A circular disc of radius $R$ is placed horizontally and coaxially inside an opaque hemispherical bowl of radius $a$, with the disc's centre on the vertical axis of the bowl. The circular rim of the bowl lies in a horizontal plane at the top, with the two ends of a rim diameter labelled A and B and its centre O; the disc sits a vertical distance $d$ below this rim plane. An observer's eye is placed at the edge (rim) of the bowl and looks across and down into it. When the bowl is empty, the far edge of the disc is just visible along the straight line of sight from the eye. The bowl is then filled to the brim with a transparent liquid of refractive index $\mu$; refraction now bends the line of sight so that the near edge of the disc becomes just visible instead. Find how far below the top of the bowl (the value of $d$) the disc is placed, in terms of $a$, $R$ and $\mu$.

A circular disc of radius R is placed horizontally and coaxially inside an opaque hemispherical bowl of radius

NCERT Exemplar · Q23

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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.

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With the eye fixed at the rim, the same viewing direction points to the disc's far edge when the bowl is empty and, after filling, to the near edge because the liquid surface refracts the ray. Setting up the two geometric angles and joining them by Snell's law at the rim gives a single equation for the depth $d$ .

Set-up

Take the liquid surface (rim plane) as height $0$ , with the eye at one rim edge, a horizontal distance $a$ from the axis. The disc lies at depth $d$ , its centre on the axis, so its far edge is a horizontal distance $a+R$ from the eye and its near edge is $a-R$ from the eye, both at depth $d$ .

Empty bowl — the line of sight to the far edge

The straight ray from the far edge to the eye makes an angle $r$ with the vertical (the surface normal):

\tan r = \frac{a+R}{d},\qquad \sin r = \frac{a+R}{\sqrt{(a+R)^2+d^2}}.

This fixes the direction in which the eye looks.

Filled bowl — the near edge becomes visible

Keeping the same viewing direction, a ray now leaves the near edge, travels up through the liquid, and refracts at the surface (right at the rim, next to the eye) into that same direction. Inside the liquid this ray makes angle $i$ with the vertical:

\tan i = \frac{a-R}{d},\qquad \sin i = \frac{a-R}{\sqrt{(a-R)^2+d^2}}.

Snell's law at the surface

Going from liquid to air, $\mu\sin i = \sin r$ :

\mu\,\frac{a-R}{\sqrt{(a-R)^2+d^2}} = \frac{a+R}{\sqrt{(a+R)^2+d^2}}.

Solve for $d$

Square and cross-multiply: …

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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)

Set-up

Empty bowl — the line of sight to the far edge

Filled bowl — the near edge becomes visible

Snell's law at the surface

Solve for $d$

Method: Locating a "Just Visible" Point Using Snell's Law at a Rim/Edge

Steps

Set-up

Empty bowl — the line of sight to the far edge

Filled bowl — the near edge becomes visible

Snell's law at the surface

Solve for ddd

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Solve for $d$

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