A concave mirror of radius $40 cm$ lies on a horizontal table and water is filled in it up to a height of $5.00 cm$ . 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. The image of the dust particle as seen from a point directly above it is $8 + \frac{a}{4} cm$ . Find the value of $a$ . [Use refractive index of water is 1.33]

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1.1

3.1

2.1

2.2

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{- 20} = \frac{1}{v} + \frac{1}{- 5} \\ v = 6.67 cm \end{array}$
The depth of the point A' from the surface is 6.67+5 = 11.67cm

Thus the final image is formed at a point (11.67 /1.33)=8.774cm below the water surface.
But given the image of the dust particle as seen from a point directly above it is $8 + \frac{a}{4}$ cm equating this with the obtained value, we get $8 + \frac{a}{4}$ cm =8.774
$\frac{a}{4}$ = 0.774
a=3.1