CUET UG Physics Booster Test 2-Refractive Index -

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CUET UG Physics Booster Test 2-Refractive Index

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2-Refractive Index

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QUESTION 1 OF 20

Consider the statements on refraction phenomena. Choose the correct statements:
1. A part of light gets reflected back into the first medium.
2. The rest of the light enters the other medium.
3. The direction of propagation of an obliquely incident ray changes at the interface.
4. The entire beam is absorbed by the transparent medium.

Answer Review

1 Consider the statements on refraction phenomena. Choose the correct statements:
1. A part of light gets reflected back into the first medium.
2. The rest of the light enters the other medium.
3. The direction of propagation of an obliquely incident ray changes at the interface.
4. The entire beam is absorbed by the transparent medium.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Refraction occurs at the boundary of two media. �� Part of light is reflected and part transmitted. �� Direction changes for oblique incidence.

✅ Correct Answer Explanation

Statements 1 and 2 are correct because when light reaches the interface of two transparent media, part is reflected and part is transmitted. Statement 3 is correct because refraction involves a change in the direction of propagation of an obliquely incident ray. Statement 4 is incorrect because transparent media do not completely absorb the incident beam.

❓ 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 which statement contradicts the definition of refraction.

Final Logic: Refraction involves partial transmission, not complete absorption.

🧠 Memory Trick

"Reflect + Refract = Interface Effect."

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

2 The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. Thus, if n21 is the refractive index of the second medium with respect to the first medium, the expression for Snell's law will be

A sin i / sin r = n21

B sin r / sin i = n21

C sin i × sin r = n21

D cos i / cos r = n21

💡 Short Explanation

�� Snell's law relates incidence and refraction angles. �� Ratio of sines remains constant. �� Constant equals relative refractive index.

✅ Correct Answer Explanation

According to Snell's law, n21 = sin i / sin r where i is the angle of incidence and r is the angle of refraction. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reciprocal expression.
�� Option C → Product relation is incorrect.
�� Option D → Cosine ratio is not Snell's law.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the mathematical form of Snell's law.

Final Logic:

n21 = sin i / sin r

🧠 Memory Trick

"Snell = Sine Ratio."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Snell's Law

3 If n32 is the refractive index of medium 3 with respect to medium 2, and n31 is the refractive index of medium 3 with respect to medium 1, what will be the value of n32 if n31 is 1.5 and n12 is 0.8?

A 0.53

B 1.20

C 1.87

D 2.30

💡 Short Explanation

�� Relative refractive indices can be multiplied. �� Use index conversion relation. �� Simple substitution.

✅ Correct Answer Explanation

n32 = n31 × n12 n32 = 1.5 × 0.8 n32 = 1.20 Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect calculation.
�� Option C → Not obtained from index relation.
�� Option D → Incorrect multiplication.

🎯 Strategy

Used

Substitution

Application: Apply relative refractive-index relation.

Final Logic:

1.5 × 0.8 = 1.20

🧠 Memory Trick

"Chain Indices Multiply."

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

4 Choose the incorrect statement about optical density

A Optical density is the ratio of the speed of light in two media.

B Mass density of an optically denser medium may be less than that of an optically rarer medium.

C Mass density of turpentine is less than that of water but its optical density is higher.

D Optical density is strictly synonymous with mass per unit volume.

💡 Short Explanation

�� Optical density differs from mass density. �� It depends on refractive index. �� Higher mass density does not guarantee higher optical density.

✅ Correct Answer Explanation

Optical density refers to the extent to which a medium slows down light and is related to refractive index. It is not the same as mass density. Therefore Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Related to refractive-index concept.
�� Option B → Correct statement.
�� Option C → Standard NCERT example.

🎯 Strategy

Used

Odd One Out

Application: Distinguish optical density from physical density.

Final Logic: Optical density ≠ mass density.

🧠 Memory Trick

"Optical = Light, Not Weight."

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

5 Choose the correct statements about a rectangular slab:
1. Refraction takes place at two interfaces.
2. The emergent ray is parallel to the incident ray.
3. There is no overall angular deviation.
4. The emergent ray exhibits no lateral shift.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Refraction occurs at both surfaces. �� Emergent ray is parallel to incident ray. �� Lateral shift exists.

✅ Correct Answer Explanation

Statements 1, 2, and 3 are correct. A rectangular slab has two parallel surfaces. Refraction occurs at both surfaces and the emergent ray remains parallel to the incident ray. Statement 4 is incorrect because a lateral displacement (shift) occurs.

❓ 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: Recall slab properties.

Final Logic: Parallel emergence occurs with lateral shift.

🧠 Memory Trick

"Glass Slab: Shifted, Not Turned."

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

6 In a rectangular slab where the emergent ray is parallel to the incident ray,

A it suffers lateral displacement or shift with respect to the incident ray

B it completely retraces its original path backwards

C it bends by exactly 90 degrees at the second interface

D it reflects totally at the first interface

💡 Short Explanation

�� Emergent ray is parallel. �� Angular deviation is zero. �� Lateral shift exists.

✅ Correct Answer Explanation

In a rectangular slab, the two refracting surfaces are parallel. The emergent ray is parallel to the incident ray but displaced sideways. This displacement is called lateral shift.

❓ Why Other Options Are Incorrect

�� Option B → Retracing occurs only in special cases.
�� Option C → No 90° bending occurs.
�� Option D → Total internal reflection does not occur.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the effect of a parallel-sided slab.

Final Logic: Parallel emergence produces lateral displacement.

🧠 Memory Trick

"Parallel Out, Sideways Shift."

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

7 For viewing near the normal direction:

A the apparent depth is real depth divided by the refractive index of the medium

B the apparent depth is real depth multiplied by the refractive index

C the bottom of the tank appears to be lowered

D the apparent depth equals the real depth exactly

💡 Short Explanation

�� Apparent depth is less than real depth. �� Applicable near normal viewing. �� Water tank bottom appears raised.

✅ Correct Answer Explanation

Apparent Depth = Real Depth / μ where μ is the refractive index. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect formula.
�� Option C → Bottom appears raised, not lowered.
�� Option D → Apparent depth differs from real depth.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall apparent-depth formula.

Final Logic: Apparent depth is reduced by refractive index.

🧠 Memory Trick

"Depth Divides by μ."

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

8 Match List I with List II for a tank filled with water

List I	List II
1. Real depth	a. h₂
2. Apparent depth	b. h₁
3. Viewing direction	c. Near the normal
4. Refractive index	d. Divides the real depth to find apparent depth

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

�� Real depth is larger. �� Apparent depth is smaller. �� Refractive index reduces apparent depth.

✅ Correct Answer Explanation

1 → a : Real depth = h₂ 2 → b : Apparent depth = h₁ 3 → c : Observation is near the normal 4 → d : Refractive index divides real depth Thus: 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 the depth relation first.

Final Logic: Apparent depth = Real depth / μ.

🧠 Memory Trick

"Real Bigger, Apparent Smaller."

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

9 Reflection at the interface from denser to rarer medium

A Partly reflected, Partly refracted

B Totally reflected, Completely transmitted

C Partly reflected, Totally transmitted

D Totally reflected, Partly refracted

💡 Short Explanation

�� Ordinary incidence produces partial reflection. �� Partial transmission also occurs. �� Energy splits between two rays.

✅ Correct Answer Explanation

At a denser-rarer interface, for angles less than the critical angle, part of the light is reflected back while the remaining part is refracted into the rarer medium. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reflection and complete transmission cannot occur simultaneously.
�� Option C → Refraction is not total.
�� Option D → Total reflection occurs only in TIR.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall ordinary refraction behavior.

Final Logic: Light energy is divided into reflected and refracted components.

🧠 Memory Trick

"Before TIR, Light Splits."

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

10 Identify the incorrect statement about light entering a rarer medium from a denser medium

A It bends away from the normal.

B The angle of refraction is larger than the angle of incidence.

C The incident ray is partially reflected and partially transmitted.

D The refracted ray bends towards the normal.

💡 Short Explanation

�� Light speeds up in a rarer medium. �� It bends away from the normal. �� Refraction angle becomes larger.

✅ Correct Answer Explanation

When light travels from a denser medium to a rarer medium, it bends away from the normal. Therefore, r > i Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct behavior.
�� Option B → Correct consequence of bending away.
�� Option C → Correct for ordinary refraction.

🎯 Strategy

Used

Odd One Out

Application: Identify the statement opposite to denser-to-rarer refraction.

Final Logic: Denser → Rarer always bends away from the normal.

🧠 Memory Trick

"Dense to Rare → Away."

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Refraction from Denser to Rarer Medium

11 Choose the correct statements about the critical angle:
1. It is the angle of incidence corresponding to an angle of refraction of 90 degrees.
2. The refracted ray grazes the surface at the interface.
3. It occurs for a given pair of media.
4. It requires light to travel from a rarer to a denser medium.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Critical angle is defined when refraction angle becomes 90°. �� Refracted ray grazes the interface. �� It depends on the pair of media involved.

✅ Correct Answer Explanation

Statement 1 is correct because the critical angle is the angle of incidence for which the angle of refraction becomes 90°. Statement 2 is correct because at this condition the refracted ray travels along the boundary. Statement 3 is correct because each pair of media has its own critical angle. Statement 4 is incorrect because the critical angle exists only when light travels from a denser medium to a rarer medium.

❓ 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 physical conditions required for the critical angle.

Final Logic: Critical angle requires denser-to-rarer propagation.

🧠 Memory Trick

"90° Refraction = Critical Angle."

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

12 The maximum value of sin r is unity at the critical angle. Thus, if the angle of incidence is ic and the refractive index of the second medium with respect to the first is n21, the formula for the critical angle will be

A sin ic = n21

B cos ic = n21

C tan ic = n21

D sin ic = 1 / n21

💡 Short Explanation

�� At the critical angle, r = 90°. �� Therefore sin r = 1. �� Substitution into Snell's law gives the result.

✅ Correct Answer Explanation

Snell's law: sin i / sin r = n21 At the critical angle, i = ic, r = 90° Since sin 90° = 1 we get sin ic = n21 Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Critical angle is not related through cosine.
�� Option C → Tangent does not appear in the critical-angle relation.
�� Option D → Reciprocal relation is not valid for the definition used here.

🎯 Strategy

Used

Substitution

Application: Substitute r = 90° into Snell's law.

Final Logic:

sin ic = n21

🧠 Memory Trick

"Critical Angle → sin ic."

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

13 Consider the Conditions for total internal reflection. Choose the correct statements:
1. The angle of incidence must be larger than the critical angle.
2. Snell's law of refraction cannot be satisfied for these angles.
3. No refraction is possible.
4. Light must travel from a rarer to a denser medium.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� TIR occurs only beyond the critical angle. �� No refracted ray is produced. �� Light must travel from denser to rarer medium.

✅ Correct Answer Explanation

Statement 1 is correct because TIR requires Statement 2 is correct because Snell's law would require (\sin r>1), which is impossible. Statement 3 is correct because no refracted ray can exist. Statement 4 is incorrect because TIR requires light to travel from denser to rarer medium.

❓ 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 necessary conditions for TIR.

Final Logic: Only Statements 1, 2, and 3 are correct.

🧠 Memory Trick

"Beyond ic → Only Reflection."

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

14 In total internal reflection,

A no transmission of light takes place

B half of the light is transmitted into the rarer medium

C the reflected ray is always less intense than the incident ray

D light is completely absorbed by the boundary

💡 Short Explanation

�� Entire light is reflected back. �� No refracted ray exists. �� Energy remains in the denser medium.

✅ Correct Answer Explanation

In total internal reflection, all the incident light is reflected back into the denser medium. Therefore no transmission occurs into the rarer medium. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → No transmitted ray exists.
�� Option C → Reflected intensity is essentially equal to incident intensity.
�� Option D → Absorption does not define TIR.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the definition of total internal reflection.

Final Logic: TIR means zero transmission.

🧠 Memory Trick

"Total Reflection = Zero Transmission."

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

15 Prisms designed to bend light by 90 degrees or 180 degrees

A make use of total internal reflection

B require a critical angle greater than 45 degrees

C invert the image while changing its size

D utilize standard refraction only

💡 Short Explanation

�� Right-angle prisms employ TIR. �� TIR produces efficient reflection. �� Image size remains unchanged.

✅ Correct Answer Explanation

Prisms used for 90° and 180° deviation employ total internal reflection instead of ordinary reflection. This provides high efficiency and brightness. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → TIR requires critical angle less than 45°.
�� Option C → Prism inversion does not alter image size.
�� Option D → TIR, not ordinary refraction alone, is responsible.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall applications of TIR in prisms.

Final Logic: Large-angle bending in prisms uses TIR.

🧠 Memory Trick

"Prism Bending → TIR."

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

16 If a prism is designed to invert images without changing their size, and it relies on total internal reflection, what must be the upper limit for the critical angle ic for the material of the prism?

A Less than 45 degrees

B Exactly 90 degrees

C Greater than 45 degrees

D Exactly 60 degrees

💡 Short Explanation

�� TIR must occur at a 45° incidence. �� Critical angle must be smaller. �� This ensures total internal reflection.

✅ Correct Answer Explanation

For TIR at a prism face, i > ic In a right-angle prism, the incidence angle is typically 45°. Therefore, ic < 45° Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not required.
�� Option C → TIR would not occur.
�� Option D → Does not satisfy prism condition.

🎯 Strategy

Used

Substitution

Application: Compare prism incidence angle with critical angle.

Final Logic: (i_c) must be less than 45°.

🧠 Memory Trick

"45° Needs Smaller ic."

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

17 Match List I with List II for optical fibres

List I	List II
1. Material	a. High quality composite glass/quartz
2. Structure	b. Core and cladding
3. Core refractive index	c. Higher than that of the cladding
4. Cladding refractive index	d. Lower than that of the core

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

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

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

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

💡 Short Explanation

�� Optical fibres contain core and cladding. �� Core has higher refractive index. �� Quartz and composite glass are commonly used.

✅ Correct Answer Explanation

1 → a: Optical fibres are made of high-quality glass or quartz. 2 → b: Structure consists of core and cladding. 3 → c: Core refractive index is higher. 4 → d: Cladding refractive index is lower. Thus, 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching.
�� Option C → Multiple mismatches.
�� Option D → Incorrect fibre structure mapping.

🎯 Strategy

Used

Option Grouping

Application: Match structure and refractive-index properties first.

Final Logic: Core > Cladding for TIR.

🧠 Memory Trick

"Core High, Cladding Low."

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

18 Light traveling inside an optical fibre:

A Repeated reflections, No appreciable loss

B Single reflection, High loss

C Repeated reflections, High loss

D Single reflection, No appreciable loss

💡 Short Explanation

�� Optical fibres use repeated TIR. �� Energy loss is minimal. �� Signals travel long distances efficiently.

✅ Correct Answer Explanation

Light entering an optical fibre undergoes repeated total internal reflections at the core-cladding boundary. Due to very low absorption and scattering, there is no appreciable loss in intensity. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reflection is not single and losses are low.
�� Option C → Losses are not high.
�� Option D → Multiple reflections occur.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the working principle of optical fibres.

Final Logic: Repeated TIR ensures low-loss transmission.

🧠 Memory Trick

"Fiber = Repeated TIR."

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

19 Choose the incorrect statement about light pipes:

A They are formed by a single massive quartz crystal.

B They are used to facilitate visual examination of internal organs.

C They consist of a bundle of optical fibres.

D They can act as an optical pipe for signals.

💡 Short Explanation

�� Light pipes are bundles of optical fibres. �� Used in medical endoscopy. �� Not made from a single crystal.

✅ Correct Answer Explanation

Light pipes are bundles of optical fibres used to transmit light and images through curved paths. They are not formed from one large quartz crystal. Therefore, Option A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct application.
�� Option C → Correct structure.
�� Option D → Correct function.

🎯 Strategy

Used

Odd One Out

Application: Identify the statement inconsistent with fibre technology.

Final Logic: Light pipes consist of many fibres, not one crystal.

🧠 Memory Trick

"Many Fibres = Light Pipe."

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

20 Choose the correct statements about silica glass fibres
1. There should be very little absorption of light as it travels.
2. They are purified to minimize absorption over long distances.
3. They can transmit more than 95% of light over a 1 km length.
4. They are unable to transmit audio and video signals.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Silica fibres are highly transparent. �� Absorption losses are minimized. �� Audio and video signals can be transmitted.

✅ Correct Answer Explanation

Statements 1, 2, and 3 are correct because optical fibres are manufactured with extremely low absorption and attenuation. Statement 4 is incorrect because optical fibres are widely used to transmit audio, video, and digital communication signals.

❓ 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 each statement against fibre-optic communication principles.

Final Logic: Audio and video transmission is a major application of optical fibres.

🧠 Memory Trick

"Pure Silica = Low Loss, Long Distance."

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

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