CUET UG Physics Booster Test 3-Refractive Index -

🚀

CUET UG Physics Booster Test 3-Refractive Index

⏱️ 30 Minutes

❓ 10 Questions

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

CUET UG Physics Booster Test 3-Refractive Index

1 / 10

QUESTION 1 OF 10

When a beam of light encounters another transparent medium,

Answer Review

1 When a beam of light encounters another transparent medium,

A a part of light gets reflected back while the rest enters the other medium

B the entire beam always enters the other medium without any loss

C the direction of propagation remains unchanged for an obliquely incident ray

D the ray represents a single wave front with no direction change

💡 Short Explanation

�� At an interface, light generally splits into reflected and transmitted parts. �� Refraction changes direction for oblique incidence. �� Both reflection and refraction occur simultaneously.

✅ Correct Answer Explanation

When light strikes the boundary between two transparent media, a portion is reflected back into the first medium and the remaining portion enters the second medium. This is the basic phenomenon associated with refraction at an interface. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Complete transmission does not generally occur.
�� Option C → Oblique rays usually change direction.
�� Option D → Does not describe refraction at an interface.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall what happens at the boundary between two media.

Final Logic: Light is partly reflected and partly transmitted.

🧠 Memory Trick

"Interface = Reflect + Refract."

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

2 Identify the incorrect statement about Snell's Law

A The ratio of the sine of the angle of incidence to the sine of angle of refraction varies with the angle of incidence.

B The refractive index is a constant characteristic of the pair of media.

C If n₂₁ > 1, the refracted ray bends towards the normal.

D The incident ray, the refracted ray and the normal lie in the same plane.

💡 Short Explanation

�� Snell's law gives a constant ratio. �� Refractive index depends on media. �� Incident ray, refracted ray, and normal are coplanar.

✅ Correct Answer Explanation

According to Snell's law, sin i / sin r = n21 For a given pair of media, this ratio remains constant and does not vary with angle of incidence. Therefore Option A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct statement.
�� Option C → Correct when (n21 > 1).
�� Option D → First law of refraction.

🎯 Strategy

Used

Odd One Out

Application: Identify the statement contradicting Snell's law.

Final Logic: Ratio of sines is constant, not variable.

🧠 Memory Trick

"Snell = Constant 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. Thus, if a light ray passes through these three media consecutively, the relationship determining n₃₂ using n₃₁ and n₁₂ will be

A n₃₂ = n₃₁ × n₁₂

B n₃₂ = n₃₁ / n₁₂

C n₃₂ = n₃₁ + n₁₂

D n₃₂ = 1 / (n₃₁ × n₁₂)

💡 Short Explanation

�� Relative refractive indices combine multiplicatively. �� Intermediate medium acts as a link. �� Standard index-conversion relation.

✅ Correct Answer Explanation

n32 = n31 × n12 This relation is obtained from the definition of relative refractive index. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect relation.
�� Option C → Indices are not added.
�� Option D → Incorrect reciprocal form.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the relation between relative refractive indices.

Final Logic:

n32 = n31 n12

🧠 Memory Trick

"Chain Rule: Multiply Indices."

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

4 Choose the correct statements about optical and mass density
1. Optical density is the ratio of the speed of light in two media.
2. A medium with higher optical density may have a lower mass density.
3. The optical density of turpentine is higher than that of water.
4. Mass density of turpentine is higher than that of water.

A 2, 3, 4

B 1, 3, 4

C 1, 2, 3

D 1, 2, 4

💡 Short Explanation

�� Optical density is related to refractive index. �� Mass density and optical density are different. �� Turpentine is optically denser than water.

✅ Correct Answer Explanation

Statements 1, 2, and 3 are correct. Statement 4 is incorrect because the mass density of turpentine is lower than that of water.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 4.
�� Option A → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application: Compare optical density and mass density.

Final Logic: Higher optical density does not imply higher mass density.

🧠 Memory Trick

"Optical ≠ Mass Density."

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

5 Identify the correct statement regarding refraction through a rectangular slab:
1. The angle of emergence equals the angle of incidence.
2. The ray undergoes lateral shift.
3. Refraction occurs twice at two interfaces.
4. The emergent ray deviates significantly in angle.

A 1, 3, 4

B 1, 2, 3

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Refraction occurs at both faces. �� Emergent ray is parallel to incident ray. �� Lateral displacement occurs.

✅ Correct Answer Explanation

Statements 1, 2, and 3 are correct. The emergent ray remains parallel to the incident ray, so there is no angular deviation, although a lateral shift exists. Statement 4 is incorrect.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application: Recall rectangular slab properties.

Final Logic: Parallel emergence with lateral shift.

🧠 Memory Trick

"Shifted, Not Turned."

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

6 When a ray of light passes through a parallel-sided slab,

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

B the emergent ray completely disperses into a spectrum without any shift

C the extent of lateral shift is zero if incident obliquely

D the angle of deviation is 90 degrees

💡 Short Explanation

�� Emergent ray is parallel. �� Lateral shift occurs. �� No 90° deviation exists.

✅ Correct Answer Explanation

A parallel-sided slab produces lateral displacement of the ray while maintaining parallelism between incident and emergent rays. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Ordinary slab does not disperse light completely.
�� Option C → Oblique incidence produces lateral shift.
�� Option D → No such deviation occurs.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall the effect of a glass slab.

Final Logic: Parallel emergence causes lateral displacement.

🧠 Memory Trick

"Parallel Out, Shifted Sideways."

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

7 For viewing near the normal direction:

A Real depth is always less than apparent depth

B Apparent depth equals real depth multiplied by refractive index

C Real depth equals apparent depth divided by refractive index

D Apparent depth equals real depth divided by refractive index

💡 Short Explanation

�� Apparent depth is less than real depth. �� Formula applies near the normal. �� Water tank bottoms appear raised.

✅ Correct Answer Explanation

Apparent Depth = Real Depth / μ Hence Option D is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect formula.
�� Option C → Real depth = μ × Apparent depth.
�� Option A → Real depth is greater.

🎯 Strategy

Used

Direct Concept Recall

Application: Recall apparent-depth formula.

Final Logic: Refractive index divides the real depth.

🧠 Memory Trick

"Depth Divides by μ."

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

8 If a tank is filled with water to a real depth of 13.3 cm and the refractive index of water is 1.33, what will be the apparent depth of the bottom for viewing near the normal direction?

A 10.0 cm

B 17.6 cm

C 12.0 cm

D 1.0 cm

💡 Short Explanation

�� Use apparent-depth formula. �� Divide real depth by refractive index. �� Apparent depth is smaller.

✅ Correct Answer Explanation

Apparent Depth = Real Depth / μ13.3 / 1.33= 10.0 cm Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Greater than real depth.
�� Option C → Incorrect calculation.
�� Option D → Unrealistically small.

🎯 Strategy

Used

Substitution

Application: Apply the apparent-depth formula.

Final Logic:

13.3 / 1.33 = 10

🧠 Memory Trick

"Apparent = Real ÷ μ."

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

9 Match List I with List II for ray transitions from denser to rarer medium

List I	List II
1. Incident ray	a. Approaches the boundary between media
2. Reflected ray	b. Bounces back into the first medium
3. Refracted ray	c. Enters the second transparent medium
4. Interface	d. The boundary where internal reflection occurs

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

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

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

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

💡 Short Explanation

�� Incident ray approaches boundary. �� Reflected ray returns. �� Refracted ray enters second medium.

✅ Correct Answer Explanation

1 → a: Incident ray approaches boundary. 2 → b: Reflected ray bounces back. 3 → c: Refracted ray enters second medium. 4 → d: Interface is the boundary. Hence: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application: Match ray definitions first.

Final Logic: Each ray is matched to its physical behavior.

🧠 Memory Trick

"Incident-In, Reflected-Back, Refracted-Across."

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

10 Choose the correct statements about a ray transitioning from denser to rarer medium
1. The ray bends away from the normal.
2. The angle of refraction is larger than the angle of incidence.
3. As the incidence angle increases, the refraction angle reaches 90 degrees at the critical angle.
4. The maximum possible value for the refraction angle is 45 degrees.

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Denser to rarer means bending away. �� Refraction angle increases faster. �� At critical angle r = 90°

✅ Correct Answer Explanation

Statements 1, 2, and 3 are correct. At the critical angle, r = 90° Therefore, Statement 4 is incorrect because the maximum refraction angle is 90°, not 45°.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 4.
�� Option C → Includes incorrect Statement 4.
�� Option A → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application: Use the definition of critical angle.

Final Logic: Maximum refraction angle is 90°.

🧠 Memory Trick

"Critical Angle → r = 90°."

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

CUET UG Physics Booster Test 3-Refractive Index

Test Complete!

Answer Review