Practice question
Question
An object is at a depth of \( 26.6 \, \text{cm} \) in a medium with refractive index \( 1.33 \). What
is the apparent depth?
Explanation
**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. Apparent depth = (real depth/n) . Real depth = 26.6 cm , n = 1.33 . Apparent depth = (26.6/1.33) = 20 cm . Substituting values gives 20 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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