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Question

An object is at a depth of \( 24 \, \text{cm} \) in a medium with refractive index \( 1.6 \). What is
the apparent depth?

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Explanation

**Refraction at plane surface** changes direction due to speed change v = c/n, n = c/v. For water n=1.33 to air, n₁>n₂, ray bends away, if θ₁ > C, sinθ₂>1 impossible, TIR occurs. Snell's law quantitative prediction of θ₂ from θ₁ and indices. Apparent depth = (real depth/n) . Real depth = 24 cm , n = 1.6 . Apparent depth = (24/1.6) = 15 cm . Substituting values gives 15 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

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