A swimming pool appears to be raised by $6 m$ when viewed normally. What is the refractive index of water if the actual depth of the swim-ming pool is $24 m$ ?

1.44

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1.33

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1.56

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1.76

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Solution

The correct option is B $1.33$
It is given that surface is rised above by 6m , hence $a p p a r e n t d e p t h = 24 - 6 = 18 m$ and real depth is given 24 m.
So by formula
$μ = \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$
Hence $μ = \frac{24}{18} = \frac{4}{3}$
$μ = 1.33$

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A swimming pool appears to be raised by 6m when viewed normally. What is the refractive index of water if the actual depth of the swim-ming pool is 24m?

The correct option is B 1.33It is given that surface is rised above by 6m , hence apparentdepth=24−6=18m and real depth is given 24 m.So by formulaμ=realdepthapparentdepthHence μ=2418=43μ=1.33

A swimming pool appears to be raised by $6 m$ when viewed normally. What is the refractive index of water if the actual depth of the swim-ming pool is $24 m$ ?

The correct option is B $1.33$
It is given that surface is rised above by 6m , hence $a p p a r e n t d e p t h = 24 - 6 = 18 m$ and real depth is given 24 m.
So by formula
$μ = \frac{r e a l d e p t h}{a p p a r e n t d e p t h}$
Hence $μ = \frac{24}{18} = \frac{4}{3}$
$μ = 1.33$