Practice question
Question
An object at a depth of \( 19.95 \, \text{cm} \) in water (\( n = 1.33 \)) is viewed normally. What is
the apparent depth?
Explanation
**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. Apparent depth = (real depth/n) . Real depth = 19.95 cm , n = 1.33 . Apparent depth = (19.95/1.33) ≈ 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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