difference.
Q20For dark condition, select the correct YDSE result.
Answer: B. Δ = (2n−1)λ/2
Explanation: The correct result is Δ = (2n−1)λ/2. Here Δ denotes path difference and φ denotes phase difference.
16
15 Case Studies
1. Laser YDSE
Case: A red laser illuminates two narrow slits and clear equally spaced fringes appear.
Complete solution: The laser supplies a nearly monochromatic coherent wavefront; β = λD/d determines the spacing.
2. Thin glass plate
Case: A glass plate is inserted before S₁.
Complete solution: Additional path Δp = (μ−1)t shifts the pattern toward S₁; β is unchanged.
3. Water tank
Case: The whole apparatus is immersed in water.
Complete solution: Wavelength becomes λ/μ, so β′ = β/μ.
4. Screen moved away
Case: The screen distance is doubled.
Complete solution: β doubles because β = λD/d; angular spacing stays λ/d.
5. Source displaced
Case: The illuminating source is moved upward.
Complete solution: A constant initial path difference shifts the full pattern; β remains unchanged.
6. Missing fifth order
Case: Slit separation is five times slit width.
Complete solution: Every fifth interference maximum coincides with a diffraction minimum.
7. Unequal illumination
Case: One slit has four times the intensity of the other.
Complete solution: Amplitude ratio is 2:1, so minima are not completely dark and visibility is 0.8.
8. White-light center
Case: White light replaces monochromatic light.
Complete solution: At Δ = 0 every wavelength is constructive, producing a white central fringe.
9. Wavelength measurement
Case: Fringe width, D and d are measured.
Complete solution: Use λ = βd/D to determine the wavelength.
10. Refractive-index measurement
Case: A sheet of known thickness produces a measured shift.
Complete solution: From Δx = D(μ−1)t/d, calculate μ.
11. Microscopic thickness
Case: A film causes N fringe shifts.
Complete solution: Optical path change equals Nλ, so t = Nλ/(μ−1).
12. Two-color source
Case: Red and blue light illuminate the slits.
Complete solution: Each wavelength creates its own fringe spacing; red fringes are wider.
13. Vibration disturbance
Case: One slit plate vibrates slightly.
Complete solution: Changing path difference moves fringes and reduces time-averaged contrast.
14. Astronomical interferometry
Case: Two separated apertures observe a star.
Complete solution: Fringe visibility and spacing reveal angular size or separation.
15. Optical alignment
Case: A central maximum is monitored while a plate is rotated.
Complete solution: Changing optical thickness changes Δ and shifts the pattern precisely.
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25 Assertion-Reason Questions
AR1Assertion: The central YDSE fringe is bright. Reason: At the center, Δ = 0.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR2Assertion: Fringe width increases with D. Reason: β = λD/d.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR3Assertion: Fringe width decreases when d increases. Reason: β is inversely proportional to d.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR4Assertion: A thin sheet shifts the entire pattern. Reason: It introduces additional path Δp = (μ−1)t.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR5Assertion: A thin sheet changes fringe width. Reason: It changes only the constant path offset.
Answer: D
Explanation: The assertion is false while the reason is true.
AR6Assertion: Immersion reduces fringe width. Reason: The wavelength in the liquid is λ/μ.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR7Assertion: Source shift changes fringe width. Reason: Source shift changes the initial path difference only.
Answer: D
Explanation: The assertion is false while the reason is true.
AR8Assertion: Screen displacement changes fringe width. Reason: β depends on D.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR9Assertion: Bright fringes satisfy Δ = nλ. Reason: Their phase difference is 2nπ.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR10Assertion: Dark fringes satisfy Δ = (2n−1)λ/2. Reason: Their phase difference is an odd multiple of π.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR11Assertion: Coherent sources require equal amplitudes. Reason: Coherence concerns frequency and stable phase.
Answer: D
Explanation: The assertion is false while the reason is true.
AR12Assertion: YDSE conserves energy. Reason: Interference redistributes energy.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR13Assertion: Equal slit intensities give zero minima. Reason: The amplitudes cancel at φ = π.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR14Assertion: Angular fringe width is λ/d. Reason: For small angles, β/D = λ/d.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR15Assertion: A plate in the upper path shifts fringes upward. Reason: The central fringe moves toward the plate.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR16Assertion: The first bright fringe is at x = β. Reason: Bright positions are xₙ = nβ.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR17Assertion: The first dark fringe is at x = β/2. Reason: Dark positions are xₙ = (2n−1)β/2.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR18Assertion: Missing orders arise from diffraction. Reason: A diffraction minimum can coincide with an interference maximum.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR19Assertion: White light gives a white central fringe. Reason: At O, Δ = 0 for every wavelength.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR20Assertion: Visibility is highest for equal intensities. Reason: Then Imin = 0.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR21Assertion: Phase difference is φ = 2πΔ/λ. Reason: A path change of one wavelength gives 2π.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR22Assertion: Closing one slit removes fringes. Reason: Two coherent waves are required.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR23Assertion: Increasing wavelength broadens fringes. Reason: β is proportional to λ.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR24Assertion: Moving the screen closer narrows fringes. Reason: D decreases.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
AR25Assertion: The pattern shifts toward an inserted plate. Reason: The plate retards the beam in that path.
Answer: A
Explanation: Both statements are true and the reason correctly explains the assertion.
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Kumar Sir Exam Tips
Bright Fringe
Δ = nλ
Dark Fringe
Δ = (2n−1)λ/2
Fringe Width
β = λD/d
Thin Sheet Shift
Δx = β(μ−1)t/λ
Immersion
β′ = β/μ
Do Not Confuse
Δ = path difference; φ = phase difference.
- Source shift changes pattern position, never fringe width.
- Thin-sheet insertion changes pattern position, never fringe width.
- Use Δ for path difference everywhere.
- Use φ only for phase difference: φ = (2π/λ)Δ.