YDSE with Thin Film Intersplit Interference: Calculating the lateral displacement of the fringe pattern (fringe shift) when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is introduced in front of one of the slits in Young's Double Slit Experiment.

YDSE with Thin Film Interference: Fringe Shift

Young's Double Slit Experiment (YDSE)

In Young's Double Slit Experiment, two coherent sources produce interference fringes on a screen.

S₁ ────────────────→ P
           ＼
            ＼ Screen

S₂ ────────────────→ P

Let:

• d = distance between two slits
• D = distance between slits and screen
• λ = wavelength of light
• t = thickness of transparent sheet
• μ = refractive index of sheet

Without Sheet

Path Difference: δ = d sin θ

For small angle:

δ ≈ dy/D

Bright fringe condition:

δ = nλ

Dark fringe condition:

δ = (2n+1)λ/2

Fringe Width

β = λD/d

When a Thin Transparent Sheet is Introduced

Suppose a transparent sheet of thickness t and refractive index μ is placed in front of one slit.

The optical path through the sheet becomes μt instead of t. Therefore the additional optical path introduced is:

Additional Path Difference = μt − t

Δ = (μ − 1)t

Therefore the entire interference pattern shifts toward the slit in front of which the sheet is placed.

Fringe Shift

Number of Fringe Shifts: N = (μ − 1)t / λ

Since one fringe corresponds to β:

Fringe Shift: x = Nβ

Therefore:

x = [(μ − 1)t/λ] × [λD/d]

⭐ FINAL RESULT: x = D(μ − 1)t/d

Important: The fringe pattern shifts toward the slit covered by the transparent sheet.

Important Observations

• Fringe width β does not change.
• Only the position of the entire fringe pattern changes.
• Number of fringe shifts = (μ−1)t/λ.
• Fringe displacement = D(μ−1)t/d.
• If μ = 1 or t = 0, there is no shift.

Q1. In YDSE, fringe width is:
A. λd/D
B. λD/d
C. dD/λ
D. λ/dD

Answer: B

Fringe width: β = λD/d.

Q2. A transparent sheet of thickness t and refractive index μ introduces additional optical path:
A. μt
B. t
C. (μ−1)t
D. (μ+1)t

Answer: C

Additional path = μt − t = (μ−1)t.

Q3. Number of fringes shifted is:
A. μt/λ
B. (μ−1)t/λ
C. λ(μ−1)t
D. λ/[(μ−1)t]

Answer: B

N = additional path / wavelength = (μ−1)t/λ.

Q4. Fringe displacement due to sheet is:
A. D(μ−1)t/d
B. d(μ−1)t/D
C. λD/d
D. Dt/d

Answer: A

x = Nβ = [(μ−1)t/λ](λD/d) = D(μ−1)t/d.

Q5. A sheet is introduced before one slit. The fringe pattern shifts:
A. Away from sheet
B. Toward sheet
C. Perpendicular to screen
D. Does not shift

Answer: B

The optical path of the covered slit increases. Therefore the pattern shifts toward that slit.

Q6. If μ = 1, fringe shift is:
A. Maximum
B. Zero
C. β
D. D

Answer: B

x = D(μ−1)t/d = 0.

Q7. If thickness of sheet is doubled, fringe shift becomes:
A. Half
B. Double
C. Four times
D. Unchanged

Answer: B

x ∝ t. Therefore doubling t doubles x.

Q8. If refractive index μ increases, fringe shift:
A. Decreases
B. Increases
C. Remains zero
D. Becomes negative always

Answer: B

x ∝ (μ−1).

Q9. Introduction of a uniform transparent sheet changes:
A. Fringe width only
B. Fringe position only
C. Slit separation only
D. Screen distance only

Answer: B

For the ideal thin-sheet case, fringe width remains β = λD/d. The whole pattern is displaced.

Q10. Fringe width is independent of:
A. λ
B. D
C. d
D. Sheet thickness t

Answer: D

β = λD/d, so t does not appear.

Q11. If d is doubled, fringe shift becomes:
A. Double
B. Half
C. Four times
D. Same

Answer: B

x ∝ 1/d.

Q12. If D is doubled, fringe shift becomes:
A. Half
B. Double
C. Same
D. Zero

Answer: B

x ∝ D.

Q13. Number of fringe shifts is independent of:
A. t
B. μ
C. λ
D. All of these

Answer: D

N = (μ−1)t/λ, so it depends on all three.

Q14. If t = 0, then:
A. x = 0
B. x = β
C. x = D
D. x = λ

Answer: A

No sheet means no additional path difference.

Q15. If additional path difference equals λ, number of fringes shifted is:
A. 0
B. 1
C. 2
D. 1/2

Answer: B

N = Δ/λ = λ/λ = 1.

Q16. If additional path difference is λ/2, the pattern shifts by:
A. 1 fringe
B. 1/2 fringe
C. 2 fringes
D. Zero

Answer: B

N = (λ/2)/λ = 1/2.

Q17. For μ = 1.5 and t = 2λ, number of fringes shifted is:
A. 0.5
B. 1
C. 2
D. 3

Answer: B

N = (1.5−1)(2λ)/λ = 1.

Q18. The central bright fringe moves toward:
A. The uncovered slit
B. The covered slit
C. Centre of screen always
D. Nowhere

Answer: B

The additional optical path is introduced in front of the covered slit, so the central fringe shifts toward that slit.

Q19. The phase difference introduced by a sheet is:
A. 2πt/λ
B. 2π(μ−1)t/λ
C. λ/t
D. μt

Answer: B

Phase difference: φ = 2πΔ/λ = 2π(μ−1)t/λ.

Q20. If μ−1 is doubled while t remains fixed, fringe shift:
A. Halves
B. Doubles
C. Same
D. Zero

Answer: B

x ∝ (μ−1).

Q21. If wavelength is doubled, number of fringe shifts:
A. Doubles
B. Halves
C. Same
D. Four times

Answer: B

N = (μ−1)t/λ, so N ∝ 1/λ.

Q22. If wavelength is doubled, ideal fringe displacement due to a fixed sheet is:
A. Doubled
B. Halved
C. Unchanged
D. Zero

Answer: C

x = D(μ−1)t/d, which is independent of λ.

Q23. A sheet causes 3 fringe shifts. Its additional optical path is:
A. λ/3
B. 3λ
C. 3/λ
D. λ

Answer: B

Δ = Nλ = 3λ.

Q24. The unit of fringe displacement is:
A. metre
B. dioptre
C. radian
D. hertz

Answer: A

Fringe displacement is a distance.

Q25. The optical path length through sheet is:
A. t/μ
B. μt
C. μ/t
D. t+μ

Answer: B

Optical path length = refractive index × geometrical path.

Q26. For a sheet with μ = 1.4 and t = 5 μm, additional path is:
A. 1 μm
B. 2 μm
C. 5 μm
D. 7 μm

Answer: B

Δ = (1.4−1)×5 = 2 μm.

Q27. A sheet introduces an optical path difference of 2λ. Number of fringe shifts is:
A. 1/2
B. 1
C. 2
D. 4

Answer: C

N = 2λ/λ = 2.

Q28. Which quantity remains unchanged after inserting an ideal uniform sheet in one path?
A. Fringe position
B. Central fringe position
C. Fringe width
D. Phase difference at centre

Answer: C

The sheet introduces a constant path difference, so fringe width remains β.

Q29. The sheet produces a constant phase shift because:
A. Its thickness is uniform
B. It changes slit separation
C. It changes screen distance
D. It blocks both slits

Answer: A

Uniform thickness and refractive index introduce a constant additional optical path.

Q30. The most useful formula for direct calculation of fringe shift is:
A. β = λD/d
B. x = D(μ−1)t/d
C. x = λD/d
D. x = μt

Answer: B

Direct fringe displacement: x = D(μ−1)t/d.

A: Both Assertion and Reason are true and Reason correctly explains Assertion.
B: Both are true but Reason is not the correct explanation.
C: Assertion is true but Reason is false.
D: Assertion is false but Reason is true.

AR1. Assertion: A transparent sheet causes a fringe shift in YDSE.
Reason: The sheet introduces an additional optical path difference.

Answer: A

AR2. Assertion: Fringe width remains unchanged when a uniform thin sheet is introduced in front of one slit.
Reason: The sheet introduces a constant path difference.

Answer: A

AR3. Assertion: Fringe displacement is independent of wavelength.
Reason: x = D(μ−1)t/d.

Answer: A

AR4. Assertion: Number of fringe shifts depends on wavelength.
Reason: N = (μ−1)t/λ.

Answer: A

AR5. Assertion: The fringe pattern shifts toward the sheet-covered slit.
Reason: The optical path of that arm increases.

Answer: A

AR6. Assertion: If μ = 1, no fringe shift occurs.
Reason: The additional optical path becomes zero.

Answer: A

AR7. Assertion: Doubling sheet thickness doubles fringe shift.
Reason: Fringe shift is directly proportional to thickness.

Answer: A

AR8. Assertion: Doubling slit separation doubles fringe shift.
Reason: Fringe shift is inversely proportional to slit separation.

Answer: D

AR9. Assertion: A sheet with larger refractive index generally produces a larger shift for the same thickness.
Reason: x ∝ (μ−1).

Answer: A

AR10. Assertion: The transparent sheet changes the wavelength inside itself.
Reason: The frequency of light remains unchanged when it enters a medium.

Answer: B

1 Mark Questions

Q1. Write the expression for fringe width in YDSE.

β = λD/d.

Q2. What additional path is introduced by a sheet?

Δ = (μ−1)t.

Q3. Write the number of fringe shifts.

N = (μ−1)t/λ.

Q4. Write the fringe displacement formula.

x = D(μ−1)t/d.

Q5. In which direction does the fringe pattern shift?

Toward the slit in front of which the sheet is placed.

2 Mark Questions

Q6. Explain why a thin sheet causes fringe shift.

The sheet replaces air path t by optical path μt. Hence additional path = μt−t = (μ−1)t, causing a phase change and displacement of the fringe pattern.

Q7. A sheet has μ = 1.5 and t = 2 μm. Find additional optical path.

Δ = (1.5−1)×2 = 1 μm.

Q8. If an additional path λ is introduced, how many fringes shift?

N = λ/λ = 1 fringe.

Q9. Does fringe width change after inserting a uniform sheet?

No. The fringe width remains β = λD/d. The whole pattern is displaced.

Q10. What happens if the sheet is removed?

The additional path difference disappears and the original fringe pattern returns to its original position.

3 Mark Questions

Q11. Derive the expression for number of fringe shifts.

Additional path: Δ = (μ−1)t.

One fringe corresponds to path difference λ. Therefore:
N = Δ/λ
N = (μ−1)t/λ.

Q12. Derive fringe displacement using fringe width.

x = Nβ
β = λD/d
N = (μ−1)t/λ

Therefore:
x = [(μ−1)t/λ](λD/d)
x = D(μ−1)t/d.

Q13. A sheet of μ = 1.5 and thickness 0.01 mm is placed before one slit. If λ = 500 nm, find number of fringes shifted.

t = 0.01 mm = 10⁻⁵ m.
λ = 5×10⁻⁷ m.

N = (0.5×10⁻⁵)/(5×10⁻⁷) = 10 fringes.

Q14. Explain why the fringe width does not change.

The sheet introduces a constant optical path difference for all rays from that slit. It shifts the phase of the interference pattern but does not change the spacing between successive fringes.

Q15. A sheet causes half-fringe displacement. Find its additional optical path.

N = 1/2.
Δ = Nλ = λ/2.

4 Mark Questions

Q16. In YDSE, D = 2 m, d = 1 mm, μ = 1.5 and t = 10 μm. Find fringe displacement.

x = D(μ−1)t/d

= 2×0.5×10×10⁻⁶ / 10⁻³
= 0.01 m
= 1 cm.

Q17. For D = 1 m, d = 0.5 mm, μ = 1.4, t = 5 μm. Find fringe shift.

x = 1×0.4×5×10⁻⁶/(0.5×10⁻³)
= 4×10⁻³ m
= 4 mm.

Q18. A sheet causes 4 fringe shifts. Find its optical path difference.

Δ = Nλ = 4λ.

Q19. Show that fringe displacement is independent of wavelength.

x = Nβ
N = (μ−1)t/λ
β = λD/d

Therefore: x = D(μ−1)t/d.
λ cancels out. Hence x is independent of λ.

Q20. A sheet with μ = 1.6 and t = 20 μm is introduced. If λ = 500 nm, find number of shifted fringes.

N = (1.6−1)(20×10⁻⁶)/(500×10⁻⁹)
= 0.6×40
= 24 fringes.

5 Mark Questions

Q21. Derive completely the formula x = D(μ−1)t/d.

Optical path without sheet = t.
Optical path with sheet = μt.

Additional path:
Δ = μt−t = (μ−1)t.

Number of fringes shifted:
N = Δ/λ = (μ−1)t/λ.

Fringe width:
β = λD/d.

Therefore displacement:
x = Nβ
= [(μ−1)t/λ][λD/d]
x = D(μ−1)t/d.

Q22. In YDSE, λ = 600 nm, D = 1.5 m and d = 0.75 mm. A sheet of refractive index 1.5 is introduced. If the fringe pattern moves by 6 mm, find sheet thickness.

x = D(μ−1)t/d Therefore: t = xd/[D(μ−1)] = (6×10⁻³)(0.75×10⁻³) /[1.5×0.5] = 6×10⁻⁶ m = 6 μm.

Q23. A transparent sheet produces 5 fringe shifts in YDSE. If μ = 1.5 and λ = 500 nm, find its thickness.

N = (μ−1)t/λ Therefore: t = Nλ/(μ−1) = 5×500×10⁻⁹/0.5 = 5×10⁻⁶ m = 5 μm.

Q24. Explain experimentally how the refractive index of a thin sheet can be determined using fringe shift.

Measure: D = slit-screen distance, d = slit separation, t = sheet thickness, and x = observed fringe displacement. Use: x = D(μ−1)t/d. Therefore: μ = 1 + xd/(Dt). Thus measuring the fringe displacement allows determination of the refractive index of the sheet.

Q25. Explain the phase interpretation of fringe shift.

Additional path: Δ = (μ−1)t. Corresponding phase difference: φ = 2πΔ/λ Therefore: φ = 2π(μ−1)t/λ. The original interference pattern therefore experiences a constant phase change, which appears as a lateral displacement of the entire fringe system.

6 Mark Questions

Q26. Give a complete derivation of fringe shift when a transparent sheet is placed in front of one slit.

Let the two slits be S₁ and S₂. Initially, for a point P on the screen: Path difference = d sinθ. For a central point, path difference is zero. Now place a sheet of thickness t and refractive index μ in front of S₁. The geometrical path t is replaced by optical path μt. Therefore additional path: Δ = μt−t Δ = (μ−1)t. For the original central bright fringe to occur, the geometrical path difference from the screen must compensate for this additional path. If the displacement of the fringe pattern is x: d sinθ ≈ dx/D. Hence: dx/D = (μ−1)t. Therefore: x = D(μ−1)t/d. This is the required lateral displacement.

Q27. A YDSE setup has λ = 500 nm, D = 2 m and d = 1 mm. A sheet of μ = 1.5 and thickness 20 μm is introduced before one slit. Find:
(i) fringe width
(ii) number of fringe shifts
(iii) fringe displacement.

(i) Fringe width: β = λD/d = (500×10⁻⁹×2)/(10⁻³) = 10⁻³ m = 1 mm.
(ii) Number of shifts: N = (μ−1)t/λ = 0.5×20×10⁻⁶/(500×10⁻⁹) = 20 fringes.
(iii) Fringe displacement: x = Nβ = 20×1 mm = 20 mm = 2 cm.

Q28. A sheet causes a fringe displacement of 4 mm. In the experiment D = 1.2 m, d = 0.6 mm and μ = 1.5. Find its thickness.

x = D(μ−1)t/d Therefore: t = xd/[D(μ−1)] = (4×10⁻³)(0.6×10⁻³) /(1.2×0.5) = 4×10⁻⁶ m = 4 μm.

Q29. Explain the difference between fringe displacement and number of fringes shifted.

Number of fringes shifted: N = (μ−1)t/λ It is dimensionless.

Fringe displacement: x = D(μ−1)t/d It is a physical distance.

They are related by: x = Nβ where β is the fringe width.

Q30. Discuss the effect of changing λ, D, d, μ and t on fringe shift.

The fringe displacement is: x = D(μ−1)t/d. Therefore:
• x ∝ D → increasing D increases shift.
• x ∝ t → increasing thickness increases shift.
• x ∝ (μ−1) → increasing refractive index increases shift.
• x ∝ 1/d → increasing slit separation decreases shift.
• x is independent of λ for fixed μ, t, D and d.

However, the number of fringe shifts is: N = (μ−1)t/λ, so N decreases when wavelength increases.

Fringe Width: β = λD/d

Additional Optical Path: Δ = (μ−1)t

Number of Fringe Shifts: N = (μ−1)t/λ

Fringe Displacement: x = Nβ

⭐ Final Formula: x = D(μ−1)t/d

Phase Difference: φ = 2π(μ−1)t/λ

Refractive Index: μ = 1 + xd/(Dt)

Thickness: t = xd/[D(μ−1)]

SHEET INSERTED
↓
Additional Optical Path
Δ = (μ−1)t
↓
Number of Fringes
N = Δ/λ
↓
Fringe Width
β = λD/d
↓
Fringe Shift
x = Nβ
↓
⭐ x = D(μ−1)t/d

Quick Steps

1. Identify μ and t.
2. Calculate additional path Δ = (μ−1)t.
3. If asked number of fringes, use N = Δ/λ.
4. If asked lateral displacement, use x = D(μ−1)t/d.
5. Remember: pattern shifts toward the sheet-covered slit.
6. Fringe width remains unchanged.

MCQ Answer Key:

1-B, 2-C, 3-B, 4-A, 5-B, 6-B, 7-B, 8-B, 9-B, 10-D, 11-B, 12-B, 13-D, 14-A, 15-B, 16-B, 17-B, 18-B, 19-B, 20-B, 21-B, 22-C, 23-B, 24-A, 25-B, 26-B, 27-C, 28-C, 29-A, 30-B.

Assertion–Reason:

1-A, 2-A, 3-A, 4-A, 5-A, 6-A, 7-A, 8-D, 9-A, 10-B.