An object is placed at a certain depth from the upper surface of a liquid. When it is seen from air, it appears that the object is raised through `(1)/(3)` of the depth. Again if water is taken instead of the liquid it appears that the object is raised above through `(1)/(4)` of the real depth. Refractive index of liquid = `(3)/(2)` and refractive index of water = `(4)/(3)`. Explain the matter.

Explanation

Suppose that the real depth of the object below the surface of the liquid is d and apparent depth is d..
`"We know", mu_(1) = ("real depth")/("apparent depth")`
`or, " " (3)/(2) = (d)/(d.) or d. = (2)/(3)d`
`therefore "Apparent change in depth"`
`= d -d. = d-(2)/(3)d = (1)/(3)d`
= apparent upward displacement of the object.
`"Again refractive index of water", mu_(w) = (4)/(3)`
So in case of water,
`mu_(w) = (d)/(d.) = (4)/(3) or, d. = (3)/(4) d`
`therefore "Apparent change in depth"`
`= d -d. = d-(3)/(4)d = (1)/(4)d`