Refraction At Plane Surfaces

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A small coin lies at the bottom of a tank filled with water to a real depth of 1.6 m. Taking the refractive index of water as 4/3, what apparent depth does an observer looking straight down perceive?

RankGuru · Accepted Answer

When viewed nearly normally through a denser medium, an object appears raised because refraction bends the emerging rays away from the normal. The apparent depth relates to the real depth through $	ext{apparent depth}=\frac{	ext{real depth}}{n}$, where $n$ is the refractive index of the medium relative to the observer's medium. Substituting the real depth of 1.6 m and $n=\frac{4}{3}$ gives apparent depth $=\frac{1.6}{4/3}=1.6	imes\frac{3}{4}=1.2$ m. The coin therefore seems to sit 0.4 m higher than it truly is. The value 1.6 m is wrong because it ignores refraction entirely. The value 0.9 m is wrong as it divides by a refractive index near 1.78 rather than 1.33. The value 2.13 m is wrong because it multiplies by $n$ instead of dividing, which would make a dense medium appear deeper, contradicting observation. This follows the NCERT normal-incidence approximation for apparent depth. A sanity check confirms the apparent depth is smaller than the real depth, as expected for a denser medium.

Refraction At Plane Surfaces

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