A fish in the pond of a water appears ata depth of 6cm . What is the actual depth of the fish if refractive index of water is 4/3?

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Solution

Dear Student,

$μ = \frac{real depth}{apparent depth} μ = \frac{4}{3} Apparent depth = 6 cm \frac{4}{3} = \frac{real depth}{6} real depth = 8 cm$

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A fish in the pond of a water appears ata depth of 6cm . What is the actual depth of the fish if refractive index of water is 4/3?

Dear Student, μ=real depthapparent depthμ=43Apparent depth=6 cm43=real depth6real depth=8 cm

Dear Student,

$μ = \frac{real depth}{apparent depth} μ = \frac{4}{3} Apparent depth = 6 cm \frac{4}{3} = \frac{real depth}{6} real depth = 8 cm$