UET UG Physics Booster Test 2 -Refraction and Reflection Laws -

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CUET UG Physics Booster Test 2 -Refraction and Reflection Laws

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2 -Refraction and Reflection Laws

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QUESTION 1 OF 20

If an incident plane wavefront AB approaches boundary surface PP', the distance BC relates to the speed v₁ in medium 1. If τ is the elapsed time, the geometric distance BC is mathematically represented as:

Answer Review

1 If an incident plane wavefront AB approaches boundary surface PP', the distance BC relates to the speed v₁ in medium 1. If τ is the elapsed time, the geometric distance BC is mathematically represented as:

A v₁τ

B v₂τ

C (v₁ - v₂)τ

D (v₁ + v₂)τ

💡 Short Explanation

�� BC is the distance travelled in medium 1. �� Distance = Speed × Time. �� Huygens construction uses BC = v₁τ.

✅ Correct Answer Explanation

In Huygens' construction for refraction, point B travels through medium 1 for a time interval τ. Therefore, the distance covered is: BC = v₁τ where v₁ is the speed of light in medium 1. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Uses speed in medium 2 instead of medium 1.
�� Option C → No such relation exists in Huygens construction.
�� Option D → Sum of speeds is physically irrelevant here.

🎯 Strategy

Used

Substitution

Application:

Apply the basic relation Distance = Speed × Time.

Final Logic:

BC is travelled in medium 1, therefore BC = v₁τ.

🧠 Memory Trick

"BC belongs to medium 1."

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

2 In Huygens' construction for refraction, the angle of incidence i is the angle between the incident wavefront and the boundary surface PP'. Geometrically, angle i is also identical to

A the angle between the normal and the boundary surface.

B the angle between the incident ray and the normal.

C the calculated critical angle of the medium.

D the angle of the resulting refracted wavefront.

💡 Short Explanation

�� Incident ray is perpendicular to the wavefront. �� Normal is perpendicular to the interface. �� Both definitions give the same angle of incidence.

✅ Correct Answer Explanation

The angle of incidence is defined as the angle between the incident ray and the normal to the interface. Since the incident ray is perpendicular to the incident wavefront, the geometrical construction leads to the same angle i. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Always equals 90°, not the incidence angle.
�� Option C → Critical angle is a separate concept.
�� Option D → Refers to the refracted wavefront, not incidence.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the definition of angle of incidence.

Final Logic:

Angle i is measured between incident ray and normal.

🧠 Memory Trick

"Incidence = Ray with Normal."

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

3 In analyzing the trigonometric ratios for refraction using Huygens principle,
1. sin i primarily relies on the tangent plane of the reflected wave.
2. sin r is mathematically calculated as AE/AC.
3. the hypotenuse AC differs fundamentally for both triangles ABC and AEC.
4. AE represents the initial incident wavefront distance.

A 1

B 2

C 3

D 4

💡 Short Explanation

�� sin r = AE/AC. �� AC is common to both triangles. �� AE is distance travelled in medium 2.

✅ Correct Answer Explanation

1. Statement 1 is incorrect because sin i is derived from triangle ABC, not from a reflected wave construction. 2. Statement 2 is correct because: sin r = AE/AC 1. Statement 3 is incorrect because AC is common to both triangles. 2. Statement 4 is incorrect because AE represents the distance travelled in medium 2 during time τ. Therefore only Statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 1 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement from the Huygens geometry.

Final Logic:

Only Statement 2 correctly describes the trigonometric relation.

🧠 Memory Trick

"Refraction ratio: sin r = AE/AC."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Derivation of Snell's Law

4 If the refractive index of medium 1 is 1.0 and medium 2 is 1.5, and the angle of incidence is 30°, what is the numerical value for the sine of the angle of refraction (sin r)?

A 0.33

B 0.50

C 0.75

D 1.50

💡 Short Explanation

�� Apply Snell's law. �� n₁ sin i = n₂ sin r. �� Solve for sin r.

✅ Correct Answer Explanation

Using Snell's law: n₁ sin i = n₂ sin r 1 × sin30° = 1.5 × sin r 0.5 = 1.5 × sin r sin r = 0.5/1.5 = 0.333 ≈ 0.33 Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Equals sin30°, not sin r.
�� Option C → Incorrect calculation.
�� Option D → Sine value cannot exceed 1.

🎯 Strategy

Used

Substitution

Application:

Substitute values into Snell's law.

Final Logic:

sin r = 0.5/1.5 = 0.33.

🧠 Memory Trick

"Snell first, solve later."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Snell's Law

5 Identify the incorrect statement about energy and speed behavior when a light wave enters a denser medium:

A The propagation speed of the wave decreases.

B The physical wavelength of the wave decreases.

C The reduction in speed implies a reduction in the energy carried by the light wave.

D The frequency of the light wave remains perfectly constant.

💡 Short Explanation

�� Frequency remains unchanged. �� Energy depends on frequency. �� Lower speed does not imply lower photon energy.

✅ Correct Answer Explanation

Photon energy is given by: E = hf Since frequency remains unchanged during refraction, photon energy remains unchanged. A decrease in speed or wavelength does not imply a decrease in energy. Therefore Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct; speed decreases in a denser medium.
�� Option B → Correct; wavelength decreases.
�� Option D → Correct; frequency remains constant.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting wave-energy principles.

Final Logic:

Energy depends on frequency, not speed.

🧠 Memory Trick

"Same f → Same E."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction and Frequency Invariance

6 When moving to a rarer medium, the speed of light ___, and the refracted wavefront physically bends ___ the normal.

A Increases, Away from

B Decreases, Towards

C Increases, Towards

D Decreases, Away from

💡 Short Explanation

�� Rarer medium means higher speed. �� Light bends away from the normal. �� Angle of refraction increases.

✅ Correct Answer Explanation

When light enters a rarer medium: v₂ > v₁ The speed increases and the refracted ray bends away from the normal. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Both statements are opposite.
�� Option C → Bending direction is wrong.
�� Option D → Speed does not decrease.

🎯 Strategy

Used

Conceptual Recall

Application:

Apply the denser-to-rarer refraction rule.

Final Logic:

Rarer medium → faster speed → away from normal.

🧠 Memory Trick

"Fast means Far."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction into a Rarer Medium

7 Match the specific condition with its corresponding wave wavelength change.

List I	List II
1. Entering a denser medium (v₂ < v₁)	a. Wavelength decreases
2. Entering a rarer medium (v₂ > v₁)	b. Wavelength increases
3. Reflection within the same medium	c. Wavelength remains exactly the same
4. Total internal reflection occurrence	d. Energy totally stays in medium 1

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

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

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

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

💡 Short Explanation

�� Wavelength follows speed. �� Reflection keeps wavelength unchanged. �� TIR keeps energy in medium 1.

✅ Correct Answer Explanation

1. → a because wavelength decreases in a denser medium. 2. → b because wavelength increases in a rarer medium. 3. → c because reflection occurs in the same medium. 4. → d because total internal reflection keeps energy in medium 1. Therefore: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect wavelength assignments.
�� Option C → Incorrect matching throughout.
�� Option D → Incorrect matching throughout.

🎯 Strategy

Used

Option Grouping

Application:

Associate each condition with its wavelength behavior.

Final Logic:

Each physical condition has a unique wavelength outcome.

🧠 Memory Trick

"Dense-Down, Rare-Up."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Wavelength Changes During Refraction

9 Correct statements applying to the relation v₁ < v₂:
1. Light is entering a rarer medium.
2. The ray bends significantly away from the normal.
3. A critical angle boundary becomes geometrically possible.
4. Wavelength inevitably decreases.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� v₁ < v₂ indicates transition to a rarer medium. �� Light bends away from the normal. �� Total internal reflection becomes possible for light traveling from medium 2 back to medium 1.

✅ Correct Answer Explanation

1. Statement 1 is correct because v₂ > v₁ means the second medium allows light to travel faster, indicating that medium 2 is optically rarer. 2. Statement 2 is correct because when light enters a rarer medium, it bends away from the normal and the angle of refraction becomes greater than the angle of incidence. 3. Statement 3 is correct because when light passes from a denser medium to a rarer medium, a critical angle can exist. Since medium 2 is rarer relative to medium 1, the medium pair supports the possibility of a critical-angle condition. 4. Statement 4 is incorrect because wavelength does not decrease when entering a rarer medium. Since speed increases and frequency remains constant, wavelength increases. Therefore, Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B (2, 3, 4) → Includes incorrect Statement 4.
�� Option C (1, 3, 4) → Includes incorrect Statement 4 and omits correct Statement 2.
�� Option D (1, 2, 4) → Includes incorrect Statement 4 and omits correct Statement 3.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using the relationships among speed, refractive index, wavelength, and critical angle.

Final Logic:

Statements 1, 2 and 3 follow directly from v₂ > v₁, while Statement 4 contradicts wavelength behavior in a rarer medium.

🧠 Memory Trick

"Rare → Fast → Far → Longer λ"

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction into a Rarer Medium and Critical Angle

10 To construct the refracted wavefront at t = τ for a plane wave, a sphere of radius v₂τ is drawn from point A in the second medium. The refracted wavefront itself is then defined as

A a straight line intersecting the center of the sphere.

B the tangent plane drawn from point C onto this sphere.

C a common tangent applied to all spheres in the first medium.

D a spherical envelope propagating directly backward.

💡 Short Explanation

�� Huygens' principle uses secondary wavelets. �� Radius of secondary wavelet = v₂τ. �� The new wavefront is the common tangent.

✅ Correct Answer Explanation

According to Huygens' principle, every point on the incident wavefront acts as a source of secondary wavelets. After a time interval τ, a secondary wavelet of radius v₂τ is drawn from point A in the second medium. The refracted wavefront is obtained by drawing a tangent from point C to this secondary wavelet. This tangent represents the envelope of all secondary wavelets and therefore forms the refracted wavefront. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → The refracted wavefront does not pass through the center of the secondary wavelet.
�� Option C → Refraction is constructed using wavelets in the second medium, not by applying a common tangent to wavelets in the first medium.
�� Option D → The refracted wavefront is not a backward-propagating spherical envelope.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the standard Huygens construction for refraction at an interface.

Final Logic:

The refracted wavefront is the tangent drawn from C to the secondary wavelet of radius v₂τ.

🧠 Memory Trick

"Wavelet First, Tangent Next."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Huygens Construction for Refraction of Plane Wavefronts

8 Identify the statements regarding frequency principles during wave refraction:
1. Frequency is determined solely by the source oscillator.
2. Frequency of scattered light inherently equals incident light frequency.
3. Frequency changes proportionally with wave speed changes.
4. The energy directly depends on varying frequency amplitude.

A 1, 2

B 3, 4

C 1, 3

D 2, 4

💡 Short Explanation

�� Frequency is fixed by the source. �� Refraction does not change frequency. �� Speed and wavelength may change, but frequency remains constant.

✅ Correct Answer Explanation

1. Statement 1 is correct because the frequency of a light wave is determined by the source producing the wave. During refraction, the source remains unchanged, so the frequency remains constant. 2. Statement 2 is correct in the context of ordinary reflection/refraction and coherent wave propagation, where the emerging wave retains the same frequency as the incident wave. 3. Statement 3 is incorrect because frequency does not change when wave speed changes during refraction. Instead, the wavelength changes according to (v=f\lambda). 4. Statement 4 is incorrect because energy depends on frequency ((E=hf)), not on "frequency amplitude." Frequency and amplitude are different physical quantities. Therefore, Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B (3, 4) → Both Statements 3 and 4 are incorrect.
�� Option C (1, 3) → Includes incorrect Statement 3 and omits correct Statement 2.
�� Option D (2, 4) → Includes incorrect Statement 4 and omits correct Statement 1.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using the principles of refraction and wave propagation.

Final Logic:

Statements 1 and 2 agree with frequency invariance, while Statements 3 and 4 contradict fundamental wave concepts.

🧠 Memory Trick

"Medium changes λ and v, never f."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Frequency Invariance During Refraction

11 For total internal reflection to occur, the incident angle i and critical angle ic must satisfy the relation i - ic:

A > 0

B < 0

C = 0

D = 90

💡 Short Explanation

�� Total internal reflection occurs beyond the critical angle. �� Incident angle must exceed critical angle. �� No refracted ray is produced.

✅ Correct Answer Explanation

For total internal reflection (TIR) to occur: i > ic Therefore: i − ic > 0 When the angle of incidence becomes greater than the critical angle, refraction ceases and complete reflection occurs within the denser medium. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents incidence below the critical angle.
�� Option C → Corresponds to i = ic, where refracted ray emerges along the interface.
�� Option D → No such condition exists for TIR.

🎯 Strategy

Used

Elimination

Application:

Recall the fundamental condition for total internal reflection.

Final Logic:

TIR occurs only when i > ic.

🧠 Memory Trick

"Beyond Critical → Total Reflection."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Total Internal Reflection

12 If the speed of light in medium 1 is 2.0 × 10⁸ m s⁻¹ and in medium 2 is 2.5 × 10⁸ m s⁻¹, the theoretical critical angle for total internal reflection is:

A sin⁻¹ (0.8)

B sin⁻¹ (1.25)

C sin⁻¹ (0.5)

D sin⁻¹ (0.64)

💡 Short Explanation

�� Critical angle exists for denser-to-rarer transition. �� sin ic = n₂/n₁. �� Using speeds: sin ic = v₁/v₂.

✅ Correct Answer Explanation

Since n = c/v Therefore, sin ic = n₂/n₁ = (c/v₂)/(c/v₁) = v₁/v₂ = (2.0 × 10⁸)/(2.5 × 10⁸) = 0.8 Thus, ic = sin⁻¹(0.8) Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → sin value cannot exceed 1.
�� Option C → Incorrect calculation.
�� Option D → Not obtained from the given speeds.

🎯 Strategy

Used

Substitution

Application:

Substitute speeds into the critical-angle relation.

Final Logic:

sin ic = 2.0/2.5 = 0.8.

🧠 Memory Trick

"Critical Angle = Slower/Faster Speed."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Critical Angle

13 Choose the incorrect statement regarding the reflection of a plane wave by a reflecting surface MN:

A The time taken from point B to C is the same as from A to E.

B AE represents the radius of the sphere drawn to find the reflected wavefront.

C The speed of the wave v changes drastically after bouncing off MN.

D The incident and reflected wavefronts form perfectly equal angles with MN.

💡 Short Explanation

�� Reflection occurs in the same medium. �� Wave speed remains unchanged. �� Law of reflection remains valid.

✅ Correct Answer Explanation

During reflection, light remains in the same medium. Therefore its speed does not change after reflection. Option C is incorrect because reflection only changes the direction of propagation, not the speed. Hence Option C is the correct answer.

❓ Why Other Options Are Incorrect

�� Option A → Correct; equal time intervals are used in Huygens construction.
�� Option B → Correct; AE = vt is the radius of the secondary wavelet.
�� Option D → Correct; incident and reflected wavefronts make equal angles.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement violating the law of reflection.

Final Logic:

Reflection changes direction, not speed.

🧠 Memory Trick

"Same Medium → Same Speed."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Reflection of Plane Wavefronts

14 In proving the law of reflection using the congruent triangles EAC and BAC,
1. Side AC is unequal across both triangles.
2. Distance AE is inherently greater than BC.
3. Angles EAC and BAC directly prove Snell's law equation.
4. AE = BC = vτ fundamentally proves the triangles are congruent.

A 1

B 2

C 3

D 4

💡 Short Explanation

�� AC is common to both triangles. �� AE = BC = vt. �� Congruency leads to i = r.

✅ Correct Answer Explanation

1. Statement 1 is incorrect because AC is common to both triangles. 2. Statement 2 is incorrect because AE = BC. 3. Statement 3 is incorrect because the derivation proves the law of reflection, not Snell's law. 4. Statement 4 is correct because AE = BC = vτ and AC is common, establishing triangle congruency. Therefore only Statement 4 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 1 is incorrect.
�� Option B → Statement 2 is incorrect.
�� Option C → Statement 3 is incorrect.

🎯 Strategy

Used

Elimination

Application:

Check each statement against the Huygens reflection derivation.

Final Logic:

Only Statement 4 correctly describes the congruency condition.

🧠 Memory Trick

"AE = BC → i = r."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Huygens Derivation of Reflection

15 Match the properties of wavefronts interacting with a thin prism.

List I	List II
1. Lower portion of incident wavefront	a. Tilted plane
2. Upper portion of incident wavefront	b. Travels through the greatest thickness
3. Speed of the wave in the prism	c. Travels through the least thickness
4. Resultant emerging wavefront shape	d. Less than the speed in air

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

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

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

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

💡 Short Explanation

�� Lower portion enters thicker glass. �� Upper portion enters thinner glass. �� Prism produces a tilted wavefront.

✅ Correct Answer Explanation

1. → b because the lower portion traverses maximum glass thickness. 2. → c because the upper portion traverses minimum thickness. 3. → d because light travels slower in glass than in air. 4. → a because the emerging wavefront becomes a tilted plane. Therefore: 1-b, 2-c, 3-d, 4-a

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly assigns wavefront and thickness relations.
�� Option C → Incorrect speed assignment.
�� Option D → Incorrect matching throughout.

🎯 Strategy

Used

Option Grouping

Application:

Associate prism geometry with wavefront behavior.

Final Logic:

Each prism property has a unique physical outcome.

🧠 Memory Trick

"More Glass → More Delay → Tilt."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Refraction Through a Thin Prism

16 In a thin convex lens, the central part of the plane wave traverses the ___ portion and is delayed the ___.

A Thickest, Most

B Thinnest, Least

C Thickest, Least

D Thinnest, Most

💡 Short Explanation

�� Convex lens is thickest at the center. �� Light travels slower in glass. �� Central portion experiences maximum delay.

✅ Correct Answer Explanation

A convex lens has maximum thickness at the center. The central portion of the wavefront therefore travels through the greatest amount of glass. Since speed in glass is lower than in air, this region is delayed the most. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Center is not thinnest.
�� Option C → Thickest portion cannot produce least delay.
�� Option D → Center is not thinnest.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall convex lens geometry.

Final Logic:

Maximum glass thickness gives maximum delay.

🧠 Memory Trick

"Thick Center → More Delay."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Wavefront Through a Convex Lens

17 When a plane wave undergoes reflection from a standard concave mirror, the resulting spherical wave

A diverges away from the geometric center of curvature.

B converges to the precise focal point F.

C travels perfectly parallel to the principal axis.

D transforms into an unpolarized physical state.

💡 Short Explanation

�� Concave mirror is converging. �� Plane wave contains parallel rays. �� Reflected wavefront converges to focus.

✅ Correct Answer Explanation

A plane wave incident on a concave mirror consists of parallel rays. After reflection, these rays meet at the focal point F. Thus the reflected wavefront becomes a converging spherical wavefront centered at F. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes divergence, not convergence.
�� Option C → Reflection changes propagation direction.
�� Option D → Polarization is unrelated here.

🎯 Strategy

Used

Conceptual Recall

Application:

Use the focusing property of a concave mirror.

Final Logic:

Parallel rays converge at focus F.

🧠 Memory Trick

"Concave Collects."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Reflection from a Concave Mirror

18 Choose the correct statements about spherical wavefronts:
1. Emanates uniformly from a point source.
2. Represents the locus of points vibrating in the same phase.
3. Formed directly after a plane wave reflects from a concave mirror.
4. Randomly bends away from normal at any denser medium.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Point sources produce spherical wavefronts. �� Wavefronts join points in equal phase. �� Concave mirrors can generate converging spherical wavefronts.

✅ Correct Answer Explanation

1. Statement 1 is correct because a point source emits spherical waves. 2. Statement 2 is correct because a wavefront is the locus of points vibrating in the same phase. 3. Statement 3 is correct because reflection from a concave mirror converts a plane wave into a converging spherical wavefront. 4. Statement 4 is incorrect because bending depends on refractive indices and direction of propagation, not random behavior. 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:

Evaluate each statement using wavefront definitions.

Final Logic:

Only Statement 4 contradicts wave optics principles.

🧠 Memory Trick

"Point Source → Sphere."

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

19 Identify the correct statements about optical paths passing through a convex lens:
1. Total time taken from object to image is definitively constant along any ray.
2. Path passing right through the center is geometrically shorter.
3. Rays traveling near the edge traverse mostly through air.
4. Short path through the center is compensated by a proportionally faster speed.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Equal optical time principle applies. �� Central path is geometrically shorter. �� Edge rays travel more through air and less through glass.

✅ Correct Answer Explanation

1. Statement 1 is correct because all rays reaching the image take the same total optical time. 2. Statement 2 is correct because the central ray follows the shortest geometric path. 3. Statement 3 is correct because marginal rays spend a greater fraction of their path in air than the central ray, which passes through the thickest glass region. 4. Statement 4 is incorrect because the compensation occurs due to slower speed in glass, not faster speed. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 4 and omits Statement 1.
�� Option C → Includes incorrect Statement 4 and omits Statement 2.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

Check each statement against the equal-time principle.

Final Logic:

Statement 4 contradicts the slower speed of light in glass.

🧠 Memory Trick

"Short Path, Slow Glass, Same Time."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Equal Time Principle and Convex Lens

20 The slower speed of light passing through a glass lens:

A forces all rays to travel geometrically longer macroscopic paths.

B causes the central ray to unequivocally take less time than edge rays.

C compensates for the geometrically shorter physical path through its center.

D inherently leads to total internal reflection within the optical lens.

💡 Short Explanation

�� Central path is geometrically shorter. �� Light is slower in glass. �� Equal optical time is maintained.

✅ Correct Answer Explanation

The central ray passes through the thickest part of the lens and therefore spends more time in glass, where light travels more slowly. This compensates for the geometrically shorter path near the center. As a result, rays from different regions reach the image simultaneously. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not all rays travel longer geometric paths.
�� Option B → Equal-time principle prevents this conclusion.
�� Option D → Slower speed alone does not cause total internal reflection.

🎯 Strategy

Used

Conceptual Recall

Application:

Apply the equal optical path-time principle.

Final Logic:

Slower speed in glass compensates for the shorter central path.

🧠 Memory Trick

"Slow Glass Balances Short Path."

📚 Class XII – Physics Part II – Chapter: Wave Optics – Topic: Convex Lens and Equal Optical Time Principle

CUET UG Physics Booster Test 2 -Refraction and Reflection Laws

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