Apparent Depth in Multi-Layered Immiscible Liquids: Formulating and solving the apparent shift and depth of an object kept at the bottom of a container filled with multiple immiscible liquid layers of different refractive indices.

Apparent Depth in Multi-Layered Immiscible Liquids

Apparent Depth in Multiple Liquid Layers

When an object is placed at the bottom of a transparent container containing several immiscible liquids, light travels successively through different liquid layers before emerging into air. Due to refraction, the object appears raised and its observed depth is less than its actual depth.

Object at Bottom ↓ Liquid Layer 1 ↓ Liquid Layer 2 ↓ Liquid Layer 3 ↓ Air ↓ Eye

Basic Formula

For a single liquid:

Apparent Depth = Real Depth / μ

Multiple Liquid Layers

For nearly normal viewing through multiple parallel layers:

Apparent Depth = h₁/μ₁ + h₂/μ₂ + h₃/μ₃ + ... + hₙ/μₙ

Here h₁, h₂, h₃ ... are the actual thicknesses of the respective layers, and μ₁, μ₂, μ₃ ... are their refractive indices with respect to air.

Apparent Shift = Real Depth − Apparent Depth

Shift = (h₁+h₂+...+hₙ) − (h₁/μ₁+h₂/μ₂+...+hₙ/μₙ)

Important Condition

The simple additive formula above applies to normal or paraxial viewing through plane-parallel interfaces. For oblique viewing, the exact apparent position must be obtained using Snell's law and ray tracing.

Snell's Law: μ₁ sin i = μ₂ sin r

Key Idea

Each layer contributes an apparent thickness equal to its actual thickness divided by its refractive index. Therefore the total apparent depth is the sum of the apparent thicknesses of all layers.

Q1. The apparent depth of an object in a liquid is generally:
A. Greater than real depth
B. Equal to real depth
C. Less than real depth
D. Zero

Answer: C

Due to refraction at the liquid-air interface, the object appears raised. Therefore apparent depth is less than real depth.

Q2. For a single liquid of refractive index μ and depth h, apparent depth is:
A. hμ
B. h/μ
C. μ/h
D. h+μ

Answer: B

Apparent depth = h/μ.

Q3. Two liquid layers have thicknesses h₁ and h₂ and refractive indices μ₁ and μ₂. Apparent depth is:
A. h₁μ₁+h₂μ₂
B. h₁/μ₁+h₂/μ₂
C. μ₁/ h₁ + μ₂/h₂
D. h₁+h₂

Answer: B

Each layer contributes h/μ to apparent depth.

Q4. If μ = 1.5 and real depth = 30 cm, apparent depth is:
A. 15 cm
B. 20 cm
C. 30 cm
D. 45 cm

Answer: B

Apparent depth = 30/1.5 = 20 cm.

Q5. Real depth is 40 cm and apparent depth is 25 cm. Apparent shift is:
A. 10 cm
B. 15 cm
C. 25 cm
D. 65 cm

Answer: B

Shift = 40 − 25 = 15 cm.

Q6. In multiple layers, the apparent thickness of a layer h having refractive index μ is:
A. hμ
B. h/μ
C. μ/h
D. h+μ

Answer: B

The contribution of each layer is h/μ.

Q7. If a liquid has refractive index 1, its apparent depth is:
A. h/2
B. h
C. 2h
D. Zero

Answer: B

Apparent depth = h/1 = h. No refraction occurs when the indices are equal.

Q8. If refractive index of a liquid increases, its apparent depth:
A. Increases
B. Decreases
C. Remains same
D. Becomes infinite

Answer: B

Apparent depth = h/μ. Increasing μ decreases apparent depth.

Q9. For two layers h₁ = 10 cm, μ₁ = 2 and h₂ = 20 cm, μ₂ = 1.5, apparent depth is:
A. 15 cm
B. 18.33 cm
C. 25 cm
D. 30 cm

Answer: B

10/2 + 20/1.5 = 5 + 13.33 = 18.33 cm.

Q10. For the above system, real depth is 30 cm. The apparent shift is:
A. 11.67 cm
B. 18.33 cm
C. 30 cm
D. 48.33 cm

Answer: A

Shift = 30 − 18.33 = 11.67 cm.

Q11. The phenomenon responsible for apparent depth is:
A. Reflection
B. Refraction
C. Diffraction
D. Polarisation

Answer: B

Apparent depth is caused by refraction at the interface.

Q12. Immiscible liquids are liquids which:
A. Mix completely
B. Do not mix
C. Have same density
D. Have same refractive index

Answer: B

Immiscible liquids remain as separate layers.

Q13. In a three-layer system, apparent depth is:
A. Sum of actual depths
B. Sum of hᵢ/μᵢ
C. Product of all μ
D. Difference of indices

Answer: B

For normal/paraxial viewing, apparent depth = Σ(hᵢ/μᵢ).

Q14. If all three layers have the same refractive index μ, total apparent depth is:
A. Hμ
B. H/μ
C. μ/H
D. H+μ

Answer: B

If total actual depth H = h₁+h₂+h₃, apparent depth = H/μ.

Q15. If all liquid layers have refractive index equal to air, apparent shift is:
A. Maximum
B. Zero
C. Infinite
D. Negative

Answer: B

No refractive bending occurs, so apparent depth equals real depth.

Q16. A 20 cm layer has μ = 1.25. Its apparent thickness is:
A. 10 cm
B. 16 cm
C. 20 cm
D. 25 cm

Answer: B

20/1.25 = 16 cm.

Q17. A 30 cm layer has μ = 1.5. Apparent thickness is:
A. 15 cm
B. 20 cm
C. 30 cm
D. 45 cm

Answer: B

30/1.5 = 20 cm.

Q18. For normal viewing through parallel liquid layers, lateral displacement is:
A. Important in the simple depth formula
B. Not included in the simple apparent-depth formula
C. Always infinite
D. Equal to depth

Answer: B

The simple formula calculates apparent axial depth. Oblique viewing requires complete ray tracing.

Q19. At normal incidence, the ray:
A. Bends strongly
B. Does not change direction
C. Reflects completely
D. Stops

Answer: B

At normal incidence i = 0°, therefore r = 0°.

Q20. If actual depth is H and apparent depth is H', apparent shift is:
A. H+H'
B. H−H'
C. HH'
D. H/H'

Answer: B

Shift = real depth − apparent depth.

Q21. A liquid layer of 12 cm and μ = 1.2 has apparent thickness:
A. 8 cm
B. 10 cm
C. 12 cm
D. 14.4 cm

Answer: B

12/1.2 = 10 cm.

Q22. A liquid layer of 15 cm and μ = 1.5 contributes apparent thickness:
A. 5 cm
B. 10 cm
C. 15 cm
D. 22.5 cm

Answer: B

15/1.5 = 10 cm.

Q23. Which layer contributes the smallest apparent thickness for equal physical thickness?
A. Lowest μ
B. Highest μ
C. μ = 1
D. All same

Answer: B

Apparent thickness = h/μ. Higher μ gives smaller contribution.

Q24. The apparent depth is always less than real depth when:
A. μ > 1 for every liquid layer
B. μ < 1 for every layer
C. μ = 0
D. No interface exists

Answer: A

For ordinary liquids relative to air, μ > 1, so each h/μ < h.

Q25. The correct general expression for multiple layers is:
A. Σhᵢμᵢ
B. Σhᵢ/μᵢ
C. Πhᵢ/μᵢ
D. Σμᵢ/hᵢ

Answer: B

Each layer contributes hᵢ/μᵢ.

Q26. If h₁ = h₂ = 10 cm and μ₁ = μ₂ = 2, apparent depth is:
A. 5 cm
B. 10 cm
C. 20 cm
D. 40 cm

Answer: B

10/2 + 10/2 = 5 + 5 = 10 cm.

Q27. If apparent depth is 12 cm and μ = 1.5, real depth is:
A. 8 cm
B. 12 cm
C. 18 cm
D. 24 cm

Answer: C

Apparent depth = real depth/μ.
Real depth = 12 × 1.5 = 18 cm.

Q28. In a multilayer liquid system, which quantity is additive for normal viewing?
A. Refractive index
B. Apparent thickness contributions
C. Optical power only
D. Angles

Answer: B

The total apparent depth is the sum of hᵢ/μᵢ contributions.

Q29. For oblique observation, the simple h/μ formula:
A. Always gives exact result
B. May not give exact apparent position
C. Is independent of angle
D. Is never useful

Answer: B

At oblique incidence, Snell's law must be applied at each interface.

Q30. The safest method for a difficult multilayer problem is:
A. Guessing
B. Ray diagram + Snell's law
C. Ignoring interfaces
D. Using only total depth

Answer: B

A ray diagram and Snell's law correctly track the light through every interface.

A: Both Assertion and Reason are true and Reason correctly explains Assertion.
B: Both are true but Reason is not the correct explanation.
C: Assertion is true but Reason is false.
D: Assertion is false but Reason is true.

AR1. Assertion: An object under liquid appears raised.
Reason: Light refracts at the liquid-air interface.

Answer: A

AR2. Assertion: Apparent depth is smaller than real depth for ordinary liquids.
Reason: Refractive index of ordinary liquids relative to air is greater than 1.

Answer: A

AR3. Assertion: For a single liquid, apparent depth = h/μ.
Reason: The apparent depth decreases as refractive index increases.

Answer: B

AR4. Assertion: In multiple layers, each layer contributes h/μ to apparent depth.
Reason: The total apparent depth is the sum of apparent thicknesses of individual layers for normal/paraxial viewing.

Answer: A

AR5. Assertion: Apparent shift is real depth minus apparent depth.
Reason: The apparent image is generally closer to the observer.

Answer: A

AR6. Assertion: Increasing liquid refractive index increases apparent depth.
Reason: Apparent depth is h/μ.

Answer: D

AR7. Assertion: Immiscible liquids form separate layers.
Reason: They do not mix appreciably under ordinary conditions.

Answer: A

AR8. Assertion: The simple multilayer formula is exact for every oblique viewing angle.
Reason: Snell's law must be considered at each interface for oblique rays.

Answer: D

AR9. Assertion: A layer with larger refractive index contributes less apparent thickness for the same physical thickness.
Reason: Apparent thickness = h/μ.

Answer: A

AR10. Assertion: Drawing a ray diagram is useful in multilayer problems.
Reason: It helps identify refraction at every interface.

Answer: A

1 Mark Questions

Q1. Define apparent depth.

The depth at which an object appears to be located when viewed through a refracting medium is called apparent depth.

Q2. Write the formula for apparent depth in a single liquid.

Apparent depth = h/μ.

Q3. Define apparent shift.

Shift = Real depth − Apparent depth.

Q4. What are immiscible liquids?

Liquids that do not mix appreciably and remain as separate layers.

Q5. State Snell's law.

μ₁ sin i = μ₂ sin r.

2 Mark Questions

Q6. Why does an object under water appear raised?

Light bends away from the normal when it travels from water to air. The backward extension of the refracted ray makes the object appear closer to the surface.

Q7. A liquid has μ = 1.5 and depth 30 cm. Find apparent depth.

d' = d/μ = 30/1.5 = 20 cm.

Q8. Define apparent shift and write its formula.

Apparent shift = actual depth − apparent depth.

Q9. What happens to apparent depth when μ increases?

Since d' = d/μ, apparent depth decreases.

Q10. Why must every liquid layer be considered separately?

Each layer has its own thickness and refractive index, so each contributes a different apparent thickness hᵢ/μᵢ.

3 Mark Questions

Q11. Derive the multilayer apparent-depth formula for two layers.

For layer 1, apparent thickness = h₁/μ₁.
For layer 2, apparent thickness = h₂/μ₂.
Therefore:

d' = h₁/μ₁ + h₂/μ₂.

Q12. A 10 cm layer has μ = 2 and a 20 cm layer has μ = 1.5. Find apparent depth.

d' = 10/2 + 20/1.5
= 5 + 13.33
= 18.33 cm.

Q13. Find apparent shift for the above system.

Real depth = 10 + 20 = 30 cm.
Apparent depth = 18.33 cm.
Shift = 30 − 18.33 = 11.67 cm.

Q14. Explain the role of refractive index in apparent depth.

Apparent thickness is inversely proportional to refractive index: h' = h/μ. Therefore higher μ produces smaller apparent thickness.

Q15. Explain why the simple formula may fail for oblique viewing.

At oblique incidence, rays undergo different angular deviations at every interface. Therefore Snell's law and complete ray tracing are required.

4 Mark Questions

Q16. Three liquid layers have: h₁ = 10 cm, μ₁ = 1.2; h₂ = 20 cm, μ₂ = 1.5; h₃ = 30 cm, μ₃ = 2. Find apparent depth.

d' = 10/1.2 + 20/1.5 + 30/2
= 8.33 + 13.33 + 15
= 36.66 cm approximately.

Q17. Find apparent shift for Q16.

Real depth = 10 + 20 + 30 = 60 cm.
Apparent depth ≈ 36.66 cm.
Shift ≈ 60 − 36.66 = 23.34 cm.

Q18. Derive the expression for apparent shift in multilayer liquids.

Real depth = Σhᵢ.
Apparent depth = Σ(hᵢ/μᵢ).
Therefore:

Shift = Σhᵢ − Σ(hᵢ/μᵢ).

Q19. Explain the ray path for an object at the bottom of a multilayer container.

The ray travels through the bottom layer, refracts at each liquid-liquid interface and finally refracts at the upper liquid-air interface. The backward extensions of the emergent rays locate the apparent image.

Q20. A 40 cm liquid layer has μ = 1.6. Find apparent depth and shift.

Apparent depth = 40/1.6 = 25 cm.
Shift = 40 − 25 = 15 cm.

5 Mark Questions

Q21. Derive apparent depth for three immiscible liquid layers.

Let the actual thicknesses be h₁, h₂ and h₃ and refractive indices μ₁, μ₂ and μ₃.

The apparent thickness of each layer is: h₁/μ₁, h₂/μ₂ and h₃/μ₃.

Hence:
d' = h₁/μ₁ + h₂/μ₂ + h₃/μ₃.

Q22. Two layers are 15 cm and 25 cm thick with refractive indices 1.5 and 1.25. Find apparent depth.

d' = 15/1.5 + 25/1.25
= 10 + 20
= 30 cm.

Real depth = 40 cm.
Shift = 40 − 30 = 10 cm.

Q23. Explain how refractive index affects apparent shift.

For one layer: Shift = h − h/μ = h(1 − 1/μ).
Thus increasing μ increases the shift for a fixed physical thickness. For multiple layers the shifts from individual layers add under the normal/paraxial approximation.

Q24. Explain the importance of immiscibility in this problem.

Immiscibility keeps the liquids separated into distinct layers. Therefore each layer can be assigned a definite thickness and refractive index, allowing the total apparent depth to be calculated layer by layer.

Q25. Explain the difference between real depth and apparent depth.

Real depth is the actual physical distance from the surface to the object. Apparent depth is the position at which the object appears due to refraction. For ordinary liquids viewed from air, apparent depth is smaller.

6 Mark Questions

Q26. Derive the general formula for n immiscible liquid layers.

Let the thicknesses be: h₁, h₂, ..., hₙ and refractive indices: μ₁, μ₂, ..., μₙ.

Each layer contributes an apparent thickness: hᵢ/μᵢ.

Therefore total apparent depth is:

d' = Σ(hᵢ/μᵢ)

Actual depth:
d = Σhᵢ

Hence apparent shift:
S = Σhᵢ − Σ(hᵢ/μᵢ).

Q27. A container contains three immiscible liquids of thickness 10 cm, 20 cm and 30 cm with refractive indices 1.2, 1.5 and 2 respectively. Find apparent depth and apparent shift.

Apparent depth:

d' = 10/1.2 + 20/1.5 + 30/2
= 8.33 + 13.33 + 15
≈ 36.66 cm.

Actual depth = 60 cm.
Shift = 60 − 36.66
≈ 23.34 cm.

Q28. Explain how to solve an oblique-view multilayer problem.

Step 1: Draw the ray from the object.
Step 2: Apply Snell's law at the first interface.
Step 3: Apply Snell's law at every subsequent interface.
Step 4: Determine the emergent ray in air.
Step 5: Extend the emergent ray backward.
Step 6: Its intersection with the appropriate apparent ray gives the apparent position.

For oblique viewing, the simple Σ(h/μ) relation should not be blindly treated as an exact general formula.

Q29. A liquid system has total actual depth 50 cm and apparent depth 32 cm. Find apparent shift and average effective refractive index if the system is represented by a single equivalent layer.

Shift = 50 − 32 = 18 cm.

For an equivalent single layer: d' = d/μ_eff.
32 = 50/μ_eff.
μ_eff = 50/32 ≈ 1.5625.

Q30. Discuss all factors affecting apparent depth in a multilayer immiscible-liquid system.

The apparent depth depends on:

1. Thickness of every liquid layer.
2. Refractive index of every liquid.
3. Refractive index of the observing medium.
4. Angle of observation.
5. Orientation and parallelism of interfaces.
6. Refraction at every interface.

For normal/paraxial viewing through parallel layers: d' = Σ(hᵢ/μᵢ).

For oblique viewing or non-parallel interfaces, complete Snell-law based ray tracing is required.

Single liquid:
d' = h/μ

Multiple parallel layers:
d' = h₁/μ₁ + h₂/μ₂ + ... + hₙ/μₙ

Real depth:
d = h₁ + h₂ + ... + hₙ

Apparent shift:
S = d − d'

Therefore:
S = Σhᵢ − Σ(hᵢ/μᵢ)

Snell's Law:
μ₁ sin i = μ₂ sin r

Quick Method

ACTUAL DEPTH ↓ DIVIDE EACH LAYER BY ITS μ ↓ ADD ALL APPARENT THICKNESSES ↓ APPARENT DEPTH ↓ REAL DEPTH − APPARENT DEPTH ↓ APPARENT SHIFT

1. Write the thickness of every layer separately.
2. Write the refractive index of every layer.
3. Calculate h₁/μ₁, h₂/μ₂, etc.
4. Add all contributions.
5. Obtain apparent depth.
6. Subtract apparent depth from real depth to obtain apparent shift.
7. For oblique viewing, use Snell's law at every interface.

MCQ Answer Key:

1-C, 2-B, 3-B, 4-B, 5-B, 6-B, 7-B, 8-B, 9-B, 10-A, 11-B, 12-B, 13-B, 14-B, 15-B, 16-B, 17-B, 18-B, 19-B, 20-B, 21-B, 22-B, 23-B, 24-A, 25-B, 26-B, 27-C, 28-B, 29-B, 30-B.

Assertion–Reason:

1-A, 2-A, 3-B, 4-A, 5-A, 6-D, 7-A, 8-D, 9-A, 10-A.