CUET UG Physics Booster Test 3-Spherical Refraction and Lenses -

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CUET UG Physics Booster Test 3-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 3-Spherical Refraction and Lenses

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QUESTION 1 OF 20

For refraction by a single spherical surface separating two media

Answer Review

1 For refraction by a single spherical surface separating two media

A the normal at the point of incidence does not pass through the centre of curvature

B the tangent plane is parallel to the principal axis

C an infinitesimal part of the spherical surface can be regarded as planar

D the laws of refraction cannot be applied at the interface

💡 Short Explanation

�� Small portions of a curved surface behave like a plane. �� Refraction laws apply locally. �� Normal passes through the centre of curvature.

✅ Correct Answer Explanation

In the derivation of refraction at a spherical surface, an infinitesimally small region around the point of incidence is treated as a plane surface. This allows Snell's law to be applied locally. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Normal always passes through the centre of curvature.
�� Option B → Tangent plane is not generally parallel to the principal axis.
�� Option D → Laws of refraction are applied at every point.

🎯 Strategy

Used

Elimination

Application:

Identify the statement consistent with local planar approximation.

Final Logic:

Small spherical region ⇒ locally plane.

🧠 Memory Trick

"Tiny Curve ≈ Plane."

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

2 In the derivation of the interface formula, the aperture is taken to be _______ compared to other distances, so that _______ angle approximation can be made.

A Large, Small

B Small, Small

C Small, Large

D Large, Large

💡 Short Explanation

�� Paraxial approximation is used. �� Aperture must be small. �� Small-angle approximation follows.

✅ Correct Answer Explanation

The derivation assumes that the aperture of the spherical surface is small compared to object and image distances. Hence rays remain close to the principal axis and make small angles. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Large aperture invalidates paraxial approximation.
�� Option C → Large-angle approximation is not used.
�� Option D → Both assumptions are incorrect.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall assumptions used in derivation.

Final Logic:

Small aperture ⇒ small angles.

🧠 Memory Trick

"Small Aperture → Small Angle."

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

3 Match List I with List II for the derivation of lens maker's formula

List I	List II
1. First interface ABC	a. n₁/OB + n₂/BI₁ = (n₂ − n₁)/BC
2. Second interface ADC	b. n₂/DI₁ − n₁/DI = (n₂ − n₁)/DC
3. Total deviation	c. Not explicitly modeled by a single spherical equation
4. Thin lens approximation	d. BI₁ is approximately equal to DI₁

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

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

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

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

💡 Short Explanation

�� Two spherical surfaces are considered separately. �� Thin lens approximation connects them. �� Final result combines both equations.

✅ Correct Answer Explanation

1 → a: Refraction at first interface 2 → b: Refraction at second interface 3 → c: Total deviation is not represented by one spherical-surface equation 4 → d: Thin lens approximation gives BI₁ ≈ DI₁ Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → First and second surface equations interchanged.
�� Option C → Incorrect matching.
�� Option D → Incorrect matching.

🎯 Strategy

Used

Option Grouping

Application:

Match interface equations first, then approximation.

Final Logic:

Only Option A correctly pairs all quantities.

🧠 Memory Trick

"First Surface → First Equation."

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

4 Identify the correct statements about optical centre in thin lens derivation
1. The points B and D are close to the optical centre.
2. The distances measured from B and D are treated as being measured from the optical centre.
3. The optical centre is completely outside the lens.
4. For a thick lens, this approximation still perfectly holds.

A 1, 2

B 3, 4

C 1, 2, 3

D 2, 3, 4

💡 Short Explanation

�� Thin lens thickness is negligible. �� B and D nearly coincide with O. �� Approximation fails for thick lenses.

✅ Correct Answer Explanation

Statements 1 and 2 are correct because the poles of the two surfaces are very close to the optical centre in a thin lens. Statement 3 is incorrect because the optical centre lies inside the lens. Statement 4 is incorrect because thick lenses require separate treatment.

❓ Why Other Options Are Incorrect

�� Option B → Includes two incorrect statements.
�� Option C → Includes Statement 3.
�� Option D → Includes Statements 3 and 4.

🎯 Strategy

Used

Elimination

Application:

Identify assumptions specific to thin lenses.

Final Logic:

Only Statements 1 and 2 are correct.

🧠 Memory Trick

"Thin Lens → B ≈ D ≈ O."

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

5 In a combination of two thin lenses of focal lengths f₁ = 30 cm and f₂ = -20 cm, the image formed by the first lens acts as an object for the second. If the effective power is to be found, what is the equivalent focal length of the combination?

A -60 cm

B +60 cm

C -12 cm

D +12 cm

💡 Short Explanation

�� Powers add algebraically. �� Use reciprocal focal lengths. �� Negative result indicates diverging combination.

✅ Correct Answer Explanation

1/F = 1/f₁ + 1/f₂ 1/F = 1/30 + (−1/20) 1/F = (2 − 3)/60 = −1/60 F = −60 cm Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Wrong sign.
�� Option C → Calculation error.
�� Option D → Calculation error.

🎯 Strategy

Used

Substitution

Application:

Apply lens-combination formula.

Final Logic:

F = −60 cm

🧠 Memory Trick

"Powers Add, Follows Inverse."

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

6 Choose the incorrect statement about image formation by two thin lenses in contact

A The image formed by the first lens becomes the object for the second.

B The first image is always virtual, irrespective of the object position.

C The total magnification is the product of individual magnifications.

D The net power of the combination is the algebraic sum of individual powers.

💡 Short Explanation

�� First image may be real or virtual. �� Depends on object position. �� Lens type also matters.

✅ Correct Answer Explanation

The image produced by the first lens can be real or virtual depending on the object distance and focal length. Hence the statement that it is always virtual is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct principle.
�� Option C → Total magnification is multiplicative.
�� Option D → Powers add algebraically.

🎯 Strategy

Used

Odd One Out

Application:

Look for the absolute statement.

Final Logic:

"Always virtual" is false.

🧠 Memory Trick

"First Image Depends on Position."

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

7 If a glass lens (n = 1.5) has focal length +20 cm in air, and its radii of curvature are equal in magnitude but opposite in sign (double convex), what is the magnitude of the radius of curvature?

A 10 cm

B 20 cm

C 40 cm

D 15 cm

💡 Short Explanation

�� Use lens maker's formula. �� Double convex: (R₁ = +R, R₂ = −R) �� Solve for R.

✅ Correct Answer Explanation

(1/f) = (μ − 1) [(1/R) − (−1/R)] 1/20 = 0.5 × (2/R) 1/20 = 1/R R = 20 cm Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Too small.
�� Option C → Too large.
�� Option D → Does not satisfy formula.

🎯 Strategy

Used

Substitution

Application:

Apply lens maker's formula directly.

Final Logic:

R = 20 cm

🧠 Memory Trick

"Equal Curvature → Simple Lens Maker."

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

8 Choose the correct statements about designing lenses using the Lens Maker's Formula
1. It handles all different cases of spherical lenses.
2. If R₁ is negative and R₂ is positive, f is negative.
3. A given lens will have different focal lengths in media of different refractive indices.
4. The formula assumes the lens is thick.

A 1, 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Lens maker's formula works for spherical lenses. �� Medium affects focal length. �� Thin lens assumption is used.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. Statement 4 is incorrect because the formula is derived for thin lenses. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify the assumption used in derivation.

Final Logic:

Lens maker's formula assumes a thin lens.

🧠 Memory Trick

"Thin Lens Designer Formula."

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

9 When applying the thin lens formula ((1/v - 1/u = 1/f)) to find the position of a virtual image formed by a concave lens

A both u and f are taken as positive

B u is taken as zero

C both u and f are taken as negative according to the Cartesian sign convention

D v will always turn out to be positive

💡 Short Explanation

�� Object lies on left side. �� Concave lens has negative focal length. �� Cartesian sign convention is used.

✅ Correct Answer Explanation

For a concave lens: f < 0 Object distance: u < 0 Thus both u and f are negative while solving the lens equation. Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Sign convention violated.
�� Option B → Object distance is not zero.
�� Option D → Virtual image gives negative v.

🎯 Strategy

Used

Direct Concept Recall

Application:

Apply Cartesian sign convention.

Final Logic:

Concave lens ⇒ u < 0, f < 0

🧠 Memory Trick

"Concave → Negative f."

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

10 Match List I with List II for lenses

List I	List II
1. Ratio v/u	a. Linear magnification (m)
2. 1/v − 1/u	b. 1/f for a single lens
3. 1/f₁ + 1/f₂	c. Effective focal length (1/f)
4. m₁ × m₂ × m₃	d. Total magnification

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

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

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

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

💡 Short Explanation

�� Magnification uses (v/u). �� Lens equation gives (1/f). �� Lens powers combine algebraically.

✅ Correct Answer Explanation

1 → a: m = v/u 2 → b: 1/v − 1/u = 1/f 3 → c: 1/f = 1/f₁ + 1/f₂ 4 → d: Product of magnifications gives total magnification. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application:

Match formulas directly with definitions.

Final Logic:

Each formula uniquely identifies its physical quantity.

🧠 Memory Trick

"v/u = m, Product = Total m."

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

11 Two thin lenses of focal lengths 10 cm and -5 cm are kept in contact. At what distance from the combination will a parallel beam of light come to a principal focus?

A +10 cm

B -10 cm

C +5 cm

D -5 cm

💡 Short Explanation

�� Use lens combination formula. �� Powers add algebraically. �� Negative focal length indicates a diverging combination.

✅ Correct Answer Explanation

1/F = 1/f₁ + 1/f₂ 1/F = 1/10 + 1/(-5) 1/F = (1 − 2)/10 1/F = -1/10 F = -10 cm The combination behaves as a concave (diverging) lens.

❓ Why Other Options Are Incorrect

�� Option A → Wrong sign.
�� Option C → Calculation error.
�� Option D → Incorrect reciprocal evaluation.

🎯 Strategy

Used

Substitution

Application:

Apply the lens combination formula directly.

Final Logic:

F = -10 cm

🧠 Memory Trick

"Add powers, then invert."

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

12 Consider the incorrect statement regarding the two foci F and F' of a lens

A A double convex lens has two foci F and F'.

B A concave lens has two foci F and F'.

C They are always at unequal distances from the optical centre.

D They lie on either side of the lens.

💡 Short Explanation

�� Thin lenses possess two principal foci. �� Foci are equidistant in a uniform medium. �� Convex and concave lenses both have two foci.

✅ Correct Answer Explanation

For a thin lens placed in the same medium on both sides, the first and second principal foci are located at equal distances from the optical centre. Therefore Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct for convex lenses.
�� Option B → Correct for concave lenses.
�� Option D → Foci lie on opposite sides of the lens.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement violating lens symmetry.

Final Logic:

F and F' are equidistant from the optical centre.

🧠 Memory Trick

"Two Foci, Equal Distance."

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

13 Consider the statements about principal ray tracing in a concave lens
1. A ray parallel to the principal axis appears to diverge from the first principal focus.
2. A ray appearing to meet the second focus emerges parallel to the principal axis.
3. A ray passing through the optical centre is undeviated.
4. Ray tracing cannot be applied to virtual images.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Standard rays apply to concave lenses. �� Optical centre ray is undeviated. �� Virtual images can be traced.

✅ Correct Answer Explanation

Statement 1 is correct because parallel rays appear to diverge from the first focus. Statement 2 is correct because rays directed toward the second focus emerge parallel. Statement 3 is correct because rays through the optical centre suffer negligible deviation. Statement 4 is incorrect because ray tracing is routinely used for virtual image formation.

❓ 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 false statement among the four.

Final Logic:

Only Statement 4 is incorrect.

🧠 Memory Trick

"Parallel–Focus–Centre."

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

14 Regarding the ray of light passing through the optical centre of a thin lens

A it undergoes a phase shift of 180 degrees

B it emerges without any deviation after refraction

C it bends towards the principal focus

D it suffers maximum lateral displacement

💡 Short Explanation

�� Optical centre ray is special. �� Refractions effectively cancel. �� Emergent ray remains undeviated.

✅ Correct Answer Explanation

For a thin lens, a ray passing through the optical centre emerges in the same direction because the deviations at the two refracting surfaces cancel each other.

❓ Why Other Options Are Incorrect

�� Option A → No such phase-shift condition is involved.
�� Option C → Applies to other principal rays.
�� Option D → No lateral displacement occurs.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall standard principal rays.

Final Logic:

Optical centre ray → undeviated.

🧠 Memory Trick

"Centre Ray = Straight Way."

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

15 Choose the correct statements about the magnification produced by a lens
1. Magnification m is the ratio of image height to object height (h'/h).
2. Magnification is also given by v/u.
3. A virtual image has a positive magnification.
4. A real image has a negative magnification.

A 1, 2

B 3, 4

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

�� Magnification relates image size and object size. �� Lens formula gives m = v/u. �� Sign indicates image orientation.

✅ Correct Answer Explanation

m = h′/h = v/u Virtual images are erect and have positive magnification. Real images are inverted and have negative magnification. Therefore all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits Statements 3 and 4.
�� Option B → Omits Statements 1 and 2.
�� Option C → Omits Statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Verify all magnification properties.

Final Logic:

All four statements are correct.

🧠 Memory Trick

"+ Virtual, − Real."

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

16 In accordance with the accepted sign convention, if the total magnification of a two-lens combination is negative, it implies that the final image is

A erect and virtual

B inverted and diminished exclusively

C inverted and real with respect to the original object

D formed at infinity only

💡 Short Explanation

�� Sign determines orientation. �� Negative magnification means inversion. �� Real images are inverted.

✅ Correct Answer Explanation

A negative magnification indicates that the image is inverted relative to the object. For lens systems, inversion corresponds to a real image.

❓ Why Other Options Are Incorrect

�� Option A → Erect images have positive magnification.
�� Option B → Negative sign does not necessarily imply diminished image.
�� Option D → Magnification sign is unrelated to infinity.

🎯 Strategy

Used

Direct Concept Recall

Application:

Use sign convention of magnification.

Final Logic:

Negative magnification ⇒ inverted image.

🧠 Memory Trick

"Minus Means Upside Down."

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

17 The sum in the equation
P = P₁ + P₂ + P₃
for a combination of lenses is

A a vector sum

B a geometric sum

C an algebraic sum

D an exponential sum

💡 Short Explanation

�� Lens powers add directly. �� Signs are retained. �� No vector treatment required.

✅ Correct Answer Explanation

For thin lenses in contact: P = P₁ + P₂ + P₃ Positive and negative powers are added algebraically.

❓ Why Other Options Are Incorrect

�� Option A → Power is not a vector.
�� Option B → Not geometric addition.
�� Option D → No exponential relation exists.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall lens combination rule.

Final Logic:

Power addition is algebraic.

🧠 Memory Trick

"Powers Simply Add."

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

18 The power of a lens is defined as the tangent of the _______ by which it converges or diverges a beam of light parallel to the principal axis falling at _______ distance from the optical centre.

A Angle, Unit

B Radius, Zero

C Curvature, Infinite

D Distance, Unit

💡 Short Explanation

�� Power measures convergence/divergence. �� Defined using angular deviation. �� Unit distance is used.

✅ Correct Answer Explanation

Power is associated with the tangent of the angle through which a parallel beam is converged or diverged when incident at unit distance from the optical centre. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Radius is not used in the definition.
�� Option C → Curvature is not used directly.
�� Option D → Distance is not the required quantity.

🎯 Strategy

Used

Direct Concept Recall

Application:

Recall NCERT definition of power.

Final Logic:

Power is related to angular convergence/divergence.

🧠 Memory Trick

"Power Measures Bending Angle."

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

19 If an optician prescribes a corrective lens of power +2.0 D, and it is combined with another lens of power -0.5 D, what is the net power of the combination?

A +2.5 D

B -1.5 D

C +1.5 D

D -2.5 D

💡 Short Explanation

�� Powers add algebraically. �� Positive and negative signs matter. �� Net power remains positive.

✅ Correct Answer Explanation

P = P₁ + P₂ P = 2.0 + (−0.5) P = 1.5 D Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Added magnitudes only.
�� Option B → Incorrect sign.
�� Option D → Incorrect calculation.

🎯 Strategy

Used

Substitution

Application:

Directly add lens powers.

Final Logic:

2.0 − 0.5 = 1.5 D

🧠 Memory Trick

"+ and − Add Algebraically."

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

20 Consider the statements about combinations of converging and diverging lenses
1. Combination of lenses helps to obtain diverging or converging lenses of desired magnification.
2. The net power can be negative if the diverging lens has higher magnitude of power.
3. It enhances sharpness of the image.
4. The powers are added algebraically.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Lens combinations modify focal length. �� Net power may be positive or negative. �� Powers add algebraically.

✅ Correct Answer Explanation

Statement 1 is correct because combinations help achieve desired magnification. Statement 2 is correct because a stronger diverging lens can make net power negative. Statement 3 is correct because combinations can improve image quality and sharpness. Statement 4 is correct because lens powers add algebraically. Therefore all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits Statement 4.
�� Option B → Omits Statement 2.
�� Option C → Omits Statement 1.

🎯 Strategy

Used

Option Grouping

Application:

Verify each statement individually.

Final Logic:

All four statements are correct.

🧠 Memory Trick

"Combine Lenses → Add Powers."

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

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