Q.Obtain the equation for apparent depth.
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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≈4/3). When viewed from air (refractive index 1) from nearly directly above, the apparent depth h′ is given by:
h′=nh
The apparent depth is the real depth divided by the refractive index of the medium the object is in.
Apparent depth=Refractive index of the mediumReal depth
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′=nh — 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:
nsini=1⋅sinr
For small angles (viewing from nearly overhead), sinθ≈tanθ≈θ (in radians). So:
n⋅i≈r
From geometry: tani=hx and tanr=h′x, where x is the horizontal distance from the point directly above O to A. For small angles:
i≈hx,r≈h′x
Substitute into ni≈r:
n⋅hx≈h′x⇒h′≈nh
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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Two paraxial rays and similar triangles give d'=(n2/n1)d, simplifying to d'=d/n for viewing from air. …
Step 1. Light from an object O at real depth d in a denser medium (n1) refracts on leaving into a rarer medium (n2), bending away from the normal.
Step 2. Using two paraxial rays and small-angle triangle geometry (triangles DOB, DIB), the apparent depth comes out as d′=n1n2d. …
Use two paraxial rays and small-angle triangle geometry acro …
- Inverting the formula and writing d'=nd, confusing apparen …
- CBSE 2026Set ANNUAL1 markMCQQ.A bucket is filled with water up to a height of 24 cm. If the refractive index of water is 4/3, then the apparent depth of the object placed at the bottom of the bucket will be(a) 32 cm(b) 24 cm(c) 12 cm(d) 18 cm
›Reveal solutionSolution
When viewed from air, an object under water appears shallower; apparent depth = real depth / n.
For near-normal viewing from a rarer medium (air) into a denser medium (water) of refractive index n, apparent depth = real depth / n. …
- CBSE 2025Set ANNUAL1 markMCQQ.The bottom of the pond appears to be slightly elevated because of(a) Interference of light(b) Reflection of light(c) Refraction of light(d) Diffraction of light
›Reveal solutionSolution
Light rays from the bottom of the pond bend at the water-air interface (refraction), and the eye extrapolates them backward in straight lines, creating an image of the bottom that appears closer to the surface than it actually is.
Water is optically denser than air, so light travelling from the pond bed to an observer's eye bends away from the normal as it crosses into air (refraction, going from denser to rarer medium).
…
- CBSE 2023Set ANNUAL1 markQ.True/False : Apparent depth in water is greater than real depth.
›Reveal solutionSolution
False — the apparent depth of an object in water is less than its real depth.
When you look at an object under water, refraction bends the light so the object appears raised. The relation is apparent depth = real depth / n, where n (about 1.33 for water) is greater than 1. Dividing by a number greater than 1 makes the apparent depth smaller than the real depth — whi …
- CBSE 2022Set ANNUAL1 markMCQQ.An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is :(a) 12 cm(b) 8 cm(c) 16 cm(d) 10 cm
›Reveal solutionSolution
Converting each apparent depth to a real depth using real depth=n×apparent depth and adding the two contributions gives the total slab thickness, 12 cm.
Working
For viewing through a refracting medium of refractive index n, apparent depth and real depth are related by
n=apparent depthreal depth ⇒ real depth=n×apparent depth
The air bubble, viewed from one face, appears 5 cm deep — this is the real distance of the bubble from that first face (through thickness n=1.5 of glass):
d1=n×5=1.5×5=7.5 cm
Viewed from the opposite face, it appears 3 cm deep — the real distance of the bubble from that second face: …
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