Practice question
Question
An object is at a depth of \( 15 \, \text{cm} \) in water (\( n = 1.5 \)). What is the apparent depth?
Explanation
**Prism formula** for small A, δ_m≈(n-1)A, for 30°, n=1.6, δ_m≈0.6×30°=18°, approximate, exact using sin formula. For 45°, n=1.6, sin[(45+δ_m)/2]=1.6×sin22.5°=1.6×0.3827=0.6123, (45+δ_m)/2=37.8°, δ_m=30.6°, showing deviation increases with A and n. Apparent depth = (real depth/n) . Real depth = 15 cm , n = 1.5 . Apparent depth = (15/1.5) = 10 cm . Substituting values gives 10 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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