The "Apparent Depth" Conundrum: Why Does a Pool Look Shallower? | JEE Physics

Why Apparent Depth Matters in JEE

Apparent depth is a fundamental concept in Ray Optics that appears frequently in JEE problems. Understanding this helps solve:

Direct formula application problems (JEE Main)
Multi-layer refraction problems (JEE Advanced)
Experimental setup questions
Real-world applications like swimming pools, aquariums

Concept 1 Essential

Normal Viewing (Perpendicular to Surface)

When you look straight down into water, the apparent depth is given by:

$$d_{apparent} = \frac{d_{real}}{\mu}$$

where $\mu$ is the refractive index of the medium

🔍 Ray Diagram - Normal Viewing:

Air ($\mu_1 = 1$)

Water ($\mu_2 = \mu$)

↓ Normal ray

Real depth = $d$

Apparent depth = $d/\mu$

📐 Derivation Using Snell's Law:

Step 1: For normal incidence, angle of incidence $i = 0$

Step 2: By Snell's Law: $\mu_1 \sin i = \mu_2 \sin r$

Step 3: Since $i = 0$, we get $r = 0$

Step 4: Using similar triangles and small angle approximation:

$\frac{d_{real}}{d_{apparent}} = \frac{\mu_2}{\mu_1}$

Step 5: For air to water: $\mu_1 = 1$, $\mu_2 = \mu$

$d_{apparent} = \frac{d_{real}}{\mu}$

🎯 JEE Application Example:

Problem: A swimming pool has actual depth 4m. If refractive index of water is 4/3, find apparent depth when viewed normally from above.

Solution: Using the formula:

$d_{apparent} = \frac{d_{real}}{\mu} = \frac{4}{4/3} = 4 \times \frac{3}{4} = 3m$

The pool appears 1m shallower than actual depth!

Concept 2 Important

Oblique Viewing (At an Angle)

When you look at an angle, the apparent depth depends on the viewing angle:

$$d_{apparent} = d_{real} \times \frac{\cos r}{\cos i}$$

where $i$ = angle in rarer medium, $r$ = angle in denser medium

🔍 Ray Diagram - Oblique Viewing:

Air ($\mu_1 = 1$)

Water ($\mu_2 = \mu$)

↗ Incident ray (angle i)

↘ Refracted ray (angle r)

Lateral shift occurs

📐 Derivation Using Snell's Law:

Step 1: Snell's Law: $\mu_1 \sin i = \mu_2 \sin r$

Step 2: From geometry: $AB = \frac{d}{\cos r}$ (actual path)

Step 3: Apparent path: $AB' = \frac{d_{apparent}}{\cos i}$

Step 4: Using optical path equality:

$\mu_2 \times AB = \mu_1 \times AB'$

Step 5: Substituting and solving:

$d_{apparent} = d_{real} \times \frac{\cos r}{\cos i}$

🎯 JEE Application Example:

Problem: A coin is at the bottom of a tank filled with water ($\mu = 4/3$) to depth 3m. Find apparent depth when viewed at 45° from vertical.

Solution:

1. Find $r$ using Snell's Law: $1 \times \sin 45° = \frac{4}{3} \times \sin r$

$\sin r = \frac{3}{4} \times \frac{1}{\sqrt{2}} = \frac{3}{4\sqrt{2}}$

2. $\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \frac{9}{32}} = \sqrt{\frac{23}{32}}$

3. $\cos i = \cos 45° = \frac{1}{\sqrt{2}}$

4. $d_{apparent} = 3 \times \frac{\sqrt{23/32}}{1/\sqrt{2}} = 3 \times \sqrt{\frac{23}{16}} \approx 2.87m$

Concept 3 Advanced

Multiple Layers & Shift Calculation

For multiple transparent layers, total apparent depth is:

$$d_{apparent} = \sum \frac{d_i}{\mu_i}$$

and lateral shift $ = t \times \sin i \left(1 - \frac{\cos i}{\sqrt{\mu^2 - \sin^2 i}}\right)$

📐 Lateral Shift Derivation:

Step 1: From Snell's Law: $\sin i = \mu \sin r$

Step 2: Lateral shift $S = t \times \sin(i - r) \times \sec r$

Step 3: Using trigonometric identities:

$S = t \times \frac{\sin(i - r)}{\cos r}$

Step 4: Expand $\sin(i - r) = \sin i \cos r - \cos i \sin r$

Step 5: Substitute $\sin r = \frac{\sin i}{\mu}$ and simplify:

$S = t \sin i \left(1 - \frac{\cos i}{\sqrt{\mu^2 - \sin^2 i}}\right)$

🎯 JEE Advanced Example:

Problem: A glass slab ($\mu = 1.5$) of thickness 6cm is placed over a mark. Find apparent position of mark when viewed normally and lateral shift when viewed at 60°.

Solution:

Normal viewing: $d_{apparent} = \frac{6}{1.5} = 4cm$ (raised by 2cm)

Oblique viewing: $i = 60°$, $\sin i = \sqrt{3}/2$

$\sin r = \frac{\sin i}{\mu} = \frac{\sqrt{3}/2}{1.5} = \frac{1}{\sqrt{3}}$

$\cos r = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}$

$S = 6 \times \frac{\sqrt{3}}{2} \left(1 - \frac{0.5}{\sqrt{2.25 - 0.75}}\right) \approx 2.68cm$

🚀 Problem-Solving Strategies

Key Formulas to Remember:

Normal viewing: $d_{app} = \frac{d_{real}}{\mu}$
Oblique viewing: $d_{app} = d_{real} \frac{\cos r}{\cos i}$
Multiple layers: $d_{app} = \sum \frac{d_i}{\mu_i}$
Lateral shift: $S = t \sin i (1 - \frac{\cos i}{\sqrt{\mu^2 - \sin^2 i}})$

JEE Exam Tips:

Always draw ray diagrams for oblique viewing
Remember $\mu_{air} \approx 1$ for calculations
For small angles, apparent depth ≈ real depth/μ
Practice multi-layer refraction problems

Advanced Applications Available

Includes prism problems, total internal reflection, lens combinations, and JEE Advanced level problems

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. A vessel of depth 2m is filled with water (μ=4/3) and oil (μ=1.2) of equal depth. Find apparent depth when viewed normally.

2. Calculate lateral shift for a light ray passing through a 10cm glass slab (μ=1.5) at 45° incidence.

3. A fish is 4m deep in water. At what depth does it appear to a bird flying directly overhead?

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