Question

A concave mirror of radius 40cm lies on a horizontal table and water is filled in it upto a height 5cm.A small dust particle floats on th water surface at a point 'P' vertically above the point of contact of the mirror with the table.Locate the image of the dust particle as seen from a point directly above it.The refractive index of water is 1.3

Answer 1

Answer on Question #43669-Physics-Optics

A concave mirror of radius $40\mathrm{cm}$ lies on a horizontal table and water is filled in it up to a height $5\mathrm{cm}$ . A small dust particle floats on the water surface at a point 'P' vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The refractive index of water is 1.3.

Solution

The ray diagram is shown in figure. Let us first locate the image formed by the concave mirror. Let us take vertically upward as the negative axis. Then $R = -40 \, \text{cm}$ .

The object distance is $u = -5 \, \text{cm}$ . Using the mirror equation,

$$\frac {1}{u} + \frac {1}{v} = \frac {2}{R}$$

or,

$$\frac {1}{v} = \frac {2}{R} - \frac {1}{u} = \frac {2}{- 4 0 c m} - \frac {1}{- 5 c m} = \frac {6}{4 0 c m}$$

or, $v = 6.7 \, \text{cm}$ .

The positive sign show that the image $P_{1}$ is formed below the mirror and hence, it is virtual. These reflected rays are refracted at the water surface and go to the observer. The depth of the point $P_{1}$ from the surface is $6.7 \, \text{cm} + 5 \, \text{cm} = 11.7 \, \text{cm}$ . Due to refraction at the water surface, the image $P_{1}$ will be shifted above by a distance

$$(1 1. 7 c m) \left(1 - \frac {1}{1 . 3}\right) = 2. 7 c m.$$

Thus, the final image is formed at a point $(11.7 - 2.7)cm = 9cm$ below the water surface. Answer: $9cm$ .

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