A concave mirror of radius 40 cm lies on a horizontal table and water is filled in it upto a height of 5cm. 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 directly above it. `mu_(w)`=2/3.

Explanation

O is the real position of the dust particle. `O_(1)` is the image of O due to reflection at concave mirror. `O_(2)` is the image of `O_(1)`
due to refraction at water-air interface.
For reflection at concave mirror, u = -PO = -5cm,
R = -40 cm
`therefore"From "1/v+1/u=2/R`, we get
`1/v=(-2)/40-1/(-5)=6/40cm`
or `v=40/6cm=6.67cm`
`O_(1)` will act as an object for refraction at water-air interface.
Hence `OO_(1)` = real depth of the object = 5 cm + 6.67 cm = 11.67 cm
`OO_(2)` = the apparent depth of `O_(1)`
`therefore` From `mu=("real depth")/("apparent depth")`
apparent depth = `("real depth")/(mu)=(11.67cm)/1.33=8.75cm`
Hence, the dust will be seen at a depth of 8.75 cm below the water surface.