Question

A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small
particle P is placed in it at a distance of 5.0 cm 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 1

Answer on Question #63449, Physics / Optics

A cylindrical vessel, whose diameter and height both are equal to $30~\mathrm{cm}$, is placed on a horizontal surface and a small particle P is placed in it at a distance of $5.0~\mathrm{cm}$ 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?

Solution:

The refractive index of water is $n = 4/3$.

$$x = 30 - h$$ $$PA = 20 - x = 20 - (30 - h) = h - 10$$

By using Snell's law, we get

$$\sin r = n \sin i$$ $$\sin i = \frac{h - 10}{\sqrt{h^2 + (h - 10)^2}}$$

So,

$$\sin 45{}^\circ = \frac{4}{3} \frac{h - 10}{\sqrt{h^2 + (h - 10)^2}}$$ $$\frac{9}{2} (h^2 + (h - 10)^2) = 16 (h - 10)^2$$ $$9 h^2 = 23 (h - 10)^2$$ $$h = \frac{\sqrt{23} \cdot 10}{\sqrt{23} - 3} = 26.7~\mathrm{cm}$$

Answer: 26.7 cm

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