CUET UG Physics Booster Test 2-Reflection and Spherical Mirrors -

🚀

CUET UG Physics Booster Test 2-Reflection and Spherical Mirrors

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-Reflection and Spherical Mirrors

1 / 20

QUESTION 1 OF 20

Choose the incorrect statement about reflection laws on spherical surfaces:

Answer Review

1 Choose the incorrect statement about reflection laws on spherical surfaces:

A The angle of reflection equals the angle of incidence at each point.

B The normal is taken as normal to the tangent to the surface at the point of incidence.

C The laws of reflection are only approximately true for spherical mirrors.

D The normal is along the radius, joining the centre of curvature to the point of incidence.

💡 Short Explanation

�� Laws of reflection are exact laws. �� They apply to both plane and curved surfaces. �� Normal at a spherical surface is along the radius.

✅ Correct Answer Explanation

The laws of reflection are valid at every point of every reflecting surface, whether plane or curved. For spherical mirrors, the normal at any point is the radius joining that point to the centre of curvature. Therefore, it is incorrect to state that the laws are only approximately true for spherical mirrors.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement of the law of reflection.
�� Option B → Correct description of the normal.
�� Option D → Correct; radius acts as the normal.

🎯 Strategy

Used

Odd One Out

Application: Identify the statement that contradicts the universal validity of reflection laws.

Final Logic: Reflection laws are exact, not approximate.

🧠 Memory Trick

"Reflection laws never change."

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

2 Choose the correct statements about the plane of reflection:
1. The incident ray, reflected ray, and normal lie in the same plane.
2. This coplanarity rule is valid at each point on a curved reflecting surface.
3. The angle between the reflected ray and the normal equals the angle of incidence.
4. Coplanarity only applies to paraxial rays.

A 1, 2, 3

B 1, 3

C 2, 3, 4

D 1, 4

💡 Short Explanation

�� Reflection occurs in one plane. �� Coplanarity applies universally. �� Angle of incidence equals angle of reflection.

✅ Correct Answer Explanation

Statement 1 is correct because incident ray, reflected ray, and normal always lie in the same plane. Statement 2 is correct because the law applies at every point on curved surfaces. Statement 3 is correct because angle of reflection equals angle of incidence. Statement 4 is incorrect because coplanarity is not restricted to paraxial rays.

❓ Why Other Options Are Incorrect

�� Option B → Omits correct Statement 2.
�� Option C → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 4 and omits correct Statements 2 and 3.

🎯 Strategy

Used

Elimination

Application: Check each statement against the laws of reflection.

Final Logic: Only Statement 4 is incorrect.

🧠 Memory Trick

"Incident–Normal–Reflected: One Plane."

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

3 Consider the definitions related to spherical mirrors. Choose the correct statements:
1. The geometric centre of a spherical mirror is its pole.
2. The geometric centre of a spherical lens is its optical centre.
3. The principal axis joins the pole and the optical centre.
4. The principal axis is the line joining the pole and the centre of curvature of the spherical mirror.

A 1, 2, and 4

B 1, 3, and 4

C 2 and 3

D 1 and 2 only

💡 Short Explanation

�� Pole is the geometric centre of a mirror. �� Optical centre is a lens concept. �� Principal axis joins pole and centre of curvature.

✅ Correct Answer Explanation

Statement 1 is correct because the geometric centre of a spherical mirror is called the pole. Statement 2 is correct because the geometric centre of a lens is called the optical centre. Statement 3 is incorrect because a mirror has no optical centre. Statement 4 is correct because the principal axis joins the pole and centre of curvature.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application: Distinguish mirror terminology from lens terminology.

Final Logic: Statement 3 incorrectly mixes lens and mirror concepts.

🧠 Memory Trick

"Mirror → Pole, Lens → Optical Centre."

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

4 In the context of the principal axis of a spherical mirror, any ray travelling exactly along this axis towards the mirror

A reflects back at a 90-degree angle to the axis.

B passes through the focal plane without deviation.

C acts as a ray passing through the centre of curvature and simply retraces its path.

D emerges parallel to the incident ray but laterally shifted.

💡 Short Explanation

�� Principal axis passes through the centre of curvature. �� Such a ray strikes normally. �� Normal incidence causes retracing.

✅ Correct Answer Explanation

A ray travelling along the principal axis passes through the centre of curvature and therefore strikes the mirror normally. The angle of incidence is zero, so the reflected ray retraces its original path.

❓ Why Other Options Are Incorrect

�� Option A → Reflection at 90° does not occur.
�� Option B → Not a ray-tracing rule.
�� Option D → Lateral shift is associated with glass slabs.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the centre-of-curvature ray rule.

Final Logic: Normal incidence leads to retracing.

🧠 Memory Trick

"Axis Ray Returns."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Ray Tracing in Mirrors

5 According to the Cartesian sign convention, an object is placed 5 cm in front of a concave mirror of radius of curvature 15 cm. What is the precise value of object distance u?

A +5 cm

B -5 cm

C +15 cm

D -15 cm

💡 Short Explanation

�� Object is in front of the mirror. �� Incident light travels toward the mirror. �� Opposite direction distances are negative.

✅ Correct Answer Explanation

Under Cartesian sign convention, distances measured opposite to the direction of incident light are negative. Since the object is placed 5 cm in front of the mirror, u = -5 cm Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Wrong sign.
�� Option C → Incorrect magnitude.
�� Option D → Incorrect magnitude.

🎯 Strategy

Used

Direct Concept Recall

Application: Apply sign convention directly.

Final Logic: Object distance in front of mirror is negative.

🧠 Memory Trick

"Object in front → Negative u."

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

6 According to the Cartesian sign convention, the heights measured upwards with respect to the x-axis are ____, and heights measured downwards are ____.

A Positive, Negative

B Negative, Positive

C Positive, Positive

D Negative, Negative

💡 Short Explanation

�� Principal axis acts as x-axis. �� Upward heights are positive. �� Downward heights are negative.

✅ Correct Answer Explanation

The Cartesian sign convention assigns positive values to heights measured upward from the principal axis and negative values to heights measured downward. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the sign convention.
�� Option C → Downward heights are not positive.
�� Option D → Upward heights are not negative.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall height-sign convention.

Final Logic: Upward +, Downward −.

🧠 Memory Trick

"Up Plus, Down Minus."

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

7 Choose the correct statements regarding a concave mirror whose lower half is covered with an opaque material:
1. The mirror will form an image of the whole object.
2. The image will only show half of the object.
3. The area of the reflecting surface has been reduced.
4. The intensity of the image will be low (in this case, half).

A 1, 3, 4

B 2, 3, 4

C 1, 2, 3

D 1, 4

💡 Short Explanation

�� Every part of the mirror forms the entire image. �� Covering part reduces brightness. �� Image size remains unchanged.

✅ Correct Answer Explanation

Statement 1 is correct because even a partial mirror can form the complete image. Statement 2 is incorrect because half the mirror does not produce half the image. Statement 3 is correct because the effective reflecting area decreases. Statement 4 is correct because fewer rays contribute to image formation, reducing brightness.

❓ Why Other Options Are Incorrect

�� Option B → Statement 2 is incorrect.
�� Option C → Statement 2 is incorrect.
�� Option D → Omits correct Statement 3.

🎯 Strategy

Used

Elimination

Application: Use image formation principles for mirrors.

Final Logic: Partial obstruction reduces intensity, not image size.

🧠 Memory Trick

"Half Mirror → Full Image, Half Brightness."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Image Formation by Mirrors

8 During the derivation of the focal length equation for a spherical mirror, the distance FD is equated to CD/2 for paraxial rays. Since D is very close to P, FD becomes f and CD becomes R, yielding the formula:

A f = R/2

B f = 2R

C f = R

D f = R²

💡 Short Explanation

�� Paraxial approximation is used. �� Radius of curvature is twice focal length. �� Fundamental mirror relation.

✅ Correct Answer Explanation

From the geometrical derivation for paraxial rays: FD = CD/2 Since D is very close to P, FD = f CD = R Therefore, f = R/2 Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverse relation.
�� Option C → Focal length is not equal to radius.
�� Option D → Dimensionally incorrect.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall derived mirror relation.

Final Logic: (R=2f).

🧠 Memory Trick

"Radius is Double Focus."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Relation Between Radius and Focal Length

9 Match List I with List II for image characteristics by a concave mirror.

List I	List II
1. Object at -10 cm, f = -7.5 cm	a. Image is magnified, real and inverted
2. Object at -5 cm, f = -7.5 cm	b. Image is magnified, virtual and erect
3. Magnification m is negative	c. Real, inverted image
4. Magnification m is positive	d. Virtual, erect image

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

�� Object beyond focus forms a real inverted image. �� Object between pole and focus forms a virtual erect image. �� Negative magnification indicates inversion. �� Positive magnification indicates erectness.

✅ Correct Answer Explanation

1 → a: For (u=-10) cm and (f=-7.5) cm, the object is beyond focus, producing a magnified real inverted image. 2 → b: For (u=-5) cm and (f=-7.5) cm, the object lies between pole and focus, producing a magnified virtual erect image. 3 → c: Negative magnification corresponds to a real inverted image. 4 → d: Positive magnification corresponds to a virtual erect image. Thus, the correct matching is 1-a, 2-b, 3-c, 4-d.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the image characteristics for Cases 1 and 2.
�� Option C → Incorrect matching of magnification signs.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Option Grouping

Application: Match magnification signs first, then image types.

Final Logic: Negative m → inverted; Positive m → erect.

🧠 Memory Trick

"Minus Inverted, Plus Erect."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Image Formation by Concave Mirrors

10 When discussing image formation by a convex mirror, the image is virtual because the rays

A actually converge to a point in front of the mirror.

B do not actually meet but appear to diverge from a point behind the mirror.

C perfectly retrace their path to the object location.

D form on a screen placed at the focal point.

💡 Short Explanation

�� Convex mirrors always form virtual images. �� Reflected rays diverge. �� Backward extensions appear to meet.

✅ Correct Answer Explanation

In a convex mirror, reflected rays diverge after reflection. When these rays are extended backward, they appear to meet behind the mirror. This apparent intersection forms a virtual image.

❓ Why Other Options Are Incorrect

�� Option A → Describes a real image.
�� Option C → Retracing occurs only in special ray paths.
�� Option D → Virtual images cannot be formed on a screen.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the definition of a virtual image.

Final Logic: Apparent intersection behind the mirror creates a virtual image.

🧠 Memory Trick

"Convex = Virtual Always."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Convex Mirror Images

11 Choose the correct statements about parallel paraxial beams:
1. They make small angles with the principal axis.
2. Reflected rays converge at a point F on the principal axis of a concave mirror.
3. They are incident at points far from the pole of the mirror.
4. For a convex mirror, they appear to diverge from point F.

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2

💡 Short Explanation

�� Paraxial rays stay close to the principal axis. �� Concave mirrors converge parallel rays. �� Convex mirrors produce apparent divergence.

✅ Correct Answer Explanation

Statement 1 is correct because paraxial rays make small angles with the principal axis. Statement 2 is correct because parallel rays converge at the principal focus of a concave mirror. Statement 3 is incorrect because paraxial rays are incident near the pole. Statement 4 is correct because reflected rays appear to diverge from the focus of a convex mirror.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Omits correct Statement 4.

🎯 Strategy

Used

Elimination

Application: Identify the incorrect statement about paraxial rays.

Final Logic: Paraxial rays are close to the pole, not far away.

🧠 Memory Trick

"Paraxial = Pole Nearby."

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

12 Choose the incorrect statement about a ray passing through the centre of curvature of a concave mirror:

A The ray simply retraces its path after reflection.

B The reflected ray is parallel to the principal axis.

C It acts as a normal to the tangent at the point of incidence.

D The angle of incidence for this ray is zero.

💡 Short Explanation

�� Centre-of-curvature rays strike normally. �� Angle of incidence is zero. �� Such rays retrace their path.

✅ Correct Answer Explanation

A ray passing through the centre of curvature travels along the normal to the mirror surface. Therefore, it is reflected back along the same path. It does not become parallel to the principal axis.

❓ Why Other Options Are Incorrect

�� Option A → Correct retracing rule.
�� Option C → Radius acts as normal.
�� Option D → Normal incidence means angle of incidence is zero.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the centre-of-curvature ray rule.

Final Logic: Centre-of-curvature rays return along their own path.

🧠 Memory Trick

"C → Comes Back."

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

13 Choose the correct statements about a ray passing through the focus of a concave mirror:
1. The reflected ray is parallel to the principal axis.
2. It retraces its path after reflection.
3. It applies to a ray directed towards the focus of a convex mirror.
4. It converges at the centre of curvature.

A 1, 3

B 2, 4

C 1, 2, 3

D 3, 4

💡 Short Explanation

�� Focus and parallel-ray rules are complementary. �� Concave and convex mirrors follow corresponding focus rules. �� Retracing does not occur.

✅ Correct Answer Explanation

Statement 1 is correct because a ray through the focus of a concave mirror reflects parallel to the principal axis. Statement 2 is incorrect because retracing occurs for centre-of-curvature rays. Statement 3 is correct because a ray directed towards the focus of a convex mirror also reflects parallel to the principal axis. Statement 4 is incorrect because the reflected ray does not converge at the centre of curvature.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are incorrect.
�� Option C → Includes incorrect Statement 2.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application: Apply standard ray-tracing rules.

Final Logic: Only Statements 1 and 3 are correct.

🧠 Memory Trick

"Focus → Parallel."

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

14 For a ray incident at any angle at the pole of a spherical mirror, the reflected ray follows the ____ symmetrically with the ____.

A Laws of reflection, Principal axis

B Laws of refraction, Normal

C Paraxial approximation, Focal plane

D Magnification formula, Centre of curvature

💡 Short Explanation

�� Principal axis acts as the normal at the pole. �� Reflection law applies. �� Angles are measured with respect to the normal.

✅ Correct Answer Explanation

At the pole, the principal axis acts as the normal to the mirror surface. Therefore, the incident and reflected rays obey the laws of reflection and are symmetric about the principal axis.

❓ Why Other Options Are Incorrect

�� Option B → Refraction is not involved.
�� Option C → Not related to pole incidence.
�� Option D → Magnification is unrelated.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the pole-incidence rule.

Final Logic: Pole reflections obey reflection laws.

🧠 Memory Trick

"Pole → Principal Axis Normal."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Reflection at the Pole

15 In the derivation of the mirror equation, the right-angled triangles A'B'F and MPF are similar.
1. MP is considered to be a straight line perpendicular to CP for paraxial rays.
2. The ratio A'B'/PM equals B'F/FP.
3. PM is equal to the object height AB.
4. The triangles A'B'P and ABP are not similar.

A 1, 2, and 3

B 1, 2, and 4

C 2 and 4

D 3 and 4

💡 Short Explanation

�� Similar triangles are used in mirror derivation. �� Paraxial approximation simplifies geometry. �� Statement 4 is incorrect.

✅ Correct Answer Explanation

Statements 1, 2, and 3 are correct and arise directly from the geometrical derivation of the mirror formula. Statement 4 is incorrect because triangles A'B'P and ABP are indeed similar, which is essential in the derivation.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application: Identify the statement contradicting the derivation.

Final Logic: Similar triangles are fundamental to the proof.

🧠 Memory Trick

"Mirror Formula = Similar Triangles."

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

16 An object is placed at (u = -39) m in front of a convex mirror of radius of curvature (R = 2) m. Find the image position (v).

A 39/40 m

B 40/39 m

C 1/39 m

D -39/40 m

💡 Short Explanation

�� Convex mirror has positive focal length. �� (f = R/2). �� Use mirror formula.

✅ Correct Answer Explanation

f = R/2 = 1 m Using 1/v + 1/u = 1/f 1/v - 1/39 = 1 1/v = 40/39 v = 39/40 m Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal error.
�� Option C → Incorrect algebra.
�� Option D → Wrong sign for convex mirror image.

🎯 Strategy

Used

Substitution

Application: Substitute values into the mirror formula.

Final Logic: (v = 39/40) m.

🧠 Memory Trick

"Convex → Positive f."

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

17 Linear magnification formula (m=-v/u) is derived using similar triangles. In this context, the ratio (h'/h) becomes:

A (v/u)

B (-v/u)

C (u/v)

D (-u/v)

💡 Short Explanation

�� Magnification relates image and object heights. �� Similar triangles are used. �� Sign convention gives the negative sign.

✅ Correct Answer Explanation

m = h'/h Using similar triangles for spherical mirrors, m = -v/u Therefore, h'/h = -v/u Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Missing sign convention.
�� Option C → Reciprocal ratio.
�� Option D → Incorrect expression.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the magnification formula.

Final Logic: (m=-v/u).

🧠 Memory Trick

"Mirror m = −v/u."

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

18 Match List I with List II regarding mirror magnifications.

List I	List II
1. m = -3, u = -10 cm	a. Image at v = -30 cm (Real)
2. m = +3, u = -5 cm	b. Image at v = +15 cm (Virtual)
3. Erect image	c. m is positive
4. Inverted image	d. m is negative

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

�� Negative magnification indicates a real inverted image. �� Positive magnification indicates a virtual erect image. �� Use (m=-v/u). �� Image nature is directly linked to the sign of magnification.

✅ Correct Answer Explanation

For 1: m = -v/u -3 = -v/(-10) v = -30 cm Hence: 1 → a (Image at v = -30 cm, Real) For 2: +3 = -v/(-5) v = +15 cm Hence: 2 → b (Image at v = +15 cm, Virtual) 2 → b (Image at v = +15 cm, Virtual) For image orientation: 3 → c: Erect image corresponds to positive magnification. 4 → d: Inverted image corresponds to negative magnification. Therefore: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Reverses real and virtual image conditions.
�� Option C → Incorrectly matches magnification signs with image orientation.
�� Option D → Multiple mismatches in both image position and orientation.

🎯 Strategy

Used

Option Grouping

Application: First determine image positions using (m=-v/u), then match sign of magnification with image orientation.

Final Logic: Negative m → Real/Inverted, Positive m → Virtual/Erect.

🧠 Memory Trick

"Plus Erect, Minus Inverted."

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

19 In establishing the focal length of a spherical mirror, the small angle approximation tan θ ≈ θ is strictly applied because the derivation is valid only for paraxial rays making very small angles with the principal axis.

A the speed of light is essentially infinite in vacuum.

B paraxial rays are incident at points close to the pole and make small angles with the principal axis.

C convex mirrors always produce diminished virtual images.

D the angle of incidence is greater than the critical angle.

💡 Short Explanation

�� Paraxial rays remain close to the principal axis. �� They make very small angles with the axis. �� Small-angle approximations simplify mirror derivations.

✅ Correct Answer Explanation

The approximation tan θ ≈ θ is valid only when θ is very small (in radians). Paraxial rays strike the mirror near the pole and make small angles with the principal axis. Therefore, this approximation can be safely used in deriving mirror relations such as f = R/2 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Small-angle approximation is unrelated to the speed of light.
�� Option C → Image size has no connection with the approximation.
�� Option D → Critical angle is a concept associated with refraction, not reflection by mirrors.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the condition under which small-angle approximations are valid.

Final Logic: Small-angle approximations require paraxial rays.

🧠 Memory Trick

"Paraxial = Small θ."

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

20 Choose the correct statements about the assumptions made for spherical surfaces:
1. The aperture (or lateral size) of the surface is taken to be small compared to other distances involved.
2. Point D on the principal axis is considered very close to the point P.
3. Small angle approximations like tan 2θ ≈ 2θ are used.
4. The entire surface acts as a plane interface.

A 1, 2, 3

B 1, 3, 4

C 2, 3

D 1, 2, 4

💡 Short Explanation

�� Mirror derivations use paraxial assumptions. �� Aperture is assumed small. �� Small-angle approximations simplify geometry.

✅ Correct Answer Explanation

Statement 1 is correct because the aperture is assumed small compared to other relevant distances, ensuring paraxial conditions. Statement 2 is correct because point D is taken very close to pole P during the derivation of the focal-length relation. Statement 3 is correct because small-angle approximations such as tan θ ≈ θ and tan 2θ ≈ 2θ are used. Statement 4 is incorrect because the reflecting surface remains spherical; it is not assumed to be a plane interface. Only paraxial approximations are applied. Therefore Statements 1, 2, and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 4.
�� Option C → Omits correct Statement 1.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application: Identify which assumption is not used in spherical mirror derivations.

Final Logic: The mirror remains spherical; only paraxial approximations are introduced.

🧠 Memory Trick

"Small Aperture → Small Angles."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Assumptions in Spherical Mirror Derivations

CUET UG Physics Booster Test 2-Reflection and Spherical Mirrors

Test Complete!

Answer Review