Apparent Depth in Multiple Liquid Layers | Class 10 Physics

Quick answer: When an object at the bottom of several layers of immiscible liquids is viewed from directly above, its total apparent depth is the sum of each layer's thickness divided by its refractive index: $d' = \frac{d_1}{n_1} + \frac{d_2}{n_2} + \dots$. The object appears raised by $\Delta d = d_1\left(1 - \frac{1}{n_1}\right) + d_2\left(1 - \frac{1}{n_2}\right) + \dots$. For example, a coin under $6\text{ cm}$ of water ($n = \frac{4}{3}$) and $4.5\text{ cm}$ of oil ($n = 1.5$) appears $7.5\text{ cm}$ deep, raised by $3\text{ cm}$.

A coin at the bottom of a glass of water looks closer to the surface than it really is. This happens because light from the coin bends away from the normal as it passes from water (optically denser) into air (optically rarer). The depth at which the coin appears is called its apparent depth.

What if the vessel contains two or more layers of liquids that do not mix, such as oil floating on water? Light bends at each boundary, and each layer makes the coin appear a little higher.

In this guide, we extend the single-liquid apparent depth formula to several layers and solve a two-layer numerical step by step.

Syllabus note: The CBSE Class 10 NCERT chapter Light – Reflection and Refraction explains why objects in water appear raised, but the apparent depth formula $d' = \frac{d}{n}$ is not part of the Class 10 NCERT textbook. These numericals are useful for Olympiad, NTSE-style and higher-level practice, and are covered in more detail in Class 12 Physics.

What You Will Learn

• The apparent depth formula for one liquid, $d' = \frac{d}{n}$
• How light behaves as it crosses several parallel liquid layers
• The formula for total apparent depth: $d' = \sum \frac{d_i}{n_i}$
• The formula for total apparent shift: $\Delta d = \sum d_i\left(1 - \frac{1}{n_i}\right)$
• A solved example with water ($n = \frac{4}{3}$) and oil ($n = 1.5$)
• Common mistakes to avoid

1. Apparent Depth in One Liquid

When an object at a real depth $d$ in a liquid of refractive index $n$ is viewed from directly above (nearly along the normal):

        eye (in air)
           |
 ~~~~~~~~~~+~~~~~~~~~~  surface
           |
           *  image (apparent depth d')
           |
           |
           *  object (real depth d)
 ====================== bottom

$\text{Apparent depth } d' = \frac{\text{real depth } d}{n}$

The apparent shift is how much the object appears to be raised:

$\Delta d = d - d' = d - \frac{d}{n} = d\left(1 - \frac{1}{n}\right)$

Since $n > 1$ for any liquid, $d' < d$: the object always appears higher than it really is.

2. Apparent Depth Through Several Liquid Layers

Suppose a vessel has several layers of liquids that do not mix, one on top of another:

Because the boundaries are parallel, each layer raises the image formed by the layer below it. The effects simply add up.

Total Apparent Depth

$d'_{\text{total}} = \frac{d_1}{n_1} + \frac{d_2}{n_2} + \frac{d_3}{n_3} + \dots = \sum \frac{d_i}{n_i}$

Total Apparent Shift

$\Delta d_{\text{total}} = (\text{total real depth}) - (\text{total apparent depth}) = \sum d_i\left(1 - \frac{1}{n_i}\right)$

3. Solved Example: Water and Oil

Question

A beaker contains water (refractive index $\frac{4}{3}$) up to a height of $6.0\text{ cm}$. An oil of refractive index $1.5$ that does not mix with water is poured on top to form a layer $4.5\text{ cm}$ thick. A coin lies at the bottom of the beaker. Viewing from directly above, find:

(a) the apparent depth of the coin

(b) how much the coin appears to be raised

Solution

Step 1: Write the given values.

• Water: $d_1 = 6.0\text{ cm}$, $n_1 = \frac{4}{3}$
• Oil: $d_2 = 4.5\text{ cm}$, $n_2 = 1.5 = \frac{3}{2}$
• Total real depth: $6.0 + 4.5 = 10.5\text{ cm}$

Step 2: Apparent depth contributed by each layer.

$\frac{d_1}{n_1} = \frac{6.0}{4/3} = \frac{6.0 \times 3}{4} = 4.5\text{ cm}$

$\frac{d_2}{n_2} = \frac{4.5}{3/2} = \frac{4.5 \times 2}{3} = 3.0\text{ cm}$

Step 3: (a) Total apparent depth.

$d'_{\text{total}} = 4.5 + 3.0 = 7.5\text{ cm}$

Answer (a): The coin appears to be $7.5\text{ cm}$ below the top surface of the oil.

Step 4: (b) Apparent shift.

$\Delta d = 10.5 - 7.5 = 3.0\text{ cm}$

Answer (b): The coin appears to be raised by $3.0\text{ cm}$.

Check with the shift formula:

$\Delta d_{\text{total}} = 6.0\left(1 - \frac{3}{4}\right) + 4.5\left(1 - \frac{2}{3}\right) = 1.5 + 1.5 = 3.0\text{ cm}$

This matches the answer above. ✓

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: Apparent depth is measured from the top surface of the liquids, down to the image.

Exam Tip: Keep refractive indices as fractions ($\frac{4}{3}$, $\frac{3}{2}$) rather than decimals ($1.33$, $1.5$) to avoid rounding errors.

Common Mistake: Averaging the refractive indices and dividing the total depth by the average. This gives the wrong answer. Divide each layer's thickness by its own refractive index, then add.

Remember: These formulas apply when the object is viewed from directly above (nearly along the normal).

Frequently Asked Questions

What is apparent depth?

Apparent depth is the depth at which an object in a liquid appears to be when viewed from above. It is less than the real depth because light bends away from the normal as it leaves the liquid.

What is the formula for apparent depth in several liquids?

The total apparent depth is $\frac{d_1}{n_1} + \frac{d_2}{n_2} + \dots$, where $d$ is the thickness and $n$ the refractive index of each layer.

How do you find the apparent shift?

Subtract the apparent depth from the real depth, or use $\Delta d = \sum d_i\left(1 - \frac{1}{n_i}\right)$.

Why does a coin in water appear raised?

Light from the coin bends away from the normal as it passes from water into air, so the rays seem to come from a point higher than the coin.

What is the apparent depth of a coin under 6 cm of water and 4.5 cm of oil?

With $n = \frac{4}{3}$ for water and $1.5$ for oil, the apparent depth is $4.5 + 3.0 = 7.5\text{ cm}$, and the coin appears raised by $3.0\text{ cm}$.