A coin placed at the bottom of a beaker appears to be raised by $4$ cm. If the refractive index of water is $\frac{4}{3}$ , find the depth of water in the beaker.

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Solution

Step-1: The given data -
Apparent depth, h =d - $4 c m$
Refractive index of water, n = $\frac{4}{3}$
Real depth, d = ?
Step-2: Calculating the real depth of the water tank -
$r e f r a c t i v e i n d e x (n) = \frac{R e a l d e p t h (d)}{A p p a r e n t d e p t h (h)} \to \frac{4}{3} = \frac{d}{d - 4 c m} \to 4 d - 16 = 3 d \Rightarrow d = 16 c m$
Hence, the depth of the water in the beaker is $16 c m .$

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Step-1: The given data - Apparent depth, h =d - 4cmRefractive index of water, n = 43Real depth, d = ?Step-2: Calculating the real depth of the water tank -refractiveindex(n)=Realdepth(d)Apparentdepth(h)→43=dd-4cm→4d-16=3d⇒d=16cmHence, the depth of the water in the beaker is 16cm.

A coin placed at the bottom of a beaker appears to be raised by $4$ cm. If the refractive index of water is $\frac{4}{3}$ , find the depth of water in the beaker.