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Geometrical Optics — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsGeometrical Optics5 Marks
Question
A cylindrical vessel, whose diameter and height both are equal to 30cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible. The particle P is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible?

✓

Answer

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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