A plane glass is placed over various coloured letters (violet, green, yellow, red). The letter which appears t

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Q. A plane glass is placed over various coloured letters (violet, green, yellow, red). The letter which appears to be raised more is,

(a) red

(b) yellow

(d) violet

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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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Step 1. For total internal reflection to occur at $i=45°$ , the medium's critical angle must be less than $45°$ : $i_c<45°$ , i.e. $\sin i_c=1/n<\sin45°=0.707$ , i.e. $n>1/0.707=1.414$ .

Step 2. Checking each option against $n>1.414$ : (a) $n=1.25$ gives $i_c=\sin^{-1}(1/1.25)=53.1°>45°$ , so no TIR at $45°$ .

(b) $n=1.33$ gives $i_c=48.6°>45°$ , no TIR.

(c) $n=1.4$ gives $i_c=\sin^{-1}(1/1.4)=45.6°$ , still (just) greater than $45°$ , so no TIR at exactly $45°$ incidence. …

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Q.A plane glass is placed over various coloured letters (violet, green, yellow, red). The letter which appears to be raised more is,

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