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Question
A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) at a depth of 16 cm. By what height the coin appears to be raised when seen from vertically above?
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Solution
Given: Refractive index of water μ = `4/3`,
Real depth of coin = 16 cm
To find: Apparent depth = ?
Formula: μ water = `"real depth"/"apparent depth"`
`4/3 = 16/"apparent depth"`
∴ Apparent depth = `16 xx 3/4 = 12` cm
∴ Apparent Rise = Real Depth − Apparent Depth
= 16 – 12
= 4 cm
∴ Coin appears to be raised by 4 cm.
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