A tank is filled with water to a height of $12.5\ \text{cm}$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4\ \text{cm}$. What is the refractive index of water? If water is replaced by a liquid of refractive index $1.63$ up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

A tank is filled with water to a height of 12.5cm. The apparent depth of a needle lying at the bottom of the t

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12.5\ \text{cm}

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Concept understanding — Apparent Depth

Apparent Depth: Why a Swimming Pool Looks Shallower Than It Is

You have seen it yourself. Stand beside a swimming pool and look down at the tile pattern on the bottom. The floor looks closer than it really is. If you reach down with your hand, you miss — the water is deeper than it appears. That is apparent depth in action.

The intuition is simple: light bends when it moves from one medium to another. When you look into water, light from the bottom travels upward through water (denser) and then into air (rarer). At the water-air surface, the light bends away from the normal. Your brain, however, assumes light travels in straight lines. So it traces the bent ray backward in a straight line, and that line meets the water at a point higher than the actual bottom. The object appears raised.

The Precise Statement

Consider an object at a real depth $h$ below the surface of a medium of refractive index $n$ (for water, $n \approx 4/3$ ). When viewed from air (refractive index $1$ ) from nearly directly above, the apparent depth $h'$ is given by:

h' = \frac{h}{n}

The apparent depth is the real depth divided by the refractive index of the medium the object is in.

\text{Apparent depth} = \frac{\text{Real depth}}{\text{Refractive index of the medium}}

For water ( $n = 4/3$ ), the apparent depth is three-quarters of the real depth. A 3 m deep pool looks only 2.25 m deep.

Why "Divided by n" and Not "Multiplied by n"?

This is the most common confusion. Light bends away from the normal when going from denser to rarer. That makes the image shift upward, so the apparent depth is smaller than the real depth. Dividing by a number greater than 1 makes the result smaller — that is exactly what we need.

If the object were in air and you looked from water (the reverse situation), the apparent depth would be $h' = n h$ — the object would appear deeper. But the standard case is looking from air into a denser medium, so the formula is $h' = h/n$ .

The Derivation (For Small Angles)

›Proof

Derivation for near-normal viewing

Draw a ray from the object O at real depth $h$ to the surface at point A. The ray makes an angle $i$ with the normal inside the water. It emerges into air at angle $r$ , where Snell's law gives:

$n \sin i = 1 \cdot \sin r$

For small angles (viewing from nearly overhead), $\sin \theta \approx \tan \theta \approx \theta$ (in radians). So:

$n \cdot i \approx r$

From geometry: $\tan i = \frac{x}{h}$ and $\tan r = \frac{x}{h'}$ , where $x$ is the horizontal distance from the point directly above O to A. For small angles:

$i \approx \frac{x}{h}, \quad r \approx \frac{x}{h'}$

Substitute into $n i \approx r$ :

$n \cdot \frac{x}{h} \approx \frac{x}{h'} \quad \Rightarrow \quad h' \approx \frac{h}{n}$

The approximation is excellent when you look nearly straight down. For large viewing angles, the apparent depth changes and the image also shifts sideways — but the formula $h' = h/n$ is the standard result for normal viewing.

A Quick Check with Numbers

A coin lies at the bottom of a beaker of water, real depth 12 cm. Refractive index of water is 4/3.

h' = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9 \text{ cm}

The coin appears 9 cm below the surface — raised by 3 cm.

Watch out

A common mistake

Students sometimes write $h' = n h$ because they remember "refractive index makes things look bigger." That is for lateral magnification in lenses. For apparent depth, the denser medium raises the object, so the depth decreases. Always check: does the answer make physical sense? If the object is in water, it should look shallower, not deeper.

The Big Picture

Apparent depth is not a trick of the eye — it is a direct consequence of how light bends at boundaries. Every time you see a fish in a pond, a pencil in a glass of water, or the bottom of a swimming pool, your brain is doing this geometry unconsciously. The formula $h' = h/n$ is the precise mathematical description of that everyday experience.

Apparent depth is one of the most frequently asked numerical topics in the CBSE Class 12 Physics ray optics unit, aligned with the NCERT syllabus, and "apparent depth formula class 12 physics" is a high-traffic revision search. It's also a quick, formula-based question type that appears often in JEE Main and NEET practice sets.

Refractive index of water $=$ real depth / apparent depth $=12.5/9.4\approx1.33$ . With the tank refilled to the same height with a liquid of refractive index $1.63$ , the new apparent depth is $12.5/1.63\approx7.67\ \text{cm}$ , so the microscope must be raised by $9.4-7.67\approx1.73\ \text{cm}$ to refocus.

Setting up — apparent depth

For near-normal viewing, the refractive index of a medium relates real depth $h$ to apparent depth $h'$ by

n = \frac{h}{h'}

Step 1 — refractive index of water

Real depth $h=12.5\ \text{cm}$ , apparent depth (as measured by the microscope) $h'=9.4\ \text{cm}$ .

n_{water} = \frac{12.5}{9.4} \approx 1.33

This matches the well-known refractive index of water — a good sanity check.

Step 2 — apparent depth in the new liquid

The liquid is replaced, keeping the same height $h=12.5\ \text{cm}$ , but now with $n_{liquid}=1.63$ .

h'_{new} = \frac{h}{n_{liquid}} = \frac{12.5}{1.63} \approx 7.67\ \text{cm}

Step 3 — distance the microscope must move

The microscope was originally focused at $9.4\ \text{cm}$ below the surface. It must now be focused at $7.67\ \text{cm}$ below the surface — a smaller depth, because the denser liquid ( $n=1.63>1.33$ ) raises the apparent position of the needle further.

\Delta d = 9.4 - 7.67 = 1.73\ \text{cm}

Since the new apparent depth is smaller, the microscope must be moved upward (toward the surface) by this amount.

Watch out

A higher refractive index means a smaller apparent depth (the object looks even closer to the surface), so the microscope must move up, not down, to refocus.

✓Final answer

\boxed{1.33}

Here are the common mistakes students make on this classic refractive index problem, and how to avoid each one.

1. Confusing Apparent Depth with Real Depth

The Mistake:

Students often swap the two values — using $12.5\ \text{cm}$ as apparent depth and $9.4\ \text{cm}$ as real depth.

Why it happens:

The problem states "apparent depth is measured to be $9.4\ \text{cm}$ ." Some students misread or assume the larger number must be the real depth.

How to avoid:

Always label clearly:

Real depth ( $h$ ) = actual physical height of water = $12.5\ \text{cm}$
Apparent depth ( $h'$ ) = what the microscope reads = $9.4\ \text{cm}$

Key formula:

$\mu = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}$

2. Forgetting That Refractive Index > 1 for Water

The Mistake:

A student computes $\frac{9.4}{12.5} = 0.752$ and writes that as the refractive index.

Why it happens:

They invert the fraction without checking if the answer makes physical sense.

How to avoid:

Remember: For a denser medium (like water), $\mu > 1$ .

If your answer is less than 1, you have swapped numerator and denominator.
Always do a sanity check: water’s $\mu \approx 1.33$ , so your answer should be close to that.

Correct calculation:

\mu = \frac{12.5}{9.4} \approx 1.33

3. Mishandling the Second Part — Sign of the Shift

The Mistake:

Students compute the new apparent depth correctly but then give the wrong direction for the microscope movement (up vs. down).

Why it happens:

They forget that a higher refractive index makes the apparent depth smaller (the bottom looks shallower).

How to avoid:

For $\mu = 1.63$ , apparent depth $h'' = \frac{12.5}{1.63} \approx 7.67\ \text{cm}$
Compare with the first apparent depth ( $9.4\ \text{cm}$ ): $7.67 < 9.4$ , so the needle appears higher (closer to the surface).
Therefore, the microscope must be raised (moved upward) to refocus.

Distance moved:

\Delta = 9.4 - 7.67 = 1.73\ \text{cm} \ (\text{upward})

4. Using the Wrong Formula for Shift

The Mistake:

Some students try to use the formula for lateral shift (for a glass slab) instead of apparent depth shift.

How to avoid:

For normal viewing (microscope looking straight down), the only shift is vertical:

\text{Shift} = h - h' = h\left(1 - \frac{1}{\mu}\right)

But here, since you already have $h'$ for water, just compute the difference between the two apparent depths.

5. Rounding Too Early

The Mistake:

Rounding $12.5/9.4$ to $1.3$ in the first part, then using $1.3$ in the second part — leading to an inaccurate shift.

How to avoid:

Keep at least 3 significant figures throughout:

$\mu = 1.3298 \approx 1.33$
$h'' = 12.5 / 1.63 = 7.6687\ \text{cm}$
Shift $= 9.4 - 7.6687 = 1.7313\ \text{cm} \approx \mathbf{1.73\ cm}$

Quick Summary Checklist

Step	Common Mistake	✓ Correct Approach
Identify depths	Swap real & apparent	Real = 12.5 cm, Apparent = 9.4 cm
Compute $\mu$	Invert fraction	$\mu = \frac{12.5}{9.4}$
Second apparent depth	Use wrong $\mu$	$h'' = \frac{12.5}{1.63}$
Direction of movement	Move down instead of up	$\mu$ larger → depth smaller → raise microscope
Final answer	Round too early	Keep 3-4 digits, round at the end

Final answers:

$\mu_{\text{water}} \approx \mathbf{1.33}$
Microscope must be moved upward by $\mathbf{1.73\ cm}$

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$\dfrac{4}{3}$
›Reveal solutionSolution
The apparent depth of a layered medium is the sum of each layer’s real depth divided by its refractive index. Using the given total apparent depth, the refractive index of oil comes out to 1.45.

When you look vertically down into a stack of transparent liquids, each layer makes the bottom appear shallower. The reason is that light bends at each interface, and the apparent depth contributed by a single layer of real depth $d$ and refractive index $n$ is $d/n$ . This is a direct consequence of Snell’s law for near-normal viewing: the apparent shift is proportional to the real depth divided by the refractive index.

Because the layers are in series, the total apparent depth is simply the sum of the individual apparent depths. That’s the key idea — no complicated ray tracing needed.

For the water layer: real depth $d_w = 8\ \text{cm}$ , refractive index $n_w = \frac{4}{3}$ . Its apparent depth is

$\frac{d_w}{n_w} = \frac{8}{4/3} = 8 \times \frac{3}{4} = 6\ \text{cm}.$

For the oil layer: real depth $d_o = 5.8\ \text{cm}$ , refractive index $n_o$ is unknown. Its apparent depth is

$\frac{d_o}{n_o} = \frac{5.8}{n_o}\ \text{cm}.$

The total apparent depth is given as $10\ \text{cm}$ . So

$\frac{8}{4/3} + \frac{5.8}{n_o} = 10.$
That is,
$6 + \frac{5.8}{n_o} = 10.$

Subtract 6 from both sides:

$\frac{5.8}{n_o} = 4.$

Solve for $n_o$ :

$n_o = \frac{5.8}{4} = 1.45.$
Watch out

A common mistake is to add the real depths and divide by an average refractive index. That doesn’t work — each layer contributes its own apparent depth independently, and you must add the fractions, not the depths.

✓Final answer
$\boxed{1.45}$
TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$32$
›Reveal solutionSolution

The mirror converges sunlight toward a point $32$ cm above it. With $20$ cm of water present, the rays aim $12$ cm above the water surface; refraction at the water–air interface shrinks this to $12\times\tfrac34=9$ cm, so the focus is $9$ cm above the water level.

Focus in the empty tank. With no water, the concave mirror focuses the parallel sunlight at its focal point, $32$ cm above the mirror.

Where the rays are heading once water is added. Water fills the tank to a height of $20$ cm. Inside the water the reflected rays are still converging toward the point $32$ cm above the mirror, which is
$32-20=12\ \text{cm}$
above the water surface. This convergence point acts as a virtual object for refraction at the flat water–air surface.

Refraction at the plane water–air surface. For a plane interface, $\dfrac{n_2}{v}=\dfrac{n_1}{u}$ . Taking distances measured from the surface, the point $12$ cm above it (in the water frame) maps to
$v=u\,\frac{n_{\text{air}}}{n_{\text{water}}}=12\times\frac{1}{4/3}=12\times\frac34=9\ \text{cm}.$
The focus therefore lies $9$ cm above the water level.

✓Final answer
$Sunlight focuses 9 cm above the water level — option (B) .$

←Previous question9.2 — A 4.5cm needle is placed 12cm away from a convex mirror of focal lengt…→Next question9.4 — Figures 9.27(a) and (b) show refraction of a ray in air incident at 60…

Concept understanding — Apparent Depth

Apparent Depth: Why a Swimming Pool Looks Shallower Than It Is

The Precise Statement

Why "Divided by n" and Not "Multiplied by n"?

The Derivation (For Small Angles)

A Quick Check with Numbers

The Big Picture

Setting up — apparent depth

Step 1 — refractive index of water

Step 2 — apparent depth in the new liquid

Step 3 — distance the microscope must move

Method: Apparent Depth Formula (Index Matching)

Step-by-step solution

Final Answer

1. Confusing Apparent Depth with Real Depth

2. Forgetting That Refractive Index > 1 for Water

3. Mishandling the Second Part — Sign of the Shift

4. Using the Wrong Formula for Shift

5. Rounding Too Early

Quick Summary Checklist

Setting up — apparent depth

Step 1 — refractive index of water

Step 2 — apparent depth in the new liquid

Step 3 — distance the microscope must move

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