A pole is vertically submerged in swimming pool, such that it gives...

Let angle $BAC$ is $\theta$.

Applying Snell's law at $E$, we get $1 \times \sin 60 ° = \dfrac{4}{3} \times sin\theta$

$⇒ sin\theta = \dfrac{3\sqrt{3}}{8} ⇒ tan\theta = \dfrac{sin\theta}{\sqrt{1 - {\left(sin\theta\right)}^{2}}} = \dfrac{3\sqrt{3}}{\sqrt{37}}$

Now, $BC = 1 . 5 \times tan\theta = 1 . 28$

Then, $AB = 2 . 15 - BC = 0 . 86m$

Required value, $DF = ABtan30 ° = 0 . 5015m \approx 50cm$