For the given cylindrical vessel, dimetre = 30cm ⇒ r = 15cm and h = 30cm Now, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{3}{4}\Big[\mu_{\text{w}}=1.33=\frac{4}{3}\Big]$ $\Rightarrow\sin\text{i}=\frac{3}{4\sqrt{2}} \ \big[\because \text{r}=45^{\circ}\big]$
The point P will be visible when the refracted ray makes angle 45° at point of refraction. Let x = distance of point P from X. Now, $\tan45^{\circ}=\frac{\text{x}+10}{\text{d}}$$\Rightarrow\text{d}=\text{x}+10 \ ...(1)$
Again, $\tan\text{i}=\frac{\text{x}}{\text{d}}$$\Rightarrow\frac{3}{\sqrt{23}}=\frac{\text{d}-10}{\text{d}} \ \Big[\text{Since,} \ \sin\text{i}=\frac{3}{4\sqrt{2}}\Rightarrow\tan\text{i}=\frac{3}{4\sqrt{2}}\Big]$
$\Rightarrow\frac{3}{\sqrt{23}}-1=-\frac{10}{\text{d}}\Rightarrow\text{d}=\frac{\sqrt{23}\times10}{\sqrt{23}-3}=26.7\text{cm}$
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