UET UG Physics Booster Test 2 -Diffraction and Polarisation -

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CUET UG Physics Booster Test 2 -Diffraction and Polarisation

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2 -Diffraction and Polarisation

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QUESTION 1 OF 20

Choose the correct statements about geometrical shadow regions and diffraction:
1. Diffraction effects are best seen when the obstacle size is much larger than the wavelength
2. Alternate dark and bright regions appear close to the geometrical shadow
3. Light appears to turn around corners into the shadow
4. The finite resolution of optical instruments is limited by diffraction

Answer Review

1 Choose the correct statements about geometrical shadow regions and diffraction:
1. Diffraction effects are best seen when the obstacle size is much larger than the wavelength
2. Alternate dark and bright regions appear close to the geometrical shadow
3. Light appears to turn around corners into the shadow
4. The finite resolution of optical instruments is limited by diffraction

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Diffraction occurs near edges. �� Light bends into shadow regions. �� Optical resolution is diffraction limited.

✅ Correct Answer Explanation

Statement 2 is correct because diffraction produces alternating bright and dark fringes near the geometrical shadow. Statement 3 is correct because diffraction makes light appear to bend around obstacles and enter shadow regions. Statement 4 is correct because diffraction limits the resolving power of microscopes and telescopes. Statement 1 is incorrect because diffraction is most pronounced when obstacle dimensions are comparable to the wavelength, not much larger.

❓ Why Other Options Are Incorrect

�� Statement 1 → Diffraction is strongest when obstacle size ≈ wavelength.
�� Option A → Contains incorrect Statement 1.
�� Option C → Contains incorrect Statement 1.
�� Option D → Contains incorrect Statement 1.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 individually.

Final Logic:

Only Statements 2, 3 and 4 are correct.

🧠 Memory Trick

Size ≈ λ → Strong Diffraction

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Diffraction Basics

2 In terms of the visibility of diffraction phenomena:

A the phenomenon becomes less pronounced as the obstacle size approaches the wavelength of light

B the phenomenon is only defined for sound waves and not for light waves

C the phenomenon is sometimes difficult to observe because the wavelength of light is much smaller than most macroscopic dimensions

D the phenomenon cannot occur if the obstacle is placed in a vacuum

💡 Short Explanation

�� Visible light has a very small wavelength. �� Macroscopic objects are much larger than λ. �� Hence diffraction is often less obvious.

✅ Correct Answer Explanation

Diffraction is observed for all waves, including light. Since visible wavelengths are extremely small compared to everyday objects, diffraction effects are usually weak and difficult to notice. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Diffraction becomes more pronounced as size approaches wavelength.
�� Option B → Diffraction occurs for light as well.
�� Option D → Diffraction occurs in vacuum.

🎯 Strategy

Used

Elimination

Application:

Remove statements contradicting wave theory.

Final Logic:

Small wavelength compared to object size makes diffraction less noticeable.

🧠 Memory Trick

Small λ → Less Visible Diffraction

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Conditions for Diffraction

3 As the slit width a increases, the spread of the diffracted light and the angular width of the central maximum become respectively:

A Smaller, Smaller

B Larger, Larger

C Smaller, Larger

D Larger, Smaller

💡 Short Explanation

�� Angular width ∝ λ/a. �� Increasing slit width reduces diffraction. �� Central maximum becomes narrower.

✅ Correct Answer Explanation

The angular width of the central maximum is approximately: 2λ/a As slit width a increases, λ/a decreases. Therefore the diffraction spread decreases and the central maximum becomes narrower.

❓ Why Other Options Are Incorrect

�� Option B → Opposite trend.
�� Option C → Central maximum does not become larger.
�� Option D → Diffraction spread does not increase.

🎯 Strategy

Used

Formula Recall

Application:

Use width ∝ λ/a.

Final Logic:

Larger slit width gives smaller diffraction spread.

🧠 Memory Trick

Big Slit → Small Spread

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Width of Central Maximum

4 In the mathematical treatment of the single slit:

A the wavefront at the slit is divided into smaller parts that act as incoherent sources

B the different parts of the wavefront at the slit are treated as secondary sources that are in phase

C the secondary sources have a phase difference of π initially

D the sources are assumed to be moving across the slit

💡 Short Explanation

�� Huygens principle is used. �� Secondary sources are coherent. �� They start in phase.

✅ Correct Answer Explanation

According to Huygens' principle, every point on the incident wavefront acts as a secondary source. Since all points belong to the same wavefront, they are initially in phase and coherent.

❓ Why Other Options Are Incorrect

�� Option A → Sources are coherent, not incoherent.
�� Option C → No initial π phase difference exists.
�� Option D → Sources are stationary points on the wavefront.

🎯 Strategy

Used

Conceptual Recall

Application:

Apply Huygens' principle.

Final Logic:

Wavefront points act as in-phase secondary sources.

🧠 Memory Trick

Same Wavefront = Same Phase

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Huygens Construction

5 Consider the statements on the central maximum of single slit diffraction:
1. It occurs at an angle θ = 0
2. It represents a region where the intensity is the weakest
3. It is bounded by the first minima at ± λ/a
4. It is an area of complete destructive interference

A 1, 2, 3

B 1, 3

C 2, 4

D 1, 2, 4

💡 Short Explanation

�� Central maximum is brightest. �� Located at θ = 0. �� Lies between first minima.

✅ Correct Answer Explanation

Statement 1 is correct because the central maximum occurs at θ = 0. Statement 3 is correct because it extends between the first minima approximately located at ±λ/a. Statement 2 is incorrect because the central maximum has maximum intensity. Statement 4 is incorrect because it is produced by constructive, not destructive, interference.

❓ Why Other Options Are Incorrect

�� Statement 2 → Central maximum is brightest.
�� Statement 4 → Destructive interference produces minima.
�� Option A → Contains incorrect Statement 2.
�� Option C → Contains incorrect Statements 2 and 4.
�� Option D → Contains incorrect Statements 2 and 4.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate each statement independently.

Final Logic:

Only Statements 1 and 3 are correct.

🧠 Memory Trick

Central = Brightest at Zero

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Central Maximum in Diffraction

6 The approximate angular position for the n-th secondary maximum in single slit diffraction is given by:

A (n + 1/2)λ / a

B nλ / a

C 2nλ / a

D (2n + 1)λ / D

💡 Short Explanation

�� Secondary maxima occur between minima. �� Approximate location uses half-integer values. �� Intensity decreases with order.

✅ Correct Answer Explanation

The approximate angular position of the nth secondary maximum is: θ ≈ (n + 1/2)λ/a This lies midway between successive minima and gives the approximate location of bright secondary fringes.

❓ Why Other Options Are Incorrect

�� Option B → Represents minima condition.
�� Option C → Not a diffraction maxima formula.
�� Option D → Uses incorrect parameter D.

🎯 Strategy

Used

Formula Recall

Application:

Use standard secondary maxima relation.

Final Logic:

Half-integer multiples locate secondary maxima.

🧠 Memory Trick

Maxima → n + 1/2

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Secondary Maxima

7 A single slit is illuminated with light of wavelength 600 nm. If the first minimum is formed at an angle of 30° (sin 30° = 0.5), what is the width of the slit?

A 1.2 × 10⁻⁶ m

B 3.0 × 10⁻⁷ m

C 1.2 × 10⁻⁷ m

D 3.0 × 10⁻⁶ m

💡 Short Explanation

�� First minimum: a sinθ = λ. �� Substitute values. �� Solve for slit width.

✅ Correct Answer Explanation

For first minimum: a sinθ = λ a = λ/sinθ = (600 × 10⁻⁹)/0.5 = 1.2 × 10⁻⁶ m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Four times smaller than required.
�� Option C → Ten times smaller.
�� Option D → Larger than calculated value.

🎯 Strategy

Used

Substitution

Application:

Apply the first-minimum condition directly.

Final Logic:

a = λ/sinθ = 1.2 × 10⁻⁶ m.

🧠 Memory Trick

First Minimum → a sinθ = λ

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Diffraction Calculations

8 Identify the incorrect statement regarding the ratio λ/a in diffraction:

A A larger λ/a ratio results in a wider central maximum

B If λ is much smaller than a, the diffraction spread is very small

C The ratio λ/a determines the angular positions of the minima

D The value of λ/a must be significantly greater than 1 for ray optics to be considered valid

💡 Short Explanation

�� Ray optics requires diffraction effects to be negligible. �� This occurs when λ/a is very small. �� Statement D reverses the condition.

✅ Correct Answer Explanation

Ray optics is valid when the wavelength is much smaller than aperture dimensions: λ/a << 1 Statement D incorrectly claims λ/a should be much greater than 1. Therefore it is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct relation.
�� Option B → Correct condition.
�� Option C → Minima positions depend on λ/a.

🎯 Strategy

Used

Extreme Word Filter

Application:

Check the condition required for ray optics.

Final Logic:

Ray optics requires λ/a << 1, not >> 1.

🧠 Memory Trick

Ray Optics → λ/a Small

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Diffraction and Ray Optics

9 Match List I with List II based on Feynman's discussion:

List I	List II
1. Double-slit experiment pattern	a. Question of usage
2. Few sources	b. Superposition of single-slit diffraction and double-slit interference
3. Large number of sources	c. Usually called interference
4. Difference between interference and diffraction	d. Word diffraction is more often used

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

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

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

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

💡 Short Explanation

�� Double slit combines interference and diffraction. �� Few sources → interference. �� Many sources → diffraction.

✅ Correct Answer Explanation

1 → b : Double-slit pattern is a superposition of interference and diffraction. 2 → c : Few coherent sources are usually associated with interference. 3 → d : Large numbers of sources are commonly associated with diffraction. 4 → a : Feynman described the distinction as largely a matter of usage.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching throughout.
�� Option C → Reverses the standard associations.
�� Option D → Does not match Feynman's description.

🎯 Strategy

Used

Option Grouping

Application:

Match each concept with its proper description.

Final Logic:

Only Option A gives all correct pairings.

🧠 Memory Trick

Few → Interference, Many → Diffraction

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

10 In distinguishing interference and diffraction based on source quantity:

A a pattern from exactly two interfering sources is usually referred to as diffraction

B a pattern from a large number of secondary sources across a slit is typically called diffraction

C a pattern from a large number of sources is always called interference

D a pattern from any number of sources cannot be distinguished

💡 Short Explanation

�� Diffraction involves many secondary sources. �� Interference usually involves a few sources. �� Terminology depends on source distribution.

✅ Correct Answer Explanation

Diffraction patterns arise due to interference among a large number of secondary wavelets originating from different parts of an aperture or slit. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Two sources are typically associated with interference.
�� Option C → Large numbers of sources are usually termed diffraction.
�� Option D → Practical distinction exists.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the conventional distinction.

Final Logic:

Many secondary sources correspond to diffraction.

🧠 Memory Trick

Many Sources → Diffraction

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Diffraction Terminology

11 In the razor blade demonstration for viewing a single-slit pattern:

A the blades must be perpendicular to each other

B the lens of the eye acts to focus the diffraction pattern onto the retina

C the bulb filament plays the role of the viewing screen

D direct sunlight is highly recommended as a light source

💡 Short Explanation

�� Razor blades form a narrow slit. �� The eye observes the diffraction pattern. �� The retina acts as the screen.

✅ Correct Answer Explanation

In the razor blade experiment, the diffraction pattern formed by the slit is focused onto the retina by the eye lens. Thus, the eye lens acts as the focusing element and the retina acts as the screen where the pattern is observed.

❓ Why Other Options Are Incorrect

�� Option A → The blade edges should be parallel, not perpendicular.
�� Option C → The filament acts as the light source, not the screen.
�� Option D → Direct sunlight is not recommended because it is too intense and inconvenient.

🎯 Strategy

Used

Elimination

Application:

Remove statements inconsistent with the experimental setup.

Final Logic:

The eye lens focuses the diffraction pattern onto the retina.

🧠 Memory Trick

Eye Lens → Focus, Retina → Screen

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Experimental Observation of Diffraction

12 Choose the correct statements about color fringes in diffraction using white light:
1. Positions of all bands except the central one depend on wavelength
2. Fringes show spectral colors due to wavelength dependence
3. Using a blue filter makes the fringes wider than when using a red filter
4. The central band does not separate into distinct colors like the secondary bands

A 1, 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Diffraction angle depends on wavelength. �� Different wavelengths spread differently. �� Red fringes are wider than blue fringes.

✅ Correct Answer Explanation

Statement 1 is correct because fringe positions depend on wavelength. Statement 2 is correct because different wavelengths diffract by different amounts, producing spectral colours. Statement 4 is correct because all wavelengths overlap at the central maximum, making it nearly white. Statement 3 is incorrect because blue light has a shorter wavelength and therefore produces narrower fringes than red light.

❓ Why Other Options Are Incorrect

�� Statement 3 → Blue fringes are narrower, not wider.
�� Option B → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 individually.

Final Logic:

Only Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Red Wide, Blue Narrow

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Diffraction with White Light

13 Identify the incorrect statement about the transverse nature of light:

A The electric field associated with light is at right angles to the direction of propagation

B Polarisation phenomena are special to transverse waves

C Sound waves in air can exhibit polarisation just like light waves

D Unpolarised light has its electric vector randomly changing in the transverse plane

💡 Short Explanation

�� Sound in air is longitudinal. �� Longitudinal waves cannot be polarised. �� Polarisation confirms transverse nature.

✅ Correct Answer Explanation

Polarisation is a property of transverse waves. Since sound waves in air are longitudinal, they cannot be polarised. Therefore Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of electromagnetic waves.
�� Option B → Correct property of polarisation.
�� Option D → Correct description of unpolarised light.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting the concept of polarisation.

Final Logic:

Longitudinal sound waves cannot be polarised.

🧠 Memory Trick

No Polarisation for Longitudinal Waves

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Polarisation and Transverse Nature of Light

14 For a wave propagating in the +x direction on a string, the displacement y(x,t) representing a y-polarised wave is given by:

A a sin(kx − ωt)

B a sin(kx + ωt)

C a cos(kx − ωy)

D a sin(ωt − kx)

💡 Short Explanation

�� Standard wave travelling in +x direction. �� Displacement occurs along y-axis. �� Represents y-polarisation.

✅ Correct Answer Explanation

A wave travelling in the positive x-direction is represented by: y(x,t) = a sin(kx − ωt) The displacement is along the y-axis while propagation occurs along the x-axis, representing a transverse y-polarised wave.

❓ Why Other Options Are Incorrect

�� Option B → Represents motion in the negative x-direction.
�� Option C → Incorrect wave expression.
�� Option D → Mathematically equivalent to −sin(kx−ωt), but the standard expression given in NCERT is Option A.

🎯 Strategy

Used

Formula Recall

Application:

Use the standard travelling wave equation.

Final Logic:

Positive x-direction corresponds to kx − ωt.

🧠 Memory Trick

+x Motion → kx − ωt

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Polarisation and Wave Motion

15 In a polaroid sheet, the alignment of the long chain molecules and the direction of the transmitted electric vector (pass-axis) are respectively:

A Parallel, Parallel

B Parallel, Perpendicular

C Perpendicular, Parallel

D Perpendicular, Perpendicular

💡 Short Explanation

�� Molecules are aligned in one direction. �� Parallel electric components are absorbed. �� Perpendicular components are transmitted.

✅ Correct Answer Explanation

In a polaroid, the long molecular chains are aligned parallel to one another. Electric field components parallel to these chains are absorbed, while the perpendicular component is transmitted. Therefore, the pass-axis is perpendicular to the molecular alignment.

❓ Why Other Options Are Incorrect

�� Option A → Pass-axis is not parallel to the molecular alignment.
�� Option C → Molecular chains are not perpendicular to each other.
�� Option D → Both relationships are incorrect.

🎯 Strategy

Used

Conceptual Recall

Application:

Apply the absorption rule of polaroids.

Final Logic:

Parallel molecules absorb parallel components and pass perpendicular components.

🧠 Memory Trick

Molecules Absorb Parallel

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

16 Choose the correct statements regarding the pass-axis of a polaroid:
1. It transmits the component of the electric vector parallel to it
2. It absorbs the electric vector oscillating parallel to the aligned molecules
3. It transmits 100% of the unpolarised incident light
4. It restricts the emerging light's electric vector to a single direction

A 1, 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Polaroids transmit only one component. �� Parallel molecular component is absorbed. �� Emergent light becomes plane polarised.

✅ Correct Answer Explanation

Statement 1 is correct because the pass-axis transmits the electric field component parallel to it. Statement 2 is correct because components parallel to aligned molecules are absorbed. Statement 4 is correct because the emerging light is restricted to one vibration direction. Statement 3 is incorrect because only half of the intensity of unpolarised light is transmitted.

❓ Why Other Options Are Incorrect

�� Statement 3 → A polaroid does not transmit all incident unpolarised light.
�� Option B → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Check each statement independently.

Final Logic:

Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Pass One Direction Only

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Pass Axis of Polaroid

17 An unpolarised light beam of intensity Iin passes through a polariser, and then through an analyzer set at an angle of 60° (cos 60° = 0.5) relative to the polariser's pass axis. What is the final transmitted intensity?

A 0.25 Iin

B 0.125 Iin

C 0.5 Iin

D 0.75 Iin

💡 Short Explanation

�� First polariser halves intensity. �� Apply Malus' Law. �� Multiply successive transmission factors.

✅ Correct Answer Explanation

After the first polariser: I₁ = Iin/2 Applying Malus' Law: I = I₁ cos²60° I = (Iin/2) × (1/2)² I = (Iin/2) × 1/4 I = Iin/8 I = 0.125 Iin Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ignores initial halving effect.
�� Option C → Represents intensity after only the first polariser.
�� Option D → Exceeds physically possible transmitted intensity.

🎯 Strategy

Used

Substitution

Application:

Apply intensity reduction step-by-step.

Final Logic:

(Iin/2) × cos²60° = 0.125 Iin.

🧠 Memory Trick

Half First, Then Cos²

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Malus' Law Problems

18 When three polaroids are arranged such that the first and third are crossed (angle is 90°), and a middle polaroid is placed at an angle θ relative to the first:

A the transmitted intensity is always zero regardless of θ

B the transmitted intensity is maximum when θ = π/4

C the transmitted intensity is equal to the initial unpolarised intensity

D the transmitted intensity depends only on the wavelength of light

💡 Short Explanation

�� Middle polaroid allows partial transmission. �� Intensity ∝ cos²θ sin²θ. �� Maximum occurs at 45°.

✅ Correct Answer Explanation

For crossed polaroids with a middle polaroid: I = (I₀/2) cos²θ sin²θ Using: sin²2θ = 4sin²θ cos²θ Intensity is maximum when: 2θ = 90° θ = 45° = π/4 Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Middle polaroid allows non-zero transmission.
�� Option C → Intensity is always less than the initial intensity.
�� Option D → Depends on angle, not only wavelength.

🎯 Strategy

Used

Formula Recall

Application:

Use three-polaroid intensity relation.

Final Logic:

Maximum transmission occurs at θ = 45°.

🧠 Memory Trick

Three Polaroids → Max at 45°

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Multiple Polaroids

19 Identify the incorrect statement concerning energy in diffraction:

A Dark fringes represent regions where energy has been destroyed

B Light energy is redistributed from dark regions to bright regions

C The total energy remains constant across the pattern

D The principle of conservation of energy holds true

💡 Short Explanation

�� Energy is conserved. �� Bright and dark regions arise from redistribution. �� No energy is destroyed.

✅ Correct Answer Explanation

Dark fringes occur because waves interfere destructively at certain points. Energy is not destroyed; it is redistributed to bright regions. Therefore Statement A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct description of redistribution.
�� Option C → Total energy remains constant.
�� Option D → Conservation of energy remains valid.

🎯 Strategy

Used

Extreme Word Filter

Application:

Focus on the word "destroyed."

Final Logic:

Interference redistributes energy; it does not destroy it.

🧠 Memory Trick

Dark ≠ Energy Loss

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Energy Conservation in Diffraction

20 Match List I with List II regarding practical uses of polaroids:

List I	List II
1. Sunglasses	a. Controls intensity of light entering a room
2. 3D movie cameras	b. Uses polarisation for depth perception recording
3. Windowpanes	c. Reduces glare and overall intensity outdoors
4. First polaroid sheet	d. Reduces unpolarised light intensity by half

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

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

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

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

💡 Short Explanation

�� Sunglasses reduce glare. �� 3D systems use polarisation. �� First polaroid halves intensity.

✅ Correct Answer Explanation

1 → c : Polaroid sunglasses reduce glare and excessive light. 2 → b : 3D movie cameras use polarisation to record depth information. 3 → a : Polaroid windowpanes help control light entering rooms. 4 → d : The first polaroid reduces unpolarised light intensity to half. Thus the correct matching is 1-c, 2-b, 3-a, 4-d.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly matches all major applications.
�� Option C → Incorrect pairing of sunglasses and camera functions.
�� Option D → Incorrect assignment of first polaroid and sunglasses.

🎯 Strategy

Used

Option Grouping

Application:

Match each application with its specific function.

Final Logic:

Only Option A gives all four correct matches.

🧠 Memory Trick

Sunglasses–Glare, Camera–3D, Window–Control, First–Half

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Applications of Polaroids

CUET UG Physics Booster Test 2 -Diffraction and Polarisation

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