UET UG Physics Booster Test 2 -Huygens Principle and Wavefronts -

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CUET UG Physics Booster Test 2 -Huygens Principle and Wavefronts

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2 -Huygens Principle and Wavefronts

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QUESTION 1 OF 20

If an electromagnetic wave has a frequency of 5.0 × 10¹⁴ Hz and travels at c = 3.0 × 10⁸ m s⁻¹, what is the spatial separation (wavelength) between two constant phase surfaces that differ in phase by 2π radians?

Answer Review

1 If an electromagnetic wave has a frequency of 5.0 × 10¹⁴ Hz and travels at c = 3.0 × 10⁸ m s⁻¹, what is the spatial separation (wavelength) between two constant phase surfaces that differ in phase by 2π radians?

A 6.0 × 10⁻⁷ m

B 1.5 × 10²³ m

C 1.6 × 10⁶ m

D 6.0 × 10⁻⁹ m

💡 Short Explanation

�� Wavelength λ = c/f �� Phase difference of 2π corresponds to one wavelength �� Direct substitution gives the answer

✅ Correct Answer Explanation

Given: c = 3.0 × 10⁸ m s⁻¹ f = 5.0 × 10¹⁴ Hz λ = c/f = (3.0 × 10⁸)/(5.0 × 10¹⁴) = 6.0 × 10⁻⁷ m A phase difference of 2π corresponds to one complete cycle, i.e., one wavelength. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect exponent calculation.
�� Option C → Unrealistically large wavelength.
�� Option D → Ten times smaller than the correct value.

🎯 Strategy

Used

Substitution

Application:

�� Apply λ = c/f directly.

Final Logic:

�� λ = 6.0 × 10⁻⁷ m.

🧠 Memory Trick

One Cycle = One λ

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Wavefront and Wavelength

2 Choose the correct statement regarding point source oscillations acting as coherent sources
1. maintain a constant phase difference with respect to each other
2. produce an interference pattern that does not change rapidly with time
3. generate unpolarised waves that travel in single directions
4. can be generated by two needles oscillating periodically in a water trough

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Coherent sources have constant phase difference. �� Stable interference requires coherence. �� Water trough needles are a standard example.

✅ Correct Answer Explanation

Statement 1 is correct because coherent sources maintain a constant phase difference. Statement 2 is correct because a stable interference pattern requires coherence. Statement 3 is incorrect because coherence is not defined by unpolarised waves traveling in single directions. Statement 4 is correct because two periodically oscillating needles in a water trough can act as coherent sources. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Remove the statement unrelated to coherence.

Final Logic:

�� Coherence requires constant phase difference.

🧠 Memory Trick

Coherent = Constant Phase

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

3 When a point source emits light in a homogeneous isotropic medium, the energy spreads out in ___ directions, resulting in wavefront loci that are ___.

A All, Spherical

B Single, Planar

C Opposite, Cylindrical

D Random, Unpredictable

💡 Short Explanation

�� Point source emits uniformly. �� Isotropic medium has identical properties in all directions. �� Wavefront becomes spherical.

✅ Correct Answer Explanation

In a homogeneous isotropic medium, a point source emits energy equally in all directions. Therefore the wavefront consists of spherical surfaces centered on the source. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Point sources do not initially produce plane waves.
�� Option C → Cylindrical waves require line sources.
�� Option D → Propagation is predictable.

🎯 Strategy

Used

Conceptual Recall

Application:

�� Recall wavefront shapes for different sources.

Final Logic:

�� Point source → Spherical wavefront.

🧠 Memory Trick

Point → Sphere

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

4 Match List I with List II

List I	List II
1. Locus of points in same phase (point source)	a. Spherical wavefront
2. Path of energy propagation	b. Ray perpendicular to wavefront
3. Wavelength λ	c. Distance between consecutive identical phase wavefronts
4. Point source behavior at large distance	d. Approximates plane wave

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

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

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

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

💡 Short Explanation

�� Point source gives spherical waves. �� Rays show energy propagation. �� Far spherical waves approximate plane waves.

✅ Correct Answer Explanation

1 → a : Same-phase locus from a point source is a spherical wavefront. 2 → b : Energy propagates along rays perpendicular to wavefronts. 3 → c : Wavelength is the distance between successive identical phase wavefronts. 4 → d : At large distances, spherical waves approximate plane waves. Therefore 1-a, 2-b, 3-c, 4-d.

❓ Why Other Options Are Incorrect

�� Option B → Multiple incorrect pairings.
�� Option C → Incorrect matching.
�� Option D → Incorrect source-wavefront relation.

🎯 Strategy

Used

Option Grouping

Application:

�� Match definitions systematically.

Final Logic:

�� Only Option A satisfies all pairings.

🧠 Memory Trick

Source→Sphere, Far→Plane

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Wavefronts and Rays

5 Choose the correct statements regarding large distance effects:
1. A small portion of a spherical wave can be treated as a plane wave at large distances
2. The rays associated with this small portion become practically parallel
3. The radius of curvature of the wavefront approaches zero at large distances
4. The amplitude of the wave perfectly vanishes at large distances

A 1, 2

B 2, 3

C 3, 4

D 1, 4

💡 Short Explanation

�� Large radius gives nearly flat wavefronts. �� Rays become parallel. �� Radius of curvature becomes very large.

✅ Correct Answer Explanation

Statement 1 is correct because a distant spherical wavefront appears plane locally. Statement 2 is correct because rays become nearly parallel. Statement 3 is incorrect because radius of curvature increases, not decreases to zero. Statement 4 is incorrect because amplitude decreases but does not become exactly zero. Therefore Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statements 3 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Remove physically incorrect statements.

Final Logic:

�� Large distance ⇒ Large radius of curvature.

🧠 Memory Trick

Far Away = Flat Wave

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

6 Choose the incorrect statement about treating a small sphere portion as a plane wave

A The approximation becomes more valid as the radial distance from the source increases

B It assumes the wavefront curvature is negligibly small over that localized region

C The geometric rays in this small portion are assumed to be converging precisely to a focal point

D It is a useful geometric simplification for light arriving from distant stars

💡 Short Explanation

�� Plane-wave rays are parallel. �� Distant sources produce nearly plane wavefronts. �� No convergence is assumed.

✅ Correct Answer Explanation

In the plane-wave approximation, the wavefront is treated as flat and rays are considered parallel. Converging rays correspond to a converging spherical wavefront, not a plane wave. Therefore Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct statement.
�� Option D → Correct astronomical application.

🎯 Strategy

Used

Elimination

Application:

�� Recall the geometry of plane waves.

Final Logic:

�� Plane wave ⇒ Parallel rays.

🧠 Memory Trick

Plane = Parallel

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

7 Choose the correct statements about the primary wavefront position at t = 0
1. It represents the boundary of the wave disturbance at the initial time reference
2. It acts as the geometric locus of all secondary wavelet sources
3. It eliminates the need for geometric tangent drawing to find future positions
4. Its known shape is the necessary starting condition for Huygens' principle

A 1, 2, 4

B 2, 3, 4

C 1, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Initial wavefront is known. �� Every point becomes a secondary source. �� Future wavefronts are derived from it.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are correct. Statement 3 is incorrect because geometric tangent construction is an essential part of Huygens' principle. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement contradicting Huygens' construction.

Final Logic:

�� Tangent drawing is necessary.

🧠 Memory Trick

Known Front → New Front

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

8 If the speed of a wave in a medium is v, the spatial distance a secondary wavelet travels to determine the later time envelope position at t = τ is dynamically given by:

A vτ

B v/τ

C τ/v

D v + τ

💡 Short Explanation

�� Distance = Speed × Time �� Secondary wavelet radius equals vτ �� Used in Huygens construction

✅ Correct Answer Explanation

The radius of every secondary wavelet after time τ is: r = vτ This radius is used to draw the envelope that forms the new wavefront. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Wrong dimensional form.
�� Option C → Wrong dimensional form.
�� Option D → Physically meaningless.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Distance must have units of length.

Final Logic:

�� Only vτ gives length.

🧠 Memory Trick

Radius = vτ

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

9 In the geometrical derivation of refraction using Huygens principle, as the incident plane wavefront strikes the boundary

A the secondary disturbance vanishes completely if it travels into a denser medium

B the continuous boundary stops emitting wavelets to conserve total incident energy

C the entire boundary reflects light perfectly irrespective of the transparent mediums

D each point on the interface between the two media successively becomes a source of secondary wavelets in the second medium

💡 Short Explanation

�� Interface points become secondary sources. �� New wavelets form refracted wavefronts. �� Basis of Huygens' derivation of refraction.

✅ Correct Answer Explanation

According to Huygens' principle, every point on the interface acts as a source of secondary wavelets in the second medium. The envelope of these wavelets forms the refracted wavefront and leads to Snell's law. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Secondary disturbances continue.
�� Option B → Boundary does not stop emitting wavelets.
�� Option C → Complete reflection does not always occur.

🎯 Strategy

Used

Conceptual Recall

Application:

�� Recall Huygens' explanation of refraction.

Final Logic:

�� Interface points become secondary sources.

🧠 Memory Trick

Boundary = New Sources

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction by Huygens Principle

10 A plane wave is refracted from vacuum (c = 3.0 × 10⁸ m s⁻¹) into a denser medium. If the refractive index of the medium is 1.5, what is the distance travelled by the secondary wavelet in the medium in 2.0 × 10⁻¹⁰ s?

A 4.0 × 10⁻² m

B 6.0 × 10⁻² m

C 9.0 × 10⁻² m

D 1.0 × 10⁻² m

💡 Short Explanation

�� Speed in medium = c/n �� Distance = vt �� Apply direct substitution

✅ Correct Answer Explanation

Given: n = 1.5 c = 3.0 × 10⁸ m s⁻¹ Speed in medium: v = c/n = (3.0 × 10⁸)/1.5 = 2.0 × 10⁸ m s⁻¹ Distance travelled: d = vt = (2.0 × 10⁸)(2.0 × 10⁻¹⁰) = 4.0 × 10⁻² m Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Uses vacuum speed directly.
�� Option C → Calculation error.
�� Option D → Incorrect power of ten.

🎯 Strategy

Used

Substitution

Application:

�� First calculate speed, then distance.

Final Logic:

�� d = (c/n)t = 4.0 × 10⁻² m.

🧠 Memory Trick

c/n → Then vt

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction and Secondary Wavelets

11 Choose the correct statement regarding the common tangent drawn to secondary spheres
1. represents the continuous envelope of the secondary wavelets
2. defines the exact new wavefront position at time t
3. touches the furthest points reached by the secondary wavelets in the forward direction
4. represents the ray path traversing perpendicular to the envelope

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Envelope forms the new wavefront. �� Common tangent determines future position. �� Rays are perpendicular to the envelope, not the envelope itself.

✅ Correct Answer Explanation

Statement 1 is correct because the common tangent acts as the envelope of all secondary wavelets. Statement 2 is correct because this envelope gives the new wavefront position after time t. Statement 3 is correct because the tangent touches the forward secondary wavelets. Statement 4 is incorrect because the tangent itself is not a ray path; rays are drawn perpendicular to the wavefront. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 4.
�� Option C → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement confusing wavefronts and rays.

Final Logic:

�� Rays are perpendicular to the envelope; the envelope is not a ray.

🧠 Memory Trick

Envelope Gives Front.

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

12 According to Huygens' construction method, the new wavefront is exclusively the ___ envelope, completely ignoring the physically non-existent ___ wave.

A Forward, Backward

B Backward, Forward

C Tangential, Normal

D Normal, Tangential

💡 Short Explanation

�� New wavefront comes from forward wavelets. �� Backwave is ignored. �� This was Huygens' ad-hoc assumption.

✅ Correct Answer Explanation

In Huygens' construction, only the forward envelope of secondary wavelets is considered. The backward wave is physically absent and was originally excluded through an ad-hoc assumption. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the correct order.
�� Option C → Not related to wavefront construction.
�� Option D → Incorrect terminology.

🎯 Strategy

Used

Conceptual Recall

Application:

�� Recall the forward-envelope construction.

Final Logic:

�� Forward envelope forms the new wavefront.

🧠 Memory Trick

Forward Forms, Backward Fades.

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

13 Match List I with List II

List I	List II
1. Ad-hoc assumption	a. Proposed by Huygens initially
2. Forward direction amplitude	b. Maximum
3. Backward direction amplitude	c. Zero
4. Phenomenon explained	d. Absence of backwave

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

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

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

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

💡 Short Explanation

�� Huygens proposed the assumption. �� Forward amplitude is maximum. �� Backward amplitude is zero.

✅ Correct Answer Explanation

1 → a because the ad-hoc assumption was proposed by Huygens. 2 → b because amplitude is maximum in the forward direction. 3 → c because amplitude is assumed zero in the backward direction. 4 → d because the assumption explains the absence of the backwave. Therefore: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Multiple incorrect matches.
�� Option C → Incorrect amplitude assignments.
�� Option D → Incorrect pairings.

🎯 Strategy

Used

Option Grouping

Application:

�� Match each concept with its associated description.

Final Logic:

�� Only Option A satisfies all pairings.

🧠 Memory Trick

Forward Max, Back Zero.

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

14 Choose the correct statements regarding rigorous theory justification:
1. Huygens' initial ad-hoc assumption regarding backwaves is mathematically unsatisfactory
2. The absence of the backwave is rigorously justified by proper wave theory (Maxwell's equations)
3. The backwave theoretically travels at twice the relative speed of light
4. The rigorous theory relies solely on traditional Newtonian corpuscular mechanics

A 1, 2

B 2, 3

C 3, 4

D 1, 4

💡 Short Explanation

�� Initial assumption lacked rigorous proof. �� Maxwell's theory gives proper justification. �� Corpuscular mechanics is not involved.

✅ Correct Answer Explanation

Statement 1 is correct because Huygens introduced the absence of the backwave as an ad-hoc assumption. Statement 2 is correct because modern electromagnetic wave theory rigorously explains this result. Statement 3 is incorrect because no such speed relation exists. Statement 4 is incorrect because the explanation comes from wave theory, not Newtonian corpuscular mechanics. Therefore Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statements 3 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Remove physically unsupported statements.

Final Logic:

�� Only Statements 1 and 2 are valid.

🧠 Memory Trick

Assumed by Huygens, Proved by Maxwell.

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

15 Identify the incorrect statement about isotropic medium propagation during refraction

A When a wave gets refracted into a denser medium, its speed of propagation decreases

B The absolute wavelength decreases when refracting into a denser medium

C The frequency of the propagating wave changes based on the medium's refractive index

D Snell's law of refraction can be completely derived using Huygens wave principle

💡 Short Explanation

�� Frequency remains constant. �� Speed and wavelength change. �� Huygens' principle derives Snell's law.

✅ Correct Answer Explanation

During refraction, the frequency of light remains unchanged because it is determined by the source. When light enters a denser medium: Speed decreases. Wavelength decreases. Frequency remains constant. Therefore Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct statement.
�� Option D → Correct application of Huygens' principle.

🎯 Strategy

Used

Conceptual Recall

Application:

�� Recall which quantity remains unchanged during refraction.

Final Logic:

�� Frequency remains constant.

🧠 Memory Trick

Source Fixes Frequency.

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction of Light

16 Choose the correct statements about ray path perpendicularity
Statements:
1. A ray is conceptually defined as the path of energy propagation in the wave limit
2. The ray is always geometrically perpendicular to the wavefront in an isotropic medium
3. The total time taken from a point on the object to the image is the same measured along any geometric ray
4. Rays aggressively bend towards the normal when entering a medium where the wave travels inherently faster

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Rays indicate energy flow. �� Rays are perpendicular to wavefronts. �� Faster medium causes bending away from normal.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. Statement 4 is incorrect because when light enters a medium where speed increases, it bends away from the normal. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement 4.
�� Option C → Includes Statement 4.
�� Option D → Includes Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect refraction statement.

Final Logic:

�� Faster medium ⇒ Bend away from normal.

🧠 Memory Trick

Faster → Away.

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Rays and Wavefronts

17 When an incoming plane wave is geometrically incident on a thin convex lens, the delayed central part alters the emerging wavefront into a:

A Diverging spherical wave

B Converging spherical wave

C Perfectly identical plane wave

D Randomly scattered wave

💡 Short Explanation

�� Convex lens converges light. �� Plane waves become converging spherical waves. �� Focus is formed at the focal point.

✅ Correct Answer Explanation

A convex lens causes parallel rays from a plane wavefront to converge toward its focus. Hence the emerging wavefront becomes a converging spherical wavefront. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Produced by a concave lens.
�� Option C → Lens changes wavefront curvature.
�� Option D → No random scattering occurs.

🎯 Strategy

Used

Conceptual Recall

Application:

�� Recall wavefront transformation by a convex lens.

Final Logic:

�� Convex lens converts plane wavefronts into converging spherical wavefronts.

🧠 Memory Trick

Convex = Converge.

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Wavefront Transformation by Lens

18 To construct the refracted wavefront geometry for a wave passing from medium 1 (speed v₁) to medium 2 (speed v₂) in time interval τ

A the required radius of the secondary sphere in medium 2 is dynamically drawn as v₂τ

B the required radius is strictly taken as v₁τ in medium 2 regardless of boundary crossing

C the radius mathematically relies on the independent distance the backwave travelled

D the sphere is always drawn with a constant radius independent of time τ

💡 Short Explanation

�� Radius depends on speed in that medium. �� Huygens construction uses vτ. �� Medium 2 requires v₂τ.

✅ Correct Answer Explanation

In medium 2, secondary wavelets propagate with speed v₂. After time τ, the radius becomes: r = v₂τ This is used to construct the refracted wavefront. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Uses incorrect speed.
�� Option C → Backwave is irrelevant.
�� Option D → Radius depends on time.

🎯 Strategy

Used

Substitution

Application:

�� Apply r = vτ in medium 2.

Final Logic:

�� Radius = v₂τ.

🧠 Memory Trick

New Medium → New v.

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction by Huygens Principle

19 A plane wave travelling uniformly at 2.0 × 10⁸ m s⁻¹ has a wavefront position x = 0 at the zero time reference (t = 0). What is the geometric position of the constant phase surface at t = 2.5 × 10⁻⁸ s?

A 5.0 m

B 0.5 m

C 50.0 m

D 0.05 m

💡 Short Explanation

�� Position = vt �� Use direct substitution �� Wavefront moves with wave speed

✅ Correct Answer Explanation

Given: v = 2.0 × 10⁸ m s⁻¹ t = 2.5 × 10⁻⁸ s x = vt = (2.0 × 10⁸)(2.5 × 10⁻⁸) = 5.0 m Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Calculation error.
�� Option C → Incorrect exponent handling.
�� Option D → Decimal error.

🎯 Strategy

Used

Substitution

Application:

�� Apply x = vt.

Final Logic:

�� x = 5.0 m.

🧠 Memory Trick

Position = Speed × Time.

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

20 Future wavefront calculation for refraction into a rarer medium (where v₂ > v₁)
Statements:
1. The radius calculation for the secondary sphere is exactly v₂τ
2. The calculated angle of refraction will be greater than the original angle of incidence
3. The temporal envelope of the refracted plane wave bends away from the geometric normal
4. The refracted wavefront successfully transmits even if incident angle i > i₍c₎ (critical angle)

A 1, 2, 3

B 1, 2, 3, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Speed increases in rarer medium. �� Refraction bends away from normal. �� Total internal reflection occurs beyond critical angle.

✅ Correct Answer Explanation

Statement 1 is correct because radius = v₂τ. Statement 2 is correct because entering a rarer medium increases the angle of refraction. Statement 3 is correct because the refracted wave bends away from the normal. Statement 4 is incorrect because for i > ic, total internal reflection occurs and refraction does not take place. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 4.
�� Option C → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement violating total internal reflection.

Final Logic:

�� i > ic ⇒ No refracted ray.

🧠 Memory Trick

Beyond ic → TIR.

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction and Huygens Principle

CUET UG Physics Booster Test 2 -Huygens Principle and Wavefronts

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