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Question

An object is at a depth of \( 16 \, \text{cm} \) in water (\( n = 1.33 \)). What is the apparent depth?

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Explanation

**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Apparent depth = (real depth/n) . Real depth = 16 cm , n = 1.33 . Apparent depth = (16/1.33) ≈ 12.03 cm . Substituting values gives 12 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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