Question

c) A coin is lying at the bottom of a beaker filled with water ( mw =4/3)upto
a depth of 16cm. Calculate apparent depth and shift in the coin.
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Answer 1

$Refractive~ index~ (\mu) = \dfrac{4}{3}$

Let the coin be placed at point O

Let the height of beaker be AO = 16 cm

Let the apparent depth be AI

$\mu \: = \dfrac{Real \: depth}{Apparent \: depth}$

$\dfrac{4}{3} = \dfrac{AO}{AI}$

$\dfrac{4}{3} = \dfrac{16}{AI}$

$AI = \dfrac{3}{4} \times 16$

$\red{AI = 12~cm}$