A diver at a depth of 12 cm, ses the sky in a cone of semi-vertical angle :

{s i n}^{- 1} (\frac{4}{3})

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{s i n}^{- 1} (\frac{2}{3})

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{s i n}^{- 1} (\frac{3}{4})

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{s i n}^{- 1} (\frac{3}{2})

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Solution

The correct option is C ${s i n}^{- 1} (\frac{3}{4})$
Given,
The refractive index water is $μ_{w} = \frac{4}{3}$
From the figure it can be seen that ray coming out of water making an angle as $r = 90^{\circ}$
Applying Snell’s law of refraction
$μ_{w} sin θ = μ_{a} sin 90^{\circ}$
Where $μ_{a} = 1$ and $μ_{w} = \frac{4}{3}$
Therefore,
$\frac{4}{3} sin θ = 1 \Rightarrow sin θ = \frac{3}{4}$

$∴ θ = {sin}^{- 1} (\frac{3}{4})$

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Similar questions

(μ = \frac{4}{3})

A diver at a depth of 12m in water ( $μ$ = $\frac{4}{3}$ ) sees the sky in a cone of semi-vertical angle

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(μ = \frac{4}{3})

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- \frac{1}{2}

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