Wave Optics — Class 12 Physics practice questions

Question 1

A school science exhibition features a demonstration of Young's Double Slit Experiment (YDSE). A laser pointer of wavelength 600 nm is directed at two narrow slits separated by 0.3 mm. The interference pattern is observed on a screen placed 1.5 m away.

(i) The demonstrator notices that when she places a thin transparent glass slab (refractive index 1.5) of thickness t in front of one of the slits, the central bright fringe shifts toward that slit by 5 mm. Find the thickness t of the glass slab.

(ii) After removing the glass slab, the demonstrator replaces the laser (λ = 600 nm) with a different source of wavelength 400 nm, keeping all other parameters the same. Find the ratio of the new fringe width to the original fringe width.

(iii) The demonstrator then asks a student: 'Why must the two slits be illuminated by the same source (or coherent sources) for a stable interference pattern?' Give a reason in one or two sentences.

(iv) If the distance between the slits is halved and the distance to the screen is doubled (laser λ = 600 nm restored), by what factor does the fringe width change?

Accepted Answer

(i) Finding thickness t of the glass slab:

By the principle of optical path shift in YDSE, inserting a slab of refractive index n and thickness t in front of one slit increases its optical path by (n − 1)t.

The shift in the central fringe position is given by:

Δy = (n − 1)t · D/d

Given: Δy = 5 mm = 5 × 10⁻³ m, n = 1.5, D = 1.5 m, d = 0.3 mm = 3 × 10⁻⁴ m

Substituting:

5 × 10⁻³ = (1.5 − 1) × t × (1.5) / (3 × 10⁻⁴)

5 × 10⁻³ = 0.5 × t × 5000

5 × 10⁻³ = 2500 t

∴ t = (5 × 10⁻³) / 2500 = 2 × 10⁻⁶ m

∴ t = 2 μm

(ii) Ratio of new fringe width to original fringe width:

The fringe width in YDSE is given by:

β = λD/d

Since D and d remain unchanged:

β₁/β₂ = λ₁/λ₂

Original: λ₁ = 600 nm; New: λ₂ = 400 nm

β₂/β₁ = λ₂/λ₁ = 400/600 = 2/3

∴ Ratio of new fringe width to original fringe width = 2 : 3

(The fringe width decreases when a shorter wavelength source is used.)

(iii) Why coherent sources are necessary:

Two independent sources have a randomly and rapidly fluctuating phase difference. For a stable, observable interference pattern, the phase difference between the two interfering beams at any point on the screen must remain constant over time. Only coherent sources (derived from the same primary source) maintain a constant phase relationship, producing a sustained pattern of bright and dark fringes. Incoherent sources produce a phase difference that changes millions of times per second, washing out the fringes to give uniform illumination.

(iv) Change in fringe width when d is halved and D is doubled:

The fringe width is:

β = λD/d

New fringe width:

β' = λ(2D)/(d/2) = λ · 2D · 2/d = 4(λD/d) = 4β

∴ The fringe width increases by a factor of 4.

Question 2

A science teacher sets up Young's double-slit experiment in the school laboratory using a monochromatic sodium lamp (λ = 589 nm). The two slits are separated by d = 0.5 mm and the screen is placed at D = 1.0 m from the slits.

(i) A student notices that when she places a thin glass slab (refractive index n = 1.5) of thickness t = 5 μm in front of slit S₁, the central bright fringe shifts. State the direction of shift and calculate the number of fringes by which the central fringe shifts.

(ii) Without the glass slab, calculate the fringe width β of the interference pattern.

(iii) The student then increases the separation between the slits to 1.0 mm, keeping D and λ unchanged. What happens to the fringe width? (Give reason in one line.)

(iv) Her classmate argues: 'If we use two separate sodium lamps instead of one lamp with two slits, we will still get an interference pattern.' Is the classmate correct? Give a reason.

Accepted Answer

MARKING SCHEME (4 marks — 1 mark each)

(i) Direction of shift and number of fringes shifted:

When a glass slab is placed in front of S₁, the optical path through S₁ increases by (n − 1)t. The central fringe (zero path difference) shifts toward S₁, i.e., toward the side of slit S₁.

By the formula, the number of fringes shifted:
       N = (n − 1)t / λ
       N = (1.5 − 1) × 5 × 10⁻⁶ / (589 × 10⁻⁹)
       N = (0.5 × 5 × 10⁻⁶) / (589 × 10⁻⁹)
       N = 2.5 × 10⁻⁶ / 589 × 10⁻⁹
       N ≈ 4.24 ≈ 4 fringes

∴ The central fringe shifts toward S₁ by approximately 4 fringes.

(ii) Fringe width β:

By the formula for fringe width in YDSE:
       β = λD / d
       β = (589 × 10⁻⁹ × 1.0) / (0.5 × 10⁻³)
       β = (589 × 10⁻⁹) / (5 × 10⁻⁴)
       β = 1.178 × 10⁻³ m

∴ β ≈ 1.18 × 10⁻³ m (= 1.18 mm)

(iii) Effect of increasing slit separation:

Since β = λD/d, fringe width β is inversely proportional to d. When d is doubled (from 0.5 mm to 1.0 mm), the fringe width halves.

∴ β becomes 0.59 mm — the fringe width decreases (fringes come closer together).

(iv) Whether two separate sodium lamps produce interference:

The classmate is incorrect. Two separate sodium lamps are independent (incoherent) sources; their phases change randomly and independently with time. Sustained interference requires coherent sources that maintain a constant phase relationship. Two separate lamps do not maintain a constant phase difference, so no stable interference pattern is observed — only uniform illumination.

Question 3

A physics teacher sets up a Young's Double Slit Experiment (YDSE) in a school laboratory. The two slits are separated by d = 0.5 mm and the screen is placed at D = 1.0 m. A student notices that when a transparent glass slab of thickness t = 1.2 × 10⁻³ mm and refractive index n = 1.5 is placed in front of one of the slits (say S₁), the central bright fringe shifts. Another student then asks: 'If the experiment is now performed with white light instead of monochromatic light (λ = 600 nm), what would happen to the fringe pattern?' Answer the following:
(i) Calculate the number of fringes by which the central bright fringe shifts when the glass slab is introduced in front of S₁.
(ii) In which direction does the central fringe shift — towards S₁ or away from S₁? Justify your answer.
(iii) After the slab is introduced, a point P on the screen is located 0.6 mm above the new central bright fringe position. Find the resultant intensity at P if each slit has intensity I₀. (Use λ = 600 nm.)
(iv) State ONE reason why white light produces a less distinct interference pattern compared to monochromatic light.

Accepted Answer

(i) Number of fringes shifted:

The extra optical path introduced by the slab placed in front of S₁ is:
       Δ = (n − 1) t
→ Δ = (1.5 − 1) × 1.2 × 10⁻³ mm
→ Δ = 0.5 × 1.2 × 10⁻³ mm = 6 × 10⁻⁴ mm = 6 × 10⁻⁷ m

Fringe width:  β = λD/d = (600 × 10⁻⁹ × 1.0) / (0.5 × 10⁻³)
→ β = 1.2 × 10⁻³ m = 1.2 mm

Number of fringes shifted = Δ/λ = (6 × 10⁻⁷) / (600 × 10⁻⁹)
∴ Number of fringes shifted = 1

(ii) Direction of shift:

The glass slab placed in front of S₁ increases the optical path from S₁ to any point on the screen. The central bright fringe (zero path difference) now shifts to a point where the geometric path from S₂ is longer by the same amount (n − 1)t, to compensate. This point lies on the same side as S₁.
∴ The central bright fringe shifts towards S₁.

(iii) Resultant intensity at point P:

After the slab is introduced, the new central bright fringe is the reference (zero path difference). Point P is 0.6 mm above this new centre.

Path difference at P due to geometry:
       Δ_geo = y·d / D  (using small angle approximation)
→ Δ_geo = (0.6 × 10⁻³ × 0.5 × 10⁻³) / 1.0
→ Δ_geo = 3 × 10⁻⁷ m = 300 nm = λ/2

Phase difference:  φ = (2π/λ) × Δ_geo = (2π / 600 nm) × 300 nm = π

Using the intensity formula:
       I = 4I₀ cos²(φ/2) = 4I₀ cos²(π/2) = 4I₀ × 0
∴ Intensity at P = 0  (P coincides with the first dark fringe)

(iv) White light and fringe distinctness:

White light is a mixture of different wavelengths (colours). Since fringe width β = λD/d depends on wavelength, each colour produces its own fringe pattern with a different fringe width. These overlapping patterns of different colours wash out all fringes except the central white bright fringe, making the overall pattern far less distinct (coloured and blurred) compared to the sharp, well-defined bright and dark fringes obtained with monochromatic light.
∴ White light gives an indistinct pattern because different wavelengths produce overlapping fringe patterns of different fringe widths.

Question 4

A research student is investigating light behaviour using a specially designed optical bench. Monochromatic light of wavelength 600 nm from a laser passes through a double-slit arrangement where the slit separation is 0.3 mm and the screen is placed 2.0 m away. The student then replaces the double slit with a single slit of width 0.3 mm (same as the slit separation before) keeping all other conditions identical.

(i) Calculate the fringe width β observed in the double-slit experiment.
(ii) The student notices that the 4th bright fringe from the centre in the double-slit pattern appears to be missing when the single slit of width a = 0.3 mm is introduced alongside the double slit of separation d. If the missing fringe condition is given by d/a = n/m (where n is the order of the double-slit bright fringe and m is the order of the single-slit minimum), find the ratio d/a that causes the 4th bright fringe to be the first missing fringe (m = 1).
(iii) The student now closes one slit completely in the double-slit setup. The intensity at the centre of the screen with both slits open was I₀. What is the intensity at the centre when only one slit is open? Give a reason for your answer.
(iv) In the single-slit diffraction pattern, the student measures the width of the central maximum on the screen. If the slit width is halved to 0.15 mm (keeping wavelength and screen distance unchanged), what happens to the width of the central maximum? Justify.

Accepted Answer

(i) Fringe Width β in double-slit experiment:

By the standard result for Young's Double Slit Experiment (YDSE):

β = λD/d

where λ = 600 nm = 600 × 10⁻⁹ m, D = 2.0 m, d = 0.3 mm = 0.3 × 10⁻³ m.

Substituting:

β = (600 × 10⁻⁹ × 2.0) / (0.3 × 10⁻³)

β = (1200 × 10⁻⁹) / (0.3 × 10⁻³)

β = 4000 × 10⁻⁶ m

∴ β = 4.0 × 10⁻³ m = 4.0 mm

(ii) Missing fringe condition and ratio d/a:

In a double-slit experiment combined with a single-slit envelope, a bright fringe of order n is missing when it coincides with a minimum of the single-slit diffraction pattern of order m.

Condition for nth bright fringe (YDSE): y_n = nλD/d

Condition for mth minimum (single slit): y_m = mλD/a

For missing fringe: y_n = y_m

→ nλD/d = mλD/a

→ d/a = n/m

Given: the 4th bright fringe (n = 4) is the first missing fringe due to the first single-slit minimum (m = 1):

d/a = 4/1

∴ d/a = 4

This means the slit separation d must be 4 times the slit width a for the 4th order bright fringe to be missing.

(iii) Intensity at centre when one slit is closed:

By the principle of superposition in YDSE, when both slits are open, the amplitude at the centre is the sum of amplitudes from each slit:

A_total = A₁ + A₂

If each slit contributes equal amplitude A₀, then with both slits open:

A_total = 2A₀ → I₀ = (2A₀)² = 4A₀²

When only one slit is open:

A = A₀ → I_single = A₀² = I₀/4

∴ Intensity at centre with one slit open = I₀/4

Reason: Intensity is proportional to the square of the amplitude. With both slits open, amplitudes add constructively at centre giving amplitude 2A₀, so intensity is 4A₀². With one slit closed, only amplitude A₀ contributes, giving intensity A₀² = I₀/4.

(iv) Effect of halving slit width on central maximum width:

The half-angular width of the central maximum in single-slit diffraction is given by:

sin θ ≈ θ = λ/a (for first minimum)

The full width of the central maximum on screen:

W = 2λD/a

If the slit width is halved: a' = a/2 = 0.15 mm

W' = 2λD/a' = 2λD/(a/2) = 4λD/a = 2W

∴ The width of the central maximum doubles (becomes 8.0 mm from the original 4λD/a).

Justification: Diffraction spreading is inversely proportional to slit width (W ∝ 1/a). A narrower slit causes greater diffraction, spreading the central maximum over a wider region on the screen. Halving the slit width therefore doubles the width of the central diffraction maximum.

Question 5

A research team is designing a compact optical sensor. In their setup, two coherent slits S₁ and S₂ are separated by d = 0.4 mm and placed 1.0 m from a screen. The slits are illuminated by monochromatic light of wavelength λ = 500 nm. To test the sensor's resolution, they introduce a thin transparent glass slab of refractive index n = 1.5 and thickness t = 0.02 mm over slit S₁ only.

(i) Calculate the fringe width β in the original (without slab) interference pattern.

(ii) Due to the introduction of the glass slab, the entire fringe pattern shifts. State the direction of the shift and derive an expression for the lateral shift Δy of the central bright fringe in terms of t, n, λ, D, and d.

(iii) Calculate the number of fringes by which the central bright fringe shifts.

(iv) The team now replaces the glass slab with one of refractive index n' and the same thickness t. They observe that the central bright fringe shifts back exactly to its original position (i.e., the shift is zero). What must be the value of n'? What does this physically mean about the slab?

Accepted Answer

(i) Fringe Width β

By the formula for fringe width in YDSE:

β = λD/d

Substituting: β = (500 × 10⁻⁹ × 1.0) / (0.4 × 10⁻³)

∴ β = 1.25 × 10⁻³ m = 1.25 mm

─────────────────────────────────────────

(ii) Direction of Shift and Expression for Lateral Shift Δy

When a glass slab of refractive index n and thickness t is placed over S₁, the optical path through S₁ increases. The extra optical path introduced by the slab is:

Extra path = (n − 1)t

This means the optical path from S₁ becomes longer, so the point that was previously the central bright fringe (zero path difference) must shift toward S₁ to compensate.

∴ Direction of shift: The entire fringe pattern shifts toward S₁ (the side where the slab is placed).

Derivation of Δy:
Let P be the new position of the central bright fringe on the screen, at a distance y from the centre O. The geometric path difference at P (taking S₂ path longer by yd/D for P on the S₂ side) must equal the extra optical path introduced by the slab:

Path difference condition for central maximum:

(Optical path from S₂ to P) − (Optical path from S₁ to P) = 0

⟹ [S₂P − S₁P] + (n − 1)t = 0

For a shift Δy toward S₁ (say S₁ is the upper slit), the geometric path difference:

S₂P − S₁P = Δy · d / D

So: Δy · d/D = (n − 1)t

∴ Δy = (n − 1) t D / d

─────────────────────────────────────────

(iii) Number of Fringes Shifted

The number of fringes by which the central bright fringe shifts is given by:

Number of fringes = Δy / β = [(n − 1)t D/d] / [λD/d]

⟹ Number of fringes = (n − 1)t / λ

Substituting values:

= (1.5 − 1) × 0.02 × 10⁻³ / (500 × 10⁻⁹)

= (0.5 × 2 × 10⁻⁵) / (5 × 10⁻⁷)

= (1 × 10⁻⁵) / (5 × 10⁻⁷)

∴ Number of fringes shifted = 20

─────────────────────────────────────────

(iv) Condition for Zero Net Shift

For the central bright fringe to return to its original position, the net extra optical path introduced by the new slab must be zero:

(n' − 1)t = 0

Since t = 0.02 mm ≠ 0, we require:

n' − 1 = 0  ⟹  ∴ n' = 1

Physical meaning: A refractive index of 1 means the slab is optically identical to air (or vacuum). In practice, this means the slab is absent or has no optical effect — it does not alter the speed of light passing through it. There is no extra optical path, so no fringe shift occurs. The slab is effectively transparent with no phase-retarding property, i.e., it behaves as if it were not present.

Question 6

A physics teacher sets up a Young's Double Slit Experiment (YDSE) in the school laboratory using a sodium lamp (λ = 589 nm). The two slits are separated by d = 0.5 mm and the screen is placed at D = 1.0 m. During the demonstration, she makes three successive changes to the setup — one at a time — and asks students to predict the new fringe width in each case:

(i) She replaces the sodium lamp with a green laser (λ = 500 nm), keeping all other parameters the same. What is the new fringe width? [1]

(ii) After restoring the original setup (λ = 589 nm, d = 0.5 mm, D = 1.0 m), she moves the screen farther away to D = 1.5 m. Calculate the percentage change in fringe width. [1]

(iii) After restoring the original setup again, a student suggests immersing the entire apparatus in water (refractive index n = 4/3). Predict the new fringe width and explain why it changes. [1]

(iv) In the original setup, the 4th bright fringe from the centre is observed at a distance y from the central maximum. If the slit separation is halved (d → d/2), at what position will the 4th bright fringe now appear? Express your answer in terms of y. [1]

Accepted Answer

(i) Using the fringe width formula:
β = λD/d
Substituting λ = 500 nm = 500×10⁻⁹ m, D = 1.0 m, d = 0.5 mm = 0.5×10⁻³ m:
β = (500×10⁻⁹ × 1.0) / (0.5×10⁻³)
∴ β = 1.0×10⁻³ m = 1.0 mm

(ii) Original fringe width:
β₁ = λD₁/d = (589×10⁻⁹ × 1.0) / (0.5×10⁻³) = 1.178×10⁻³ m
New fringe width with D₂ = 1.5 m:
β₂ = λD₂/d = (589×10⁻⁹ × 1.5) / (0.5×10⁻³) = 1.767×10⁻³ m
Percentage change = [(β₂ − β₁)/β₁] × 100 = [(1.5 − 1.0)/1.0] × 100
∴ Percentage increase in fringe width = 50%

(iii) When the apparatus is immersed in water (n = 4/3), the wavelength of light decreases to:
λ' = λ/n = 589/(4/3) = 589 × 3/4 = 441.75 nm
New fringe width:
β' = λ'D/d = (441.75×10⁻⁹ × 1.0) / (0.5×10⁻³)
∴ β' = 0.8835×10⁻³ m ≈ 0.88 mm
Explanation: Because the speed of light in water is v = c/n, the wavelength decreases (λ' = λ/n) while frequency remains constant. Since β = λD/d, a smaller wavelength gives a smaller fringe width. ∴ Fringes become more closely spaced in water.

(iv) Position of 4th bright fringe in original setup:
y = y₄ = 4λD/d
When slit separation is halved (d → d/2), the new position of the 4th bright fringe:
y₄' = 4λD/(d/2) = 2 × (4λD/d) = 2y
∴ The 4th bright fringe appears at position 2y from the central maximum.

Question 7

A student is performing a single-slit diffraction experiment in a school laboratory. She uses a laser of wavelength 600 nm and a slit of width 0.3 mm. The screen is placed 1.5 m from the slit. She notices that the central bright fringe is much wider than the secondary maxima on either side.

Based on this experimental scenario, answer the following questions:

(i) What is the width of the central maximum observed on the screen?

(ii) If the student now doubles the slit width to 0.6 mm (keeping all other parameters the same), how does the width of the central maximum change? Give a reason.

(iii) The student's friend claims: "Since dark fringes appear in the diffraction pattern, light energy is destroyed at those positions, which violates conservation of energy." Is this claim correct? Justify your answer.

(iv) What will happen to the diffraction pattern if the student replaces the monochromatic laser with white light? State ONE observation she will make.

Accepted Answer

(i) Width of Central Maximum:

By the principle of single-slit diffraction, the angular position of the first minimum on each side of the central maximum is given by:

a sinθ = λ  →  sinθ ≈ λ/a  (for small θ)

Width of central maximum = 2λD/a

Substituting values: λ = 600 nm = 600 × 10⁻⁹ m, D = 1.5 m, a = 0.3 mm = 0.3 × 10⁻³ m

Width = 2 × (600 × 10⁻⁹) × 1.5 / (0.3 × 10⁻³)

Width = (2 × 600 × 10⁻⁹ × 1.5) / (3 × 10⁻⁴)

Width = (1800 × 10⁻⁹) / (3 × 10⁻⁴)

∴ Width of central maximum = 6 × 10⁻³ m = 6 mm

(ii) Effect of Doubling Slit Width:

Since width of central maximum = 2λD/a, it is inversely proportional to slit width a.

When a is doubled (0.3 mm → 0.6 mm), the width of the central maximum becomes half, i.e., it reduces to 3 mm.

Reason: A wider slit diffracts light less (the bending effect decreases as the slit size becomes larger compared to the wavelength), so the central fringe becomes narrower.

(iii) Claim Assessment — Conservation of Energy:

The student's friend's claim is INCORRECT.

In single-slit diffraction, energy is redistributed, not destroyed. At positions of dark fringes, destructive interference occurs and the intensity is zero; however, this energy is compensated by the enhanced intensity at bright fringe positions. The total energy incident on the slit equals the total energy distributed across the entire diffraction pattern. Thus, the law of conservation of energy is strictly obeyed in diffraction.

(iv) Effect of Replacing Monochromatic Light with White Light:

When white light is used, each wavelength produces its own diffraction pattern with a slightly different fringe width (since β ∝ λ). The central maximum remains white (all wavelengths superpose at the centre), but the secondary maxima appear as coloured bands (spectrum), with violet colour closer to the centre and red colour farther out on each side.

Question 8

A physics teacher sets up a Young's Double Slit Experiment (YDSE) in a laboratory using a sodium lamp (λ = 589 nm). The two slits are separated by d = 0.5 mm and the screen is placed at D = 1.0 m from the slits. During the demonstration, a student notices that when a thin transparent glass slab of refractive index n = 1.5 is introduced in front of one of the slits, the central bright fringe shifts toward that slit.

(i) State the condition for constructive interference (bright fringe) in YDSE in terms of path difference.

(ii) Calculate the fringe width β observed on the screen before the glass slab is introduced.

(iii) A student argues: 'Introducing the glass slab increases the optical path on one side, which shifts the central fringe toward the slab.' Is the student correct? Justify with one reason based on the principle of path difference.

(iv) If the glass slab of thickness t = 0.1 mm is introduced in front of slit S₁, calculate the number of fringes by which the central bright fringe shifts.

Accepted Answer

(i) Condition for Constructive Interference:
In YDSE, constructive interference (bright fringe) occurs when the path difference between the two waves is an integral multiple of the wavelength.
∴ Path difference Δ = nλ, where n = 0, ±1, ±2, …

(ii) Calculation of Fringe Width:
The fringe width in YDSE is given by:
β = λD/d

Substituting values:
β = (589 × 10⁻⁹ m × 1.0 m) / (0.5 × 10⁻³ m)
β = (589 × 10⁻⁹) / (5 × 10⁻⁴)
∴ β = 1.178 × 10⁻³ m ≈ 1.18 mm

(iii) Justification:
Yes, the student is correct.
Because the glass slab of refractive index n > 1 increases the optical path length in front of slit S₁ by (n − 1)t. The central bright fringe forms at the point where the net path difference is zero. To compensate for the extra optical path introduced on the S₁ side, the central fringe shifts toward S₁ (the slab side), so that the geometrical path from S₂ becomes correspondingly larger and equalises the total optical path from both slits.
∴ The central fringe shifts toward the slit covered by the slab.

(iv) Number of Fringes Shifted:
The extra optical path introduced by the slab = (n − 1)t
The number of fringes shifted is given by:
Number of fringes = (n − 1)t / λ

Substituting values:
Number of fringes = (1.5 − 1) × 0.1 × 10⁻³ / (589 × 10⁻⁹)
= (0.5 × 10⁻⁴) / (589 × 10⁻⁹)
= (0.5 × 10⁻⁴) / (5.89 × 10⁻⁷)
∴ Number of fringes shifted ≈ 85

Question 9

In Young's double-slit experiment, if the distance between the two slits is halved while all other quantities remain unchanged, the fringe width β will become:
(a) β/4
(b) β/2
(c) 2β
(d) 4β

Accepted Answer

Option (c) is correct.

Explanation: In YDSE, fringe width is given by β = λD/d, where λ is wavelength, D is slit-to-screen distance, and d is slit separation.

Since β ∝ 1/d, when d is halved (d → d/2), the fringe width doubles.

∴ New fringe width = 2β.

Question 10

In a single-slit diffraction experiment, the width of the slit is doubled. What happens to the width of the central maximum?

Accepted Answer

Option (C) is correct.

Explanation: The width of the central maximum in single-slit diffraction is given by:

Width = 2λD/a

where a is the slit width, λ is the wavelength, and D is the distance to the screen.

∴ Width ∝ 1/a. When a is doubled, the width of the central maximum becomes half.

Question 11

Assertion (A): In Young's double slit experiment, when the separation between the two slits is decreased, the fringe width increases.
Reason (R): Fringe width in Young's double slit experiment is directly proportional to the distance between the slits and the screen.

Accepted Answer

Option (c) is correct.

Explanation: The fringe width in Young's double slit experiment is given by:

β = λD/d

where λ is the wavelength of light, D is the distance between the slits and the screen, and d is the separation between the two slits.

Assertion (A) is TRUE: Since β ∝ 1/d, decreasing the slit separation d causes the fringe width β to increase.

Reason (R) is FALSE: Fringe width β is directly proportional to D (the slit-to-screen distance), NOT to the slit separation d. In fact, β is INVERSELY proportional to d.

∴ Assertion (A) is true but Reason (R) is false.

Question 12

In Young's double-slit experiment, when the distance between the two slits is halved and the distance of the screen from the slits is doubled, the fringe width β becomes:

Accepted Answer

Option (d) is correct.

Explanation: Fringe width in YDSE is given by β = λD/d.

New fringe width β′ = λ(2D)/(d/2) = 4λD/d = 4β.

∴ The fringe width becomes 4β.

Question 13

Assertion (A) : In Young's double slit experiment, when the two slits are replaced by a single slit of the same width, the central bright fringe becomes wider.
Reason (R) : In single slit diffraction, the angular width of the central maximum is 2λ/a, which is larger than the fringe width λD/d in double slit interference (when a < d).

Accepted Answer

Option (a) is correct.

Explanation: In Young's double slit experiment, the fringe width is β = λD/d, so individual bright fringes are narrow. When both slits are replaced by a single slit of width a (where a < d), the central maximum of single slit diffraction has an angular half-width θ = λ/a, giving a total angular width of 2λ/a. Since a < d, we have 2λ/a > λD/d (in angular terms), so the central bright fringe is indeed wider — Assertion (A) is TRUE.

The Reason (R) correctly states that the central maximum half-width in single slit diffraction is 2λ/a (total angular width), which is greater than the double slit fringe spacing λ/d when a < d — Reason (R) is TRUE and directly explains why the central fringe becomes wider.

∴ Both A and R are true, and R is the correct explanation of A.

Question 14

Light waves from two pinholes illuminated by two separate sodium lamps do not produce a sustained interference pattern. Explain why.

Accepted Answer

For a sustained interference pattern, the two sources must be coherent — they must emit light waves of the same frequency with a constant (or zero) phase difference.

Two separate sodium lamps, even though they emit light of the same wavelength, are independent sources. The atoms in each lamp emit light in random, independent bursts of finite duration (~10⁻⁸ s). As a result, the phase difference between the waves arriving from the two pinholes changes randomly and rapidly with time.

∴ The condition of coherence is NOT satisfied, the phase difference averages out, and no sustained (stable) bright and dark fringes are observed on the screen.

Question 15

Light waves from two pinholes illuminated by two separate sodium lamps do not produce a sustained interference pattern. Explain why, giving two reasons.

Accepted Answer

For a sustained interference pattern to be observed, the two sources must be coherent — they must emit light waves of the same frequency with a constant (or zero) phase difference.

Two separate sodium lamps, even though they emit light of nearly the same wavelength, are incoherent sources because:

(i) The phase difference between the waves emitted by the two independent lamps changes randomly and rapidly (of the order of 10⁻⁸ s) due to the random and independent emission of photons by excited atoms in each lamp. Since the phase difference fluctuates randomly, the positions of bright and dark fringes shift so rapidly that the eye perceives only a uniform illumination — no stable fringe pattern is formed.

(ii) Each lamp emits light of slightly different frequencies (due to Doppler broadening and independent thermal conditions of the atoms), so the condition of equal frequency for coherence is not strictly satisfied.

∴ Two separate sodium lamps acting as independent sources cannot produce a sustained interference pattern.

Question 16

In Young's double-slit experiment, the two slits are separated by a distance d and the screen is placed at a distance D from the slits. If the wavelength of light used is λ, write the expression for (i) fringe width β, and (ii) the distance of the 3rd bright fringe from the central maximum.

Accepted Answer

In Young's Double Slit Experiment (YDSE), constructive interference (bright fringe) occurs when the path difference equals a whole number multiple of wavelength.

Path difference for n-th bright fringe: Δx = nλ

(i) Fringe Width β:

The position of the n-th bright fringe is yₙ = nλD/d and the position of the (n+1)-th bright fringe is yₙ₊₁ = (n+1)λD/d.

Fringe width β = yₙ₊₁ − yₙ

∴  β = λD/d

(ii) Distance of 3rd bright fringe from central maximum:

Using the general formula for position of n-th bright fringe: yₙ = nλD/d

For n = 3:

y₃ = 3 × λD/d

∴  y₃ = 3λD/d

Question 17

Two independent sodium lamps are used to illuminate two pinholes in a Young's double-slit experiment. No interference pattern is observed on the screen. Give reason.

Accepted Answer

Two independent sodium lamps are NOT coherent sources — they emit light waves independently, so the phase difference between the waves reaching the screen changes randomly and rapidly with time.

For a sustained interference pattern, the two sources must be coherent — they must emit light waves of the same frequency with a constant (or zero) phase difference.

Two separate sodium lamps, even though they emit light of the same wavelength, are independent sources. The atoms in each lamp emit light in random, independent bursts of finite duration (~10⁻⁸ s). As a result, the phase difference between the waves arriving from the two pinholes changes randomly and rapidly with time.

∴ The condition of coherence is NOT satisfied, the phase difference averages out, and no sustained (stable) bright and dark fringes are observed on the screen.

∴ Interference pattern is NOT observed because the two independent sodium lamps are incoherent sources (no constant phase difference between them).

Question 18

In a Young's double-slit experiment, the two slits are separated by a distance d = 0.5 mm and the screen is placed at a distance D = 1.0 m from the slits. Light of wavelength λ = 500 nm is used.
(a) Calculate the fringe width β.
(b) If the intensity of each slit is I₀, write the expression for the resultant intensity at any point on the screen and hence find the intensity at a point where the path difference between the two waves is λ/4.

Accepted Answer

(a) Fringe Width:

By the formula for fringe width in YDSE:

β = λD/d

Substituting values:

β = (500 × 10⁻⁹ m × 1.0 m) / (0.5 × 10⁻³ m)

β = (500 × 10⁻⁹) / (0.5 × 10⁻³)

∴ β = 1.0 × 10⁻³ m = 1.0 mm

(b) Resultant Intensity Expression and Value at Δ = λ/4:

When each slit has intensity I₀, the resultant intensity at any point on the screen is given by:

I = 4I₀ cos²(φ/2)

where φ is the phase difference between the two waves at that point.

The relation between path difference (Δ) and phase difference (φ) is:

φ = (2π/λ) × Δ

For Δ = λ/4:

φ = (2π/λ) × (λ/4) = π/2

Substituting into the intensity formula:

I = 4I₀ cos²(π/4)

I = 4I₀ × (1/√2)²

I = 4I₀ × (1/2)

∴ I = 2I₀

Question 19

In Young's double slit experiment, the two slits are separated by a distance d = 0.5 mm and the screen is placed at a distance D = 1.0 m from the slits. Monochromatic light of wavelength λ = 600 nm is used.
(a) Calculate the fringe width β.
(b) If the intensity of each coherent source is I₀, find the resultant intensity at a point P on the screen where the path difference between the two waves is λ/6.
(c) What happens to the fringe width if the entire experimental setup is immersed in water of refractive index n = 4/3?

Accepted Answer

By the principle of superposition of coherent waves (Huygens' wave theory), two coherent sources S₁ and S₂ separated by d produce an interference pattern on a screen at distance D.

(a) Fringe Width:

The fringe width in Young's Double Slit Experiment is given by:

β = λD/d

Substituting values — λ = 600 nm = 600 × 10⁻⁹ m, D = 1.0 m, d = 0.5 mm = 0.5 × 10⁻³ m:

β = (600 × 10⁻⁹ × 1.0) / (0.5 × 10⁻³)

β = (600 × 10⁻⁹) / (5 × 10⁻⁴)

∴ β = 1.2 × 10⁻³ m  =  1.2 mm

(b) Resultant Intensity Expression:

When two coherent waves of equal individual intensity I₀ superpose with a phase difference φ, the resultant intensity at any point is:

I = 4I₀ cos²(φ/2)

At the centre of the screen, the path difference between waves from S₁ and S₂ is zero, so:

φ = 0

I_centre = 4I₀ cos²(0/2) = 4I₀ cos²(0) = 4I₀ × 1

∴ I_centre = 4I₀

[This confirms energy conservation — intensity is redistributed, not created or destroyed.]

Question 20

A research team is designing an optical instrument that uses a double-slit arrangement illuminated by a monochromatic source of wavelength 600 nm. The slits are separated by 0.3 mm and the screen is placed 1.5 m away. During testing, one of the slits is covered with a thin transparent mica sheet of refractive index 1.5, which introduces an extra optical path. The central bright fringe is observed to shift by 5 fringe widths towards the slit covered by the mica sheet.

(i) Calculate the fringe width β of the original (unshifted) pattern.
(ii) Find the thickness t of the mica sheet.
(iii) If the intensity at each slit (without the mica sheet) is I₀, write the expression for the resultant intensity at any point P on the screen in terms of the phase difference φ between the two waves. Hence state the condition for a dark fringe.
(iv) After the mica sheet is inserted, the team notices that the visibility (contrast) of the fringes does NOT change. Give a reason for this observation in terms of the properties of the two interfering waves.

Accepted Answer

(i) Fringe Width β

The fringe width in Young's double-slit experiment is given by:

β = λD / d

where λ = 600 nm = 600 × 10⁻⁹ m, D = 1.5 m, d = 0.3 mm = 0.3 × 10⁻³ m

β = (600 × 10⁻⁹ × 1.5) / (0.3 × 10⁻³)

β = (9 × 10⁻⁷) / (3 × 10⁻⁴)

∴ β = 3 × 10⁻³ m = 3 mm

(ii) Thickness of Mica Sheet

When a transparent sheet of refractive index μ and thickness t is placed over one slit, the extra optical path introduced is:

Extra path = (μ − 1) t

The central fringe shifts by a distance Δy = n × β, where n is the number of fringes shifted.

Given: shift = 5β, so Δy = 5β = 5 × 3 × 10⁻³ m = 15 × 10⁻³ m

The shift formula: Δy = (μ − 1) t × D / d

Alternatively, the extra optical path equals n × λ:

(μ − 1) t = n × λ = 5 × λ

(1.5 − 1) × t = 5 × 600 × 10⁻⁹

0.5 × t = 3000 × 10⁻⁹

t = 3000 × 10⁻⁹ / 0.5

∴ t = 6 × 10⁻⁶ m = 6 μm

(iii) Resultant Intensity and Dark Fringe Condition

By the principle of superposition, when two coherent waves each of intensity I₀ superpose with a phase difference φ, the resultant intensity is:

I = I₀ + I₀ + 2√(I₀ · I₀) cos φ

∴ I = 4I₀ cos²(φ/2)

Condition for a dark fringe: I = 0

⇒ cos²(φ/2) = 0 ⇒ φ/2 = (2n+1)π/2

∴ φ = (2n + 1)π, where n = 0, 1, 2, …

In terms of path difference: path difference = (2n + 1) λ/2, i.e., an odd multiple of half-wavelength.

(iv) Reason for Unchanged Visibility

The mica sheet introduces only an additional optical path (phase lag) to the wave passing through it; it does NOT alter the amplitude of that wave. Since both slits still emit waves of equal amplitude (√I₀), the ratio of maximum to minimum intensity — and hence the visibility V = (I_max − I_min)/(I_max + I_min) — depends only on the amplitudes of the two interfering waves. Because the amplitudes remain equal after insertion of the sheet, I_min = 0 and I_max = 4I₀ as before, giving V = 1 (perfect visibility). The entire fringe pattern simply shifts bodily; the contrast (visibility) is unchanged.

Wave Optics: Class 12 Physics Practice Questions

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