UET UG Physics Booster Test 2 - Interference and Superposition -

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CUET UG Physics Booster Test 2 - Interference and Superposition

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2 - Interference and Superposition

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QUESTION 1 OF 20

If two coherent light waves arriving at a point produce displacements y₁ = a cos(ωt) and y₂ = a cos(ωt + φ), the superposition principle yields that
1. the resulting displacement is identical to the arithmetic product of y₁ and y₂
2. the resultant displacement is 2a cos(φ/2) cos(ωt + φ/2)
3. the resultant intensity drops to exactly zero regardless of phase
4. they will not superpose if the amplitude is the same

Answer Review

1 If two coherent light waves arriving at a point produce displacements y₁ = a cos(ωt) and y₂ = a cos(ωt + φ), the superposition principle yields that
1. the resulting displacement is identical to the arithmetic product of y₁ and y₂
2. the resultant displacement is 2a cos(φ/2) cos(ωt + φ/2)
3. the resultant intensity drops to exactly zero regardless of phase
4. they will not superpose if the amplitude is the same

A 1

B 2

C 3

D 4

💡 Short Explanation

�� Superposition requires addition of displacements. �� Trigonometric addition formula is used. �� Resultant amplitude depends on phase difference.

✅ Correct Answer Explanation

Using: y = y₁ + y₂ = a cos(ωt) + a cos(ωt + φ) Applying: cos A + cos B = 2 cos[(A+B)/2] cos[(A−B)/2] y = 2a cos(φ/2) cos(ωt + φ/2) Thus Statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Superposition involves addition, not multiplication.
�� Option C → Intensity depends on phase difference and is not always zero.
�� Option D → Equal amplitudes do not prevent superposition.

🎯 Strategy

Used

Substitution

Application:

Apply the trigonometric identity to the superposition equation.

Final Logic:

Addition of two coherent waves gives 2a cos(φ/2) cos(ωt + φ/2).

🧠 Memory Trick

"Cos + Cos → 2CosCos."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Principle of Superposition

2 When applying the vector addition from the superposition principle to optical interference

A it demands that the interfering light waves act as strictly longitudinal waves

B the resultant displacement equation is fully valid for determining both constructive and destructive conditions

C each source must emit varying frequencies to validate the algebraic sum

D it works only when the phase difference is rapidly fluctuating

💡 Short Explanation

�� Superposition predicts maxima and minima. �� Phase difference determines the outcome. �� Applicable to all wave interference.

✅ Correct Answer Explanation

The resultant displacement obtained from superposition allows determination of: • Constructive interference (maximum intensity) • Destructive interference (minimum intensity) Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Light is transverse, not longitudinal.
�� Option C → Coherent sources require same frequency.
�� Option D → Rapid phase fluctuations destroy stable interference.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the role of superposition in interference.

Final Logic:

The superposition equation predicts both maxima and minima.

🧠 Memory Trick

"Superposition Predicts Fringes."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Superposition and Interference

3 Identify the incorrect statement about the phase relationship of coherent sources

A Coherent sources ensure that any random phase shift in the primary source occurs simultaneously in the secondary sources

B The phase difference φ at any given point on the screen must remain constant over time

C A zero phase difference is the singular, exclusive requirement for sources to be considered coherent

D Two sources are locked in phase when they are derived from the exact same wave origin

💡 Short Explanation

�� Coherence requires constant phase difference. �� The phase difference need not be zero. �� Same frequency is also required.

✅ Correct Answer Explanation

Coherent sources must maintain a constant phase difference. The phase difference may be: 0, π/2, π, or any fixed value. Therefore zero phase difference is not the exclusive requirement. Hence Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct for sources derived from the same origin.
�� Option B → Correct definition of coherence.
�� Option D → Correct description of coherent source generation.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement violating the definition of coherence.

Final Logic:

Constant phase difference is required, not necessarily zero phase difference.

🧠 Memory Trick

"Coherent = Constant φ, Not Zero φ."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Coherent Sources

4 If two vibrating sources fail to maintain a constant phase difference, their interference pattern shifts instantly, and the viewer perceives a _______ intensity rather than a _______ geometric pattern.

A Time-averaged, Stable

B Maximum, Minimum

C Minimum, Unstable

D Zero, Visible

💡 Short Explanation

�� Random phase changes destroy fixed fringes. �� The eye records an average intensity. �� Stable interference disappears.

✅ Correct Answer Explanation

When phase difference changes rapidly: • Maxima and minima continuously shift. • No stable fringe pattern exists. • The observer sees only the time-averaged intensity. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not generally true.
�� Option C → Average intensity is not necessarily minimum.
�� Option D → Light does not vanish.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall incoherent addition of light.

Final Logic:

Rapid phase fluctuations produce time-averaged intensity.

🧠 Memory Trick

"Changing Phase → Average Brightness."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Incoherent Addition

5 Choose the statements about finding a point P perfectly equidistant from two identical oscillating needles
1. The physical waves from both needles arrive at point P out of phase.
2. The distance relation S₁P = S₂P forces the waves to arrive in phase.
3. The sources effectively mimic optical coherent sources.
4. The combined resultant amplitude at point P reaches a minimum.

A 1, 2

B 2, 3

C 1, 4

D 3, 4

💡 Short Explanation

�� Equal distances imply zero path difference. �� Waves arrive in phase. �� Constructive interference occurs.

✅ Correct Answer Explanation

1. Statement 1 is incorrect because equal path lengths produce phase agreement. 2. Statement 2 is correct because S₁P = S₂P gives path difference zero. 3. Statement 3 is correct because the oscillating needles behave as coherent sources. 4. Statement 4 is incorrect because the resultant amplitude is maximum. Therefore Statements 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 1.
�� Option C → Includes incorrect Statements 1 and 4.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

Check phase relationship at an equidistant point.

Final Logic:

Zero path difference produces constructive interference.

🧠 Memory Trick

"Equal Distance → Equal Phase."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Water Wave Analogy

6 Correct statements about observing a point Q where the path difference is precisely 2λ in a ripple tank
1. Waves from source S₁ will arrive exactly two cycles earlier than those from S₂.
2. The differing travel times cause the waves to arrive completely out of phase.
3. The point falls on an antinodal line indicating maximum physical displacement.
4. The point Q consistently exhibits a state of destructive interference.

A 2, 4

B 1, 2, 4

C 1, 3

D 3, 4

💡 Short Explanation

�� 2λ is an integral wavelength difference. �� Integral wavelength difference gives constructive interference. �� Antinodal lines correspond to maxima.

✅ Correct Answer Explanation

1. Statement 1 is correct because a path difference of 2λ corresponds to a lead of two complete cycles. 2. Statement 2 is incorrect because two-cycle difference still means the waves are in phase. 3. Statement 3 is correct because constructive interference produces an antinode. 4. Statement 4 is incorrect because destructive interference requires half-integral wavelength difference. Therefore Statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 2.
�� Option B → Includes incorrect Statements 2 and 4.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

Use the constructive interference condition Δx = nλ.

Final Logic:

2λ corresponds to constructive interference and an antinode.

🧠 Memory Trick

"Whole λ → Antinode."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Constructive Interference

7 Match List I (Given Path Difference Δx) with List II (Resulting Phase Difference φ)

List I	List II
1. Δx = 0	a. φ = 4π
2. Δx = λ	b. φ = 6π
3. Δx = 2λ	c. φ = 0
4. Δx = 3λ	d. φ = 2π

A 1-c, 2-d, 3-a, 4-b

B 1-d, 2-c, 3-a, 4-b

C 1-c, 2-a, 3-d, 4-b

D 1-a, 2-b, 3-c, 4-d

💡 Short Explanation

�� Phase difference φ = 2πΔx/λ. �� Substitute each path difference. �� Match the values.

✅ Correct Answer Explanation

Using: φ = 2πΔx/λ 1. Δx = 0 → φ = 0 → c 2. Δx = λ → φ = 2π → d 3. Δx = 2λ → φ = 4π → a 4. Δx = 3λ → φ = 6π → b Therefore: 1-c, 2-d, 3-a, 4-b Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching for first two entries.
�� Option C → Incorrect matching for λ and 2λ.
�� Option D → Completely incorrect sequence.

🎯 Strategy

Used

Substitution

Application:

Apply φ = 2πΔx/λ to each entry.

Final Logic:

Direct substitution gives the correct mapping.

🧠 Memory Trick

"One λ = 2π."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Path Difference and Phase Difference

8 In a double-slit setup, if the intensity of light from each individual slit is 10 W/m², what will be the resultant intensity specifically at a point where the phase difference is exactly 4π?

A 0 W/m²

B 10 W/m²

C 20 W/m²

D 40 W/m²

💡 Short Explanation

�� φ = 4π corresponds to constructive interference. �� I = 4I₀ cos²(φ/2). �� cos²(2π) = 1.

✅ Correct Answer Explanation

Given: I₀ = 10 W/m² φ = 4π I = 4I₀ cos²(φ/2) = 4(10)cos²(2π) = 40 × 1 = 40 W/m² Hence Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Destructive interference value.
�� Option B → Single-source intensity.
�� Option C → Incoherent addition result.

🎯 Strategy

Used

Substitution

Application:

Substitute values into the intensity formula.

Final Logic:

Constructive interference gives 4I₀.

🧠 Memory Trick

"φ = 2nπ ⇒ Maximum Intensity."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Intensity Distribution

9 If a designated point R has a physical path difference S₂R − S₁R = −2.5λ, what will be the exact phase of the displacement y₂ relative to y₁ if y₁ = a cos(ωt)?

A ωt − 4π

B ωt + 5π

C ωt − 2.5π

D ωt + 2π

💡 Short Explanation

�� Phase difference φ = 2πΔx/λ. �� Δx = −2.5λ. �� Phase can differ by integral multiples of 2π and represent the same physical state.

✅ Correct Answer Explanation

Using: φ = −2π(Δx/λ) φ = −2π(−2.5) = 5π Therefore: y₂ = a cos(ωt + 5π) Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to −4π phase difference.
�� Option C → Incorrect conversion from path difference to phase difference.
�� Option D → Corresponds to constructive interference.

🎯 Strategy

Used

Substitution

Application:

Convert path difference into phase difference.

Final Logic:

Δx = −2.5λ gives φ = +5π.

🧠 Memory Trick

"Phase = 2π × Path Difference/λ."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Phase Difference and Path Difference

10 Identify the incorrect statement regarding the parameters for absolute zero resultant intensity

A The governing phase difference must be ±π, ±3π, ±5π, etc.

B The interacting waves must arrive at the specific point completely out of phase.

C The path difference defining the point must be an exact integer multiple of the wavelength.

D The analytical cancellation relies on the individual intensities being equal to I₀.

💡 Short Explanation

�� Zero intensity requires destructive interference. �� Path difference must be half-integral λ. �� Integer λ gives constructive interference.

✅ Correct Answer Explanation

For complete cancellation: Path Difference = (n + 1/2)λ Phase Difference = (2n + 1)π Therefore Option C is incorrect because integer multiples of λ produce maxima.

❓ Why Other Options Are Incorrect

�� Option A → Correct destructive condition.
�� Option B → Correct description.
�� Option D → Equal amplitudes/intensities are needed for complete cancellation.

🎯 Strategy

Used

Odd One Out

Application:

Compare constructive and destructive interference conditions.

Final Logic:

Integer λ gives maxima, not minima.

🧠 Memory Trick

"Whole λ Bright, Half λ Dark."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Destructive Interference

11 The significance of the formula I = 4I₀ cos²(φ/2) is that it implies
1. intensity is perfectly linearly proportional to the phase difference.
2. the peak intensity achievable is only double the independent source intensity.
3. light energy is dynamically redistributed, establishing stationary bright and dark fringes.
4. fully incoherent light sources will naturally yield this exact periodic intensity pattern.

A 1

B 2

C 3

D 4

💡 Short Explanation

�� Intensity varies with phase difference. �� Bright and dark fringes appear. �� Energy redistribution occurs.

✅ Correct Answer Explanation

1. Statement 1 is incorrect because intensity varies as cos²(φ/2), not linearly. 2. Statement 2 is incorrect because maximum intensity equals 4I₀. 3. Statement 3 is correct because interference redistributes light energy into maxima and minima. 4. Statement 4 is incorrect because incoherent sources do not produce stable interference patterns. Therefore only Statement 3 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 1 is incorrect.
�� Option B → Statement 2 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

Analyze physical meaning of the intensity equation.

Final Logic:

The equation explains bright and dark fringe formation.

🧠 Memory Trick

"Cos² Creates Fringes."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Intensity Distribution in Interference

12 For an arbitrary location G where the static phase difference between waves is φ, the true resultant displacement amplitude scales as

A a cos(φ)

B 2a cos(φ/2)

C 4a cos²(φ/2)

D a cos(ωt + φ/2)

💡 Short Explanation

�� Result obtained from superposition. �� Amplitude depends on phase difference. �� Basis of intensity variation.

✅ Correct Answer Explanation

For two equal-amplitude coherent waves: A = 2a cos(φ/2) This is the resultant amplitude. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect amplitude relation.
�� Option C → Related to intensity, not amplitude.
�� Option D → Represents displacement, not amplitude.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the resultant amplitude formula.

Final Logic:

Amplitude = 2a cos(φ/2).

🧠 Memory Trick

"Resultant Amplitude = 2aCos(φ/2)."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Resultant Amplitude

13 When analyzing the behavior of two isolated, independent sodium lamps illuminating a screen

A their phase difference stays perfectly fixed due to electromagnetic interactions

B they fail to interfere because their phase difference drifts abruptly in extremely short time spans

C the resulting intensity forms a spatial pattern strictly described by 4I₀ cos²(φ/2)

D the waves cancel out perfectly yielding a uniformly dark display

💡 Short Explanation

�� Independent lamps are incoherent. �� Phase difference fluctuates rapidly. �� Stable fringes do not form.

✅ Correct Answer Explanation

Independent light sources do not maintain a constant phase difference. Rapid random phase changes destroy stable interference patterns. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No phase locking exists.
�� Option C → Formula applies to coherent sources.
�� Option D → Light does not cancel everywhere.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall why independent lamps fail to produce interference.

Final Logic:

Random phase drift prevents stable fringes.

🧠 Memory Trick

"Independent Lamps = Random Phase."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Incoherent Sources

14 Identify the correct statements about time-averaged intensities resulting from incoherent light addition
1. The physical intensity randomly cycles, preventing stationary fringes.
2. The mathematically evaluated average intensity over time becomes I = 2I₀.
3. The cos²(φ/2) factor averages out to 1/2 over significant periods.
4. The coherent addition rules perfectly describe the resulting visible fringes.

A 2, 4

B 1, 2, 3

C 1, 3, 4

D 3, 4

💡 Short Explanation

�� Phase fluctuates randomly. �� No stationary fringes form. �� Average intensity equals 2I₀.

✅ Correct Answer Explanation

1. Statement 1 is correct because random phase variations destroy stable fringes. 2. Statement 2 is correct because average intensity becomes: I = I₀ + I₀ = 2I₀ 1. Statement 3 is correct because the interference term averages out over time. 2. Statement 4 is incorrect because coherent addition rules do not describe incoherent sources. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 4.
�� Option C → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 4 and omits correct Statements 1 and 2.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using incoherent addition principles.

Final Logic:

Only Statement 4 is incorrect.

🧠 Memory Trick

"Random Phase → Average Intensity."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Incoherent Addition

15 Young cleverly bypassed the inherent incoherence of normal light by allowing a single pinhole to dictate the wavefronts, thus making the secondary slits operate in a _______ manner, generating _______ fringes.

A Random, Unstable

B Incoherent, Uniform

C Coherent, Stable

D Fluctuating, Faint

💡 Short Explanation

�� Single source produces coherence. �� Secondary sources remain phase-locked. �� Stable fringes are observed.

✅ Correct Answer Explanation

Young used a single source to illuminate both slits. This made S₁ and S₂ coherent secondary sources with a fixed phase difference, producing stable interference fringes. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite of the actual setup.
�� Option B → Slits become coherent, not incoherent.
�� Option D → Fringes are stable.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect coherence with fringe stability.

Final Logic:

Coherent sources produce stable interference.

🧠 Memory Trick

"One Source → Stable Fringes."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Young's Double Slit Experiment

16 Choose the correct statements about the phase locking sequence in Young's experiment
1. A primary source brightly illuminates a single pinhole S.
2. The wavefront from S splits identically to fall upon S₁ and S₂.
3. An abrupt phase shift occurring at S disrupts S₁ and S₂ unevenly.
4. The technique successfully establishes S₁ and S₂ as coherent secondary radiators.

A 1, 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Single source illuminates both slits. �� Phase changes affect both equally. �� Coherent secondary sources are produced.

✅ Correct Answer Explanation

1. Statement 1 is correct. 2. Statement 2 is correct. 3. Statement 3 is incorrect because any phase change affects both slits equally. 4. Statement 4 is correct because coherence is established. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Includes incorrect Statement 3.

🎯 Strategy

Used

Elimination

Application:

Check each statement using Young's setup.

Final Logic:

Only Statement 3 is incorrect.

🧠 Memory Trick

"Same Source → Same Phase Change."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Coherent Source Generation

17 If a Young's double-slit experiment runs with λ = 5 × 10⁻⁷ m, D = 1 m, and d = 10⁻³ m, what is the calculated distance x from the central bright axis to the very first bright fringe?

A 5 × 10⁻⁴ m

B 2.5 × 10⁻⁴ m

C 10 × 10⁻⁴ m

D 5 × 10⁻⁷ m

💡 Short Explanation

�� First bright fringe: n = 1. �� x = nλD/d. �� Substitute values.

✅ Correct Answer Explanation

For the first bright fringe: x = λD/d = (5 × 10⁻⁷ × 1)/(10⁻³) = 5 × 10⁻⁴ m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Half the correct value.
�� Option C → Double the correct value.
�� Option D → Equals wavelength only.

🎯 Strategy

Used

Substitution

Application:

Use the bright fringe position formula.

Final Logic:

x = λD/d = 5 × 10⁻⁴ m.

🧠 Memory Trick

"First Bright: x = λD/d."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Bright Fringe Position

18 Match List I (Fringe order n for Destructive Minima) with List II (Position formula scalar for λD/d)

List I	List II
1. n = 0	a. −0.5
2. n = 1	b. 0.5
3. n = 2	c. 1.5
4. n = −1	d. 2.5

A 1-a, 2-b, 3-c, 4-d

B 1-b, 2-c, 3-d, 4-a

C 1-d, 2-c, 3-b, 4-a

D 1-c, 2-b, 3-a, 4-d

💡 Short Explanation

�� Dark fringe position: x = (n + 1/2) λD/d �� Substitute each n.

✅ Correct Answer Explanation

1. n = 0 → 0.5 → b 2. n = 1 → 1.5 → c 3. n = 2 → 2.5 → d 4. n = −1 → −0.5 → a Therefore: 1-b, 2-c, 3-d, 4-a Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Options A, C and D contain incorrect substitutions.

🎯 Strategy

Used

Substitution

Application:

Substitute values into x = (n + 1/2)λD/d.

Final Logic:

Direct substitution gives the correct mapping.

🧠 Memory Trick

"Dark Fringe = n + 1/2."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Dark Fringe Positions

19 Identify the incorrect statement regarding the derived position equations xₙ = nλD/d and xₙ = (n+1/2)λD/d

A The spacing of the fringes directly scales up if the wavelength is increased.

B Bringing the slits closer together (decreasing d) physically widens the fringes.

C Moving the screen farther away (increasing

D The analytical calculations show that dark fringes possess a smaller width than bright fringes.

💡 Short Explanation

�� Bright and dark fringes have equal width. �� Fringe width = λD/d. �� Width is independent of fringe type.

✅ Correct Answer Explanation

The fringe width is: β = λD/d Both bright and dark fringes are equally spaced and have equal width. Therefore Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct dependence on λ.
�� Option B → Correct dependence on d.
�� Option C → Correct dependence on D.

🎯 Strategy

Used

Odd One Out

Application:

Compare each statement with the fringe-width formula.

Final Logic:

Dark and bright fringes have equal width.

🧠 Memory Trick

"Same β for Bright and Dark."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Fringe Width

20 By evaluating the mathematical difference between sequential orders in the formula xₙ = nλD/d, we confirm that

A the separation gap between consecutive bright fringes heavily depends on the order n

B the spacing between adjacent dark bands grows non-linearly outward

C the strict distance between any two adjacent bright fringes resolves identically to λD/d

D visual fringes become progressively condensed near the outermost edges of the screen

💡 Short Explanation

�� Consecutive bright fringes differ by one fringe width. �� Fringe width remains constant. �� Independent of fringe order.

✅ Correct Answer Explanation

For bright fringes: xₙ = nλD/d xₙ₊₁ = (n+1)λD/d Difference: xₙ₊₁ − xₙ = λD/d Thus every pair of adjacent bright fringes is separated by the same distance. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Separation does not depend on n.
�� Option B → Spacing remains constant.
�� Option D → Fringes do not become condensed.

🎯 Strategy

Used

Substitution

Application:

Subtract successive fringe positions.

Final Logic:

β = λD/d remains constant.

🧠 Memory Trick

"Adjacent Brights = One β Apart."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Equal Spacing of Fringes

UET UG Physics Booster Test 2 – Interference and Superposition

CUET UG Physics Booster Test 2 - Interference and Superposition

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