Skip to content

Question

A fish is at a depth of \( 40 \, \text{cm} \) in water (\( n = 1.33 \)). What is its apparent depth
when viewed from above?

Options

Choose one · Correct answer highlighted

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 = 40 cm , n = 1.33 . Apparent depth = (40/1.33) ≈ 30.08 cm . Substituting values gives 30 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.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.