Practice question
Question
An object is at a depth of \( 19.95 \, \text{cm} \) in a medium with refractive index \( 1.5 \). What
is the apparent depth?
Explanation
**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. Apparent depth = (real depth/n) . Real depth = 19.95 cm , n = 1.5 . Apparent depth = (19.95/1.5) = 13.3 cm . Substituting values gives 13.3 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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