UET UG Physics Booster Test 3 -Lens Combinations and Prisms -

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CUET UG Physics Booster Test 3 -Lens Combinations and Prisms

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3 -Lens Combinations and Prisms

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QUESTION 1 OF 20

Match the combinations of two thin lenses in contact (List I) with their effective nature (List II)

List I	List II
1. Convex (+20 cm) & Concave (-20 cm)	a. Diverging (f = -20 cm)
2. Convex (+20 cm) & Concave (-10 cm)	b. Converging (f = +20 cm)
3. Convex (+10 cm) & Concave (-20 cm)	c. Plane glass (f = ∞)
4. Convex (+10 cm) & Convex (+10 cm)	d. Converging (f = +5 cm)

Answer Review

1 Match the combinations of two thin lenses in contact (List I) with their effective nature (List II)

List I	List II
1. Convex (+20 cm) & Concave (-20 cm)	a. Diverging (f = -20 cm)
2. Convex (+20 cm) & Concave (-10 cm)	b. Converging (f = +20 cm)
3. Convex (+10 cm) & Concave (-20 cm)	c. Plane glass (f = ∞)
4. Convex (+10 cm) & Convex (+10 cm)	d. Converging (f = +5 cm)

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

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

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

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

💡 Short Explanation

�� Power of lenses in contact adds algebraically. �� P = P1 + P2 �� Effective focal length determines whether the combination is converging, diverging, or neutral.

✅ Correct Answer Explanation

For lenses in contact: P = P1 + P2 For Pair 1: 1/f = 1/20 + 1/(-20) = 0 Therefore f = ∞ 1 → c For Pair 2: 1/f = 1/20 + 1/(-10) 1/f = -1/20 f = -20 cm 2 → a For Pair 3: 1/f = 1/10 + 1/(-20) 1/f = 1/20 f = +20 cm 3 → b For Pair 4: 1/f = 1/10 + 1/10 1/f = 1/5 f = +5 cm 4 → d Hence the correct matching is: 1-c, 2-a, 3-b, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of 2 and 3.
�� Option C → Multiple incorrect focal length assignments.
�� Option D → Does not satisfy lens combination formula.

🎯 Strategy

Used

Substitution

Application:

�� Calculate the effective focal length for each pair and match accordingly.

Final Logic:

�� Direct substitution into the lens combination formula gives 1-c, 2-a, 3-b, 4-d.

🧠 Memory Trick

Contact lenses = Powers Add.

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

2 Three thin lenses are placed in contact. The first is a convex lens of focal length 40 cm, the second is a concave lens of focal length 25 cm, and the third is a convex lens of focal length 50 cm. What is the net power and effective focal length of this combination?

A Net Power = +0.5 D, f = +200 cm

B Net Power = -0.5 D, f = -200 cm

C Net Power = +1.5 D, f = +66.7 cm

D Net Power = -2.5 D, f = -40 cm

💡 Short Explanation

�� Power is measured in dioptres. �� Convex lens power is positive. �� Concave lens power is negative.

✅ Correct Answer Explanation

Power of first lens: P1 = 1/0.4 = +2.5 D Power of second lens: P2 = 1/(-0.25) = -4 D Power of third lens: P3 = 1/0.5 = +2 D Net power: P = 2.5 - 4 + 2 P = +0.5 D Effective focal length: f = 1/P f = 1/0.5 f = 2 m f = 200 cm Therefore: Net Power = +0.5 D Effective focal length = +200 cm

❓ Why Other Options Are Incorrect

�� Option B → Wrong sign of net power.
�� Option C → Incorrect addition of powers.
�� Option D → Net power is not negative.

🎯 Strategy

Used

Substitution

Application:

�� Convert focal lengths into powers and add them algebraically.

Final Logic:

�� Net power is +0.5 D, giving focal length +200 cm.

🧠 Memory Trick

Contact lenses = Add Powers.

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

3 Choose the correct statements regarding the benefits of multi-lens systems in precision optical instruments:
1. They enhance the overall sharpness of the final image.
2. They can be designed to obtain diverging or converging systems of desired magnification.
3. They are used to entirely eliminate the need for an objective lens in telescopes.
4. They improve image quality by minimizing various optical aberrations present in single lenses.

A 1, 2, and 4

B 2, 3, and 4

C 1 and 4 only

D 1, 2, 3, and 4

💡 Short Explanation

�� Multiple lenses improve image quality. �� Aberrations can be reduced. �� Objective lenses remain necessary in telescopes.

✅ Correct Answer Explanation

Statement 1 is correct because lens combinations improve image sharpness. Statement 2 is correct because combinations can be designed to produce required focal lengths and magnifications. Statement 3 is incorrect because telescopes still require an objective lens. Statement 4 is correct because combinations help reduce spherical and chromatic aberrations. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Eliminate all options containing Statement 3.

Final Logic:

�� Only Option A contains all correct statements.

🧠 Memory Trick

More Lenses = Better Image.

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

4 Choose the correct statements for a multi-lens combination:
1. The total magnification is strictly the algebraic sum of the individual magnifications.
2. The total magnification is m1 × m2 × m3.
3. The image formed by the first lens serves as the object for the second lens.
4. A combination of lenses is commonly used in designing camera lenses.

A 1, 3, 4

B 2, 3, 4

C 1, 2, 4

D 2 and 4 only

💡 Short Explanation

�� Magnifications multiply. �� Intermediate image acts as next object's source. �� Cameras use multiple lenses.

✅ Correct Answer Explanation

Statement 1 is incorrect because magnifications are not added. Statement 2 is correct because: m = m1 × m2 × m3 Statement 3 is correct because the image produced by one lens becomes the object for the next. Statement 4 is correct because modern camera systems use multiple lens elements. Therefore Statements 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 1.
�� Option C → Includes incorrect Statement 1.
�� Option D → Omits correct Statement 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect statement first.

Final Logic:

�� Statement 1 is false; Option B remains.

🧠 Memory Trick

Magnifications Multiply.

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

5 If a ray of light is incident on a prism such that it undergoes minimum deviation Dm, the relationship between the angle of incidence i, prism angle A, and minimum deviation Dm evaluates to:

A /2

B i = (A + Dm)/2

C i = A + Dm

D i = (Dm -

💡 Short Explanation

�� At minimum deviation, i = e. �� The path is symmetric. �� Standard prism relation is used.

✅ Correct Answer Explanation

At minimum deviation: i = e Also: Dm = i + e - A Substituting i = e: Dm = 2i - A Therefore: i = (A + Dm)/2 Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect sign.
�� Option C → Not obtained from prism formula.
�� Option D → Incorrect rearrangement.

🎯 Strategy

Used

Substitution

Application:

�� Apply the minimum deviation condition i = e.

Final Logic:

�� Simplification gives i = (A + Dm)/2.

🧠 Memory Trick

Minimum Deviation ⇒ i = e.

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

6 Identify the incorrect statement about the emergent ray from a glass prism in air:

A It bends away from the normal as it exits the glass into the rarer air medium.

B The total deviation depends on the angles of incidence and emergence.

C It will always be parallel to the incident ray regardless of the angle of incidence.

D At minimum deviation, the angle of emergence is exactly equal to the angle of incidence.

💡 Short Explanation

�� A prism deviates light. �� Parallel emergence occurs in glass slabs. �� Emergent and incident rays are generally not parallel.

✅ Correct Answer Explanation

Statement A is correct because light bends away from the normal while moving from glass to air. Statement B is correct because deviation depends on incidence and emergence conditions. Statement C is incorrect because a prism causes angular deviation and the emergent ray is generally not parallel to the incident ray. Statement D is correct because at minimum deviation: i = e Hence Option C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct statement.
�� Option D → Correct property of minimum deviation.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the statement that describes a glass slab instead of a prism.

Final Logic:

�� Parallel emergence is characteristic of a glass slab.

🧠 Memory Trick

Slab = Parallel, Prism = Deviation.

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

7 Prism deviation profile: The plot of the angle of deviation versus the angle of incidence for a prism takes the shape of a ____________ curve, indicating that any given value of deviation (except minimum) corresponds to ____________ possible values of the angle of incidence.

A Linear, Infinite

B U-shaped, Two

C Exponential, Single

D Hyperbolic, Three

💡 Short Explanation

�� Deviation first decreases then increases. �� A minimum deviation exists. �� Two incidence angles can produce the same deviation.

✅ Correct Answer Explanation

The deviation-incidence graph for a prism is U-shaped. At the lowest point, deviation becomes minimum. For every deviation greater than the minimum value, there are two corresponding angles of incidence due to symmetry. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Graph is not linear.
�� Option C → Not exponential and not single-valued.
�� Option D → Not hyperbolic.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the standard prism deviation graph.

Final Logic:

�� U-shaped graph gives two incidence angles for one deviation value.

🧠 Memory Trick

Prism Graph = U Curve.

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

8 A ray of light passes through an equilateral prism (A = 60°). If the angle of incidence is 45° and the angle of emergence is observed to be 55°, what is the total angle of deviation produced by the prism?

A 30°

B 40°

C 50°

D 100°

💡 Short Explanation

�� Use prism deviation formula. �� D = i + e - A. �� Substitute the values directly.

✅ Correct Answer Explanation

Given: i = 45° e = 55° A = 60° Using: D = i + e - A D = 45 + 55 - 60 D = 40° Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect calculation.
�� Option C → Arithmetic error.
�� Option D → Not obtained using prism formula.

🎯 Strategy

Used

Substitution

Application:

�� Directly substitute the given values.

Final Logic:

�� D = 45 + 55 − 60 = 40°.

🧠 Memory Trick

D = i + e − A

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

9 When a prism is placed exactly in the position of minimum deviation, a unique internal geometry is established. In this specific state:

A the ray inside the prism bends directly toward the apex

B the angle of refraction at the first face is twice the angle of incidence

C the refracted ray inside the prism becomes parallel to its base

D total internal reflection occurs at the second interface

💡 Short Explanation

�� Minimum deviation gives symmetric path. �� Internal ray becomes parallel to the prism base. �� Refraction angles become equal.

✅ Correct Answer Explanation

At minimum deviation: i = e r1 = r2 = A/2 Under this condition, the refracted ray inside the prism becomes parallel to the base of the prism. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the defining condition.
�� Option B → No such prism relation exists.
�� Option D → Minimum deviation does not require total internal reflection.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the standard geometry at minimum deviation.

Final Logic:

�� Symmetric refraction makes the internal ray parallel to the base.

🧠 Memory Trick

Minimum Deviation = Base Parallel Ray.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Minimum Deviation Position

10 In the context of the deviation of light through a prism, if the angle of incidence is carefully adjusted such that it equals the angle of emergence:

A the prism must be undergoing total internal reflection at both faces

B the deviation angle has reached its minimum possible value Dm

C the emergent ray will be directed straight back towards the source

D the angle of deviation will automatically drop to zero

💡 Short Explanation

�� Minimum deviation occurs when i = e. �� The light path becomes symmetric. �� Deviation becomes minimum, not zero.

✅ Correct Answer Explanation

For a prism at minimum deviation: i = e and r1 = r2 This corresponds to a symmetrical path through the prism. Under this condition, the deviation becomes the minimum deviation Dm. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No total internal reflection is implied.
�� Option C → Light does not retrace its path.
�� Option D → Minimum deviation is generally non-zero.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the condition for minimum deviation.

Final Logic:

�� i = e directly indicates minimum deviation.

🧠 Memory Trick

i = e ⇒ D = Dm

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Minimum Deviation Condition

11 An equilateral glass prism (A = 60°) produces a minimum deviation of 60°. What is the precise refractive index of the glass material?

A 1.414

B 1.500

C 1.732

D 2.000

💡 Short Explanation

�� Use the prism formula for minimum deviation. �� A = 60°, Dm = 60°. �� Evaluate refractive index directly.

✅ Correct Answer Explanation

For a prism: n = sin[(A + Dm)/2] / sin(A/2) Substituting: n = sin[(60 + 60)/2] / sin(60/2) n = sin 60° / sin 30° n = (√3/2) / (1/2) n = √3 n = 1.732 Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Value corresponds to √2, not obtained here.
�� Option B → Incorrect calculation.
�� Option D → Refractive index is overestimated.

🎯 Strategy

Used

Substitution

Application:

�� Substitute A and Dm into the standard prism formula.

Final Logic:

�� n = sin60°/sin30° = 1.732.

🧠 Memory Trick

60–60 Prism ⇒ n = √3.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Refractive Index of a Prism

12 The refractive index formula: n21 = sin[(A + Dm)/2] / sin(A/2), is highly practical in experimental physics because:

A it does not require a light source to determine the index accurately

B the angles A and Dm can be easily and accurately measured using experimental setups

C it is universally valid for thick lenses as well as thin prisms

D it functions identically regardless of the surrounding medium's refractive index

💡 Short Explanation

�� Prism angle and minimum deviation are measurable. �� Experimental determination becomes easy. �� Widely used in laboratories.

✅ Correct Answer Explanation

The formula depends only on two measurable quantities: • Prism angle (A) • Minimum deviation (Dm) Both can be measured accurately using spectrometers. Therefore refractive index can be determined experimentally with good precision. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → A light source is required.
�� Option C → Formula is for prisms, not lenses.
�� Option D → Relative refractive index depends on surrounding medium.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate statements that contradict experimental optics.

Final Logic:

�� Only Option B correctly explains the practical usefulness of the formula.

🧠 Memory Trick

Measure A and Dm → Get n.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Experimental Determination of Refractive Index

13 Consider the statements regarding a thin prism. Choose the correct statements:
1. The refracting angle A is extremely large.
2. The minimum deviation Dm produced is negligible or very small.
3. The approximation tan θ ≈ θ or sin θ ≈ θ is utilized in deriving its deviation.
4. Thin prisms do not deviate light much.

A 1, 2, 3

B 2, 3, 4

C 1, 3, 4

D 1 and 4 only

💡 Short Explanation

�� Thin prisms have small refracting angles. �� Small-angle approximations are used. �� Deviation produced is small.

✅ Correct Answer Explanation

Statement 1 is incorrect because a thin prism has a small refracting angle. Statement 2 is correct because thin prisms produce small deviations. Statement 3 is correct because small-angle approximations are used in derivation. Statement 4 is correct because deviation is very small. Therefore Statements 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 1.
�� Option C → Includes incorrect Statement 1.
�� Option D → Includes incorrect Statement 1 and omits correct Statements 2 and 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect statement first.

Final Logic:

�� Thin prisms have small A, making Statement 1 false.

🧠 Memory Trick

Thin Prism = Small A = Small D.

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

14 If a thin prism of angle A and refractive index ng is immersed in a liquid of refractive index nl (where ng > nl), the relative refractive index n21 changes. The new angle of minimum deviation Dm' is given by:

A Dm' = (ng - 1)A

B Dm' = ((ng/nl) - 1)A

C Dm' = (nl - ng)A

D Dm' = ((nl/ng) - 1)A

💡 Short Explanation

�� Relative refractive index must be used. �� Thin prism deviation depends on relative index. �� Medium affects deviation.

✅ Correct Answer Explanation

For a thin prism: Dm = (nrel − 1)A When immersed in a liquid: nrel = ng/nl Therefore: Dm' = ((ng/nl) − 1)A Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Valid only in air.
�� Option C → Wrong expression.
�� Option D → Incorrect relative refractive index.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall thin prism deviation in a medium.

Final Logic:

�� Replace n by relative refractive index ng/nl.

🧠 Memory Trick

Medium Present → Use Relative Index.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Thin Prism in a Medium

15 Choose the correct statements regarding material interaction with light:
1. Silica glass fibers can transmit more than 95% of light over 1 km due to specific material purification and preparation.
2. A monochromatic light may produce an entirely different color perception compared to white light on the same object.
3. Thick glass lenses perfectly focus all constituent colors of white light to a single, identical focal point.
4. The core of an optical fibre has a higher refractive index than its cladding.

A 1, 2, and 4

B 2, 3, and 4

C 1, 3, and 4

D 1, 2, and 3

💡 Short Explanation

�� Optical fibres have low loss. �� Colour perception depends on illumination. �� Core index exceeds cladding index.

✅ Correct Answer Explanation

Statement 1 is correct because modern silica fibres have extremely low transmission loss. Statement 2 is correct because object colour depends on incident light. Statement 3 is incorrect because chromatic aberration causes different colours to focus at different points. Statement 4 is correct because total internal reflection requires: ncore > ncladding Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect statement regarding chromatic aberration.

Final Logic:

�� Statement 3 is false because different colours have different focal lengths.

🧠 Memory Trick

Fibre: Core > Cladding.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Optical Fibres and Chromatic Aberration

16 When applying the universal laws of reflection or refraction to a highly curved spherical interface:

A the entire surface is broadly treated as a single uniform plane

B an infinitesimal part of the surface is regarded as planar, and the normal passes through the center of curvature

C the angles must always be measured relative to the principal axis rather than the local normal

D Snell's law must be fundamentally altered to account for the continuous curvature

💡 Short Explanation

�� Small surface element behaves like a plane. �� Normal passes through center of curvature. �� Snell's law remains valid.

✅ Correct Answer Explanation

For a spherical surface, a very small portion near the point of incidence can be approximated as a plane surface. The normal at that point is the radius joining the point to the center of curvature. Thus reflection and refraction laws remain applicable. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Entire curved surface cannot be treated as a plane.
�� Option C → Angles are measured from the normal.
�� Option D → Snell's law remains unchanged.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall treatment of spherical refracting surfaces.

Final Logic:

�� Local plane approximation makes Snell's law valid.

🧠 Memory Trick

Small Patch = Plane Surface.

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

17 Identify the incorrect statement about observing a real image formed by a lens:

A An image can only be verified to exist if a screen is present to capture it.

B The screen's primary physical role is simply to diffuse the focused rays so they can reach the observer's eye.

C A real image consists of rays actually converging to a distinct point in space.

D When a screen is removed, the rays still converge to and then diverge from the exact image location.

💡 Short Explanation

�� Real images exist even without screens. �� Rays physically converge. �� Screen only helps viewing.

✅ Correct Answer Explanation

A real image forms wherever light rays actually converge. The image exists even when a screen is absent. The screen merely scatters the converged light toward the observer. Therefore Statement A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct function of a screen.
�� Option C → Correct definition of a real image.
�� Option D → Rays still converge even after screen removal.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the physical meaning of a real image.

Final Logic:

�� Real image existence does not depend on a screen.

🧠 Memory Trick

Real Image Exists First, Screen Later.

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

18 Match List I (Image Scenario) with List II (Explanation)

List I	List II
1. Real image without screen	a. No distinct geometric image point is formed
2. Virtual image	b. Dust particles in air act as diffusers to reveal the image
3. Irregular reflection (e.g. book page)	c. Rays actually converge in space but need diffusion to be easily seen
4. Laser show in air	d. Rays appear to diverge from a point, not actually meeting there

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

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

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

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

💡 Short Explanation

�� Real images involve actual convergence. �� Virtual images involve apparent divergence. �� Dust particles reveal light paths.

✅ Correct Answer Explanation

1 → c because real image rays actually meet. 2 → d because virtual image rays only appear to originate from a point. 3 → a because irregular reflection destroys a distinct image. 4 → b because dust particles scatter laser light making the path visible. Hence: 1-c, 2-d, 3-a, 4-b

❓ Why Other Options Are Incorrect

�� Option B → Multiple incorrect pairings.
�� Option C → Incorrect matching of 3 and 4.
�� Option D → Multiple incorrect assignments.

🎯 Strategy

Used

Concept Matching

Application:

�� Match each optical phenomenon with its defining property.

Final Logic:

�� Only Option A gives all correct pairings.

🧠 Memory Trick

Real Meet, Virtual Seem.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Real Images, Virtual Images and Diffuse Reflection

19 The phenomenon of dispersion implies that the refractive index of a medium is a ____________ property, which directly dictates the ____________ of white light as it passes through a prism.

A Constant, Bending

B Wavelength-dependent, Splitting

C Density-independent, Reflection

D Temperature-dependent, Absorption

💡 Short Explanation

�� Different colours have different wavelengths. �� Refractive index varies with wavelength. �� This causes dispersion.

✅ Correct Answer Explanation

Dispersion occurs because refractive index depends on wavelength. Different colours travel with different speeds in a medium and refract by different amounts. As a result, white light splits into constituent colours. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Refractive index is not constant for all wavelengths.
�� Option C → Reflection is not dispersion.
�� Option D → Absorption is unrelated to colour splitting.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the cause of dispersion.

Final Logic:

�� Wavelength dependence causes splitting of white light.

🧠 Memory Trick

Different λ → Different n → Dispersion.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Dispersion Through Prism

20 If a thick lens defect is modeled mathematically by a combination of two thin lenses in contact to achieve a net focal length of 40 cm, and one of the lenses has a focal length of -60 cm, what must be the focal length of the second lens to achieve this combined focus?

A 24 cm

B -24 cm

C 120 cm

D 100 cm

💡 Short Explanation

�� Use lens combination formula. �� One focal length is known. �� Solve for the unknown focal length.

✅ Correct Answer Explanation

For lenses in contact: 1/F = 1/f1 + 1/f2 Given: F = 40 cm f1 = -60 cm Therefore: 1/40 = -1/60 + 1/f2 1/f2 = 1/40 + 1/60 1/f2 = (3 + 2)/120 1/f2 = 5/120 1/f2 = 1/24 f2 = 24 cm Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Wrong sign.
�� Option C → Does not satisfy lens combination equation.
�� Option D → Incorrect focal length calculation.

🎯 Strategy

Used

Substitution

Application:

�� Substitute known focal lengths into the lens combination formula.

Final Logic:

�� Solving 1/40 = -1/60 + 1/f2 gives f2 = 24 cm.

🧠 Memory Trick

Contact Lenses = Add Reciprocals.

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

CUET UG Physics Booster Test 3 -Lens Combinations and Prisms

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