CUET UG Physics Booster Test 2-Spherical Refraction and Lenses -

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CUET UG Physics Booster Test 2-Spherical Refraction and Lenses

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-Spherical Refraction and Lenses

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QUESTION 1 OF 20

Identify the incorrect statement about spherical interfaces

Answer Review

1 Identify the incorrect statement about spherical interfaces

A An infinitesimal part of a spherical surface can be regarded as planar.

B The same laws of refraction apply at every point on the spherical surface.

C The normal at the point of incidence is parallel to the tangent plane.

D The normal passes through the centre of curvature.

💡 Short Explanation

�� Normal is perpendicular to the tangent plane. �� Radius acts as the normal. �� Normal passes through the centre of curvature.

✅ Correct Answer Explanation

For a spherical surface, the normal at any point is along the radius joining that point to the centre of curvature. Since the normal is perpendicular to the tangent plane, it cannot be parallel to it. Therefore Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct approximation used in derivations.
�� Option B → Laws of refraction apply at every point.
�� Option D → Radius passes through the centre of curvature.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting the definition of a normal.

Final Logic:

Normal ⟂ Tangent Plane, not ∥ Tangent Plane.

🧠 Memory Trick

"Normal = 90° to Tangent."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Refraction at Spherical Surfaces

2 Choose the correct statements about refraction at a spherical surface separating media of refractive indices n₁ and n₂
1. The formula n₂/v − n₁/u = (n₂ − n₁)/R uses the Cartesian sign convention.
2. u, v, and R represent the magnitudes of distances before sign convention is applied in the derivation.
3. It assumes the aperture of the surface is small compared to other distances.
4. It is only valid when n₂ is greater than n₁.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Uses Cartesian sign convention. �� Derived using paraxial approximation. �� Valid for any pair of refractive indices.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. The spherical surface formula is derived using magnitudes initially and later applying Cartesian signs. The derivation assumes a small aperture and paraxial rays. Statement 4 is incorrect because the formula is valid for both (n_2>n_1) and (n_2<n_1).

❓ 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:

Check the validity conditions of the spherical refraction formula.

Final Logic:

Only Statement 4 is incorrect.

🧠 Memory Trick

"Paraxial + Small Aperture."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Refraction at Spherical Surface

3 In a thin lens approximation, the distance between the points B and D (poles of the two spherical surfaces) is considered to be negligible, allowing them to be assumed to coincide with the

A centre of curvature

B principal focus

C optical centre

D focal plane

💡 Short Explanation

�� Thin lens thickness is negligible. �� Both poles are assumed to coincide. �� Coinciding point is the optical centre.

✅ Correct Answer Explanation

For a thin lens, the separation between the two refracting surfaces is very small compared to other distances. Hence the poles B and D are assumed to coincide at the optical centre O. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Centre of curvature is different.
�� Option B → Focus is not the coincident point.
�� Option D → Focal plane is a plane, not a point.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall assumptions of thin lens approximation.

Final Logic:

Negligible thickness ⇒ poles coincide at optical centre.

🧠 Memory Trick

"Thin Lens → One Centre."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Thin Lens Approximation

4 When defining the focal properties of a thin lens,

A the optical centre is ignored for measuring object distances

B all distances are measured from the optical centre of the lens

C the principal axis is perpendicular to the line joining the optical centre and the principal focus

D the optical centre shifts depending on the surrounding medium

💡 Short Explanation

�� Optical centre is the reference point. �� Distances use Cartesian convention. �� Focal length is measured from optical centre.

✅ Correct Answer Explanation

In thin lens analysis, object distance, image distance and focal length are measured from the optical centre. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Optical centre is the reference point.
�� Option C → Principal focus lies on the principal axis.
�� Option D → Optical centre remains fixed.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall the measurement convention used for lenses.

Final Logic:

All lens distances are measured from O.

🧠 Memory Trick

"Lens Distances Start at O."

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

5 Light from a point source in air (n₁ = 1) falls on a convex spherical glass surface (n₂ = 1.5) of radius of curvature 20 cm. If the source is 100 cm away from the surface, what is the position of the image formed by this single surface?

A +100 cm

B +50 cm

C -100 cm

D +20 cm

💡 Short Explanation

�� Use spherical refraction formula. �� Apply Cartesian signs. �� Solve for image distance.

✅ Correct Answer Explanation

Using n₂/v − n₁/u = (n₂ − n₁)/R Given: n₁ = 1, n₂ = 1.5 u = −100 cm, R = +20 cm Substituting: 1.5/v − 1/(−100) = 0.5/20 1.5/v = 0.025 − 0.01 1.5/v = 0.015 v = 1.5/0.015 = 100 cm Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect calculation.
�� Option C → Wrong sign.
�� Option D → Does not satisfy the formula.

🎯 Strategy

Used

Substitution

Application:

Apply spherical refraction formula directly.

Final Logic:

v = +100 cm

🧠 Memory Trick

"Refraction Surface → Use n₂/v − n₁/u."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Refraction at Spherical Surface

6 For the second surface of a thin double convex lens
1. The first image I₁ acts as a real object.
2. The medium on the right side of the surface has refractive index n₁.
3. The distance DI₁ is taken as positive in the derivation.
4. The radius of curvature is taken as positive.

A 1 and 2

B 2 only

C 2 and 4

D 1, 2, 3, 4

💡 Short Explanation

�� I₁ acts as a virtual object. �� Right side medium is air (n₁). �� Sign conventions must be applied carefully.

✅ Correct Answer Explanation

For the second refracting surface of a thin double convex lens: • I₁ acts as a virtual object. • The medium outside the lens on the right side has refractive index (n_1). • (DI_1) is negative under sign convention. • Radius of curvature for the second surface is negative. Hence only Statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Statement 1 → I₁ is a virtual object.
�� Statement 3 → DI₁ is not positive.
�� Statement 4 → (R_2) is negative.

🎯 Strategy

Used

Elimination

Application:

Apply sign convention for the second surface.

Final Logic:

Only Statement 2 survives.

🧠 Memory Trick

"Second Surface → Virtual Object."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Lens Maker's Formula Derivation

7 For a double convex lens, R₁ is _______ and R₂ is _______ according to the Cartesian sign convention.

A Positive, Positive

B Negative, Positive

C Positive, Negative

D Negative, Negative

💡 Short Explanation

�� First surface centre lies right of pole. �� Second surface centre lies left of pole. �� Cartesian signs apply.

✅ Correct Answer Explanation

For a double convex lens: R₁ > 0 R₂ < 0 Hence the correct combination is Positive, Negative.

❓ Why Other Options Are Incorrect

�� Option A → R₂ is not positive.
�� Option B → R₁ is not negative.
�� Option D → Both signs incorrect.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall radius sign convention for lenses.

Final Logic:

Double convex lens → (+R_1,-R_2).

🧠 Memory Trick

"Convex Lens: + , −."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Radius Sign Convention

8 Match List I with List II for the Lens Maker's Formula components

List I	List II
1. (n₂₁ − 1)	a. Refractive index factor
2. (1/R₁ − 1/R₂)	b. Depends on the radii of curvature
3. Focal length f	c. Positive for a converging lens
4. Refractive index n₂₁	d. Ratio of n₂ to n₁

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

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

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

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

💡 Short Explanation

�� Lens maker's formula contains two factors. �� One depends on refractive index. �� Other depends on radii.

✅ Correct Answer Explanation

1 → a: Refractive index factor 2 → b: Curvature factor 3 → c: Positive for converging lens 4 → d: n₂₁ = n₂/n₁ Therefore: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching.
�� Option C → Incorrect matching.
�� Option D → Incorrect matching.

🎯 Strategy

Used

Option Grouping

Application:

Match formula components first.

Final Logic:

Only Option A gives all correct correspondences.

🧠 Memory Trick

"Index Factor × Curvature Factor."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Lens Maker's Formula

9 Consider the thin lens formula derivation statements. Choose the correct statements:
1. It applies the formula for a single spherical surface successively.
2. It assumes the lens is a thick transparent optical medium.
3. The object is placed at infinity to define the focus.
4. It adds equations for the first and second interfaces.

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Derived from two spherical surfaces. �� Thin lens approximation is used. �� Focus defined using object at infinity.

✅ Correct Answer Explanation

Statements 1, 3 and 4 are correct. The thin lens formula is obtained by applying the spherical refraction formula at each surface and adding the resulting equations. Statement 2 is incorrect because the derivation assumes a thin lens, not a thick lens.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 2.
�� Option C → Includes incorrect Statement 2.
�� Option D → Includes incorrect Statement 2.

🎯 Strategy

Used

Elimination

Application:

Identify the assumption used in the derivation.

Final Logic:

Thin lens approximation makes Statement 2 false.

🧠 Memory Trick

"Two Surfaces → Add Equations."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Thin Lens Formula Derivation

10 Identify thencorrect statement about distance measurements for lenses

A BO = -u is applied in the thin lens approximation.

B DI = +v is applied for real images formed to the right of the lens.

C The ratio v/u gives the linear magnification.

D Distances measured opposite to incident light are taken as positive.

💡 Short Explanation

�� Cartesian sign convention is used. �� Opposite direction is negative. �� Real image distances are positive.

✅ Correct Answer Explanation

According to the Cartesian sign convention: • Direction of incident light → Positive • Opposite direction → Negative Therefore Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct sign usage.
�� Option B → Correct for real images.
�� Option C → Correct lens magnification relation.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall Cartesian sign convention.

Final Logic:

Opposite to incident light ⇒ negative, not positive.

🧠 Memory Trick

"With Light = +, Against Light = −."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Cartesian Sign Convention for Lenses

11 Choose the correct statements about the principal focus of a lens
1. The focus on the side of the original source of light is called the first focal point.
2. The point where image of an object at infinity is formed is the second focal point.
3. A concave lens appears to diverge parallel rays from the first focal point.
4. A lens has only one principal focus.

A 1, 2, 3

B 1, 2

C 2, 3, 4

D 1, 3

💡 Short Explanation

�� A lens has two principal foci. �� First focus is on the object side. �� Second focus is on the image side.

✅ Correct Answer Explanation

Statement 1 is correct because the focus on the side from which light originates is called the first principal focus. Statement 2 is correct because parallel rays from an object at infinity converge (or appear to diverge) at the second principal focus. Statement 3 is correct because a concave lens causes parallel rays to diverge as if they originated from the first principal focus. Statement 4 is incorrect because every thin lens has two principal foci. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits Statement 3, which is correct.
�� Option C → Includes Statement 4, which is incorrect.
�� Option D → Omits Statement 2, which is correct.

🎯 Strategy

Used

Elimination

Application:

Identify the incorrect statement first. Since a lens has two principal foci, Statement 4 is false.

Final Logic:

Only Statements 1, 2 and 3 are correct.

🧠 Memory Trick

"Lens → Two Foci, Not One."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Principal Foci of Lenses

12 For a thin double convex lens placed in air, the first focal point and second focal point are equidistant from the optical centre. This symmetry fundamentally arises because

A the lens has a constant thickness everywhere

B the light rays always pass un-deviated

C the surrounding medium is the same on both sides

D the magnification is always equal to 1

💡 Short Explanation

�� Symmetry depends on surrounding medium. �� Air exists on both sides. �� Hence focal lengths are equal in magnitude.

✅ Correct Answer Explanation

For a thin lens placed in the same medium on both sides (typically air), the optical behaviour is symmetrical. Therefore: OF = OF′ This makes the first and second focal points equidistant from the optical centre. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Constant thickness is not the reason for focal symmetry.
�� Option B → Only rays through the optical centre emerge undeviated.
�� Option D → Magnification depends on object position.

🎯 Strategy

Used

Conceptual Elimination

Application:

Check which property directly produces focal symmetry.

Final Logic:

Identical surrounding medium on both sides gives equal focal distances.

🧠 Memory Trick

"Same Medium → Same Focal Distance."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Principal Foci and Optical Centre

13 In practice, to find the image of an object by a lens

A we must trace all the infinite rays emanating from the object

B it is convenient to choose any two of the standard principal rays

C the ray passing through the first focus is never used

D the laws of reflection must strictly replace the laws of refraction

💡 Short Explanation

�� Infinite rays emerge from each object point. �� Any two principal rays locate the image. �� Ray diagrams use this simplification.

✅ Correct Answer Explanation

Although infinitely many rays emerge from a point object, all rays intersect at the image point. Therefore, it is sufficient to trace any two standard principal rays to determine image location. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not practically required.
�� Option C → Rays through the first focus are commonly used.
�� Option D → Lens image formation uses refraction, not reflection.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall standard ray construction rules.

Final Logic:

Two principal rays are sufficient to locate the image.

🧠 Memory Trick

"Two Rays Find the Image."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Ray Diagrams for Lenses

14 If a ray of light passes through the optical centre of a lens at an angle of 5 degrees to the principal axis, its angle of deviation after emergence will be

A 5 degrees

B 10 degrees

C 0 degrees

D 2.5 degrees

💡 Short Explanation

�� Optical centre ray passes undeviated. �� Thin lens approximation is used. �� Direction remains unchanged.

✅ Correct Answer Explanation

One of the principal rays used in lens construction is the ray passing through the optical centre. For a thin lens, such a ray emerges without deviation. Therefore: Deviation = 0° Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → No deviation occurs.
�� Option B → Not supported by lens ray rules.
�� Option D → No partial deviation exists here.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall the optical-centre ray rule.

Final Logic:

Optical centre → undeviated path.

🧠 Memory Trick

"Through Centre = No Bending."

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

15 The total magnification (m) of a combination of lenses

A is the sum of their individual magnifications

B is the product of their individual magnifications

C is the ratio of their individual focal lengths

D depends only on the power of the first lens

💡 Short Explanation

�� Each lens produces its own magnification. �� Overall effect multiplies. �� Used for lens combinations.

✅ Correct Answer Explanation

For multiple lenses: m = m₁ × m₂ × m₃ × ⋯ Therefore, the total magnification equals the product of the individual magnifications. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Magnifications do not add.
�� Option C → No such relation exists.
�� Option D → All lenses contribute.

🎯 Strategy

Used

Direct Formula Recall

Application:

Recall the magnification relation for lens combinations.

Final Logic:

Overall magnification is multiplicative.

🧠 Memory Trick

"Lens Chain → Multiply m."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Combination of Lenses

16 For an erect and virtual image formed by a lens, m is _______, while for an inverted and real image, m is _______.

A Negative, Positive

B Positive, Negative

C Positive, Positive

D Negative, Negative

💡 Short Explanation

�� Sign indicates orientation. �� Erect image → positive magnification. �� Inverted image → negative magnification.

✅ Correct Answer Explanation

For lenses: m = h′/h Erect virtual image: m > 0 Inverted real image: m < 0 Therefore: Positive, Negative Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Signs reversed.
�� Option C → Real inverted image has negative magnification.
�� Option D → Erect image cannot have negative magnification.

🎯 Strategy

Used

Direct Concept Recall

Application:

Use sign convention for image orientation.

Final Logic:

Erect → positive, Inverted → negative.

🧠 Memory Trick

"+ Erect, − Inverted."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Lens Magnification Sign Convention

17 Match List I with List II for the power of a lens

List I	List II
1. P = 1/f	a. Formula for power of a single lens
2. P = P₁ + P₂ + P₃	b. Combination of thin lenses in contact
3. Tangent of angle d	c. Definition measure of convergence
4. Dioptre	d. SI unit for power

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

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

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

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

💡 Short Explanation

�� Power relates to focal length. �� Powers add for thin lenses in contact. �� Dioptre is the SI unit.

✅ Correct Answer Explanation

1 → a: (P=\frac1f) 2 → b: Power addition rule 3 → c: Definition of convergence/divergence measure 4 → d: SI unit is dioptre Hence: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching.
�� Option C → Incorrect matching.
�� Option D → Incorrect matching.

🎯 Strategy

Used

Option Grouping

Application:

Match known formulas first.

Final Logic:

Power formula, unit and combination rule uniquely determine the answer.

🧠 Memory Trick

"Power: 1/f, Unit: D."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Power of Lens

18 Consider the statements about Dioptre unit. Choose the correct statements:
Statements
1. It is the SI unit for the power of a lens.
2. 1 D = 1 cm⁻¹.
3. 1 D = 1 m⁻¹.
4. A lens of 1 metre focal length has a power of 1 D.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 4

💡 Short Explanation

�� Dioptre is the unit of power. �� Defined using metre. �� Not based on centimetre.

✅ Correct Answer Explanation

Statement 1 is correct. 1 D = 1 m⁻¹ Hence Statement 3 is correct. For f = 1 m P = 1/f = 1 D Hence Statement 4 is correct. Statement 2 is incorrect because: 1 D ≠ 1 cm⁻¹. Therefore Statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 2.
�� Option C → Includes incorrect Statement 2.
�� Option D → Omits Statement 3.

🎯 Strategy

Used

Elimination

Application:

Check the definition of dioptre.

Final Logic:

Only Statement 2 is incorrect.

🧠 Memory Trick

"Dioptre Uses Metre."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Dioptre Unit

19 Identify the incorrect statement about converging lenses

A A converging lens has positive power.

B A double convex lens is an example of a converging lens.

C It bends the incident parallel light away from the principal axis.

D A lens of power +2.5 D is a converging lens.

💡 Short Explanation

�� Converging lenses bring rays together. �� Parallel rays move toward the axis. �� Positive power indicates convergence.

✅ Correct Answer Explanation

A converging lens causes parallel rays to move toward the principal axis and meet at the focus. Therefore Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct property.
�� Option B → Double convex lens is converging.
�� Option D → Positive power indicates converging lens.

🎯 Strategy

Used

Odd One Out

Application:

Compare the behavior of converging lenses.

Final Logic:

Converging lenses bend rays toward, not away from, the axis.

🧠 Memory Trick

"Converging = Come Together."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Converging Lens

20 Choose the correct statements about diverging lenses
1. A concave lens diverges a beam of light parallel to the principal axis.
2. The power of a diverging lens is negative.
3. A lens of power -4.0 D is a diverging lens.
4. A diverging lens has a positive focal length.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Concave lenses diverge light. �� Diverging lenses have negative focal length. �� Therefore power is negative.

✅ Correct Answer Explanation

Statement 1 is correct because a concave lens spreads parallel rays. Statement 2 is correct because P = 1/f and (f) is negative. Statement 3 is correct because a negative power lens is diverging. Statement 4 is incorrect because a diverging lens has negative focal length. Hence 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:

Check focal length sign of a diverging lens.

Final Logic:

Negative focal length ⇒ negative power ⇒ diverging lens.

🧠 Memory Trick

"Concave = Negative f = Negative P."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Diverging Lens and Power Sign Convention

CUET UG Physics Booster Test 2-Spherical Refraction and Lenses

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