A man standing on the bottom. The depth o fthe swimming pool looks at a stone lying on the bootom. The depth of the swimming pool is equal to `h`. At what distance from the surafce of water is the image of the stone formed if the line of vision makes an angle `theta `with the normal to the surface?

Explanation

From the Fig.
`sin d alpha = (MP)/(OM) = (MN cos alpha)/(h sec (alpha + dalpha))`
As `d alpha` is very small, so
`d alpha = (MN cos alpha)/(H sec alpha) = (MN cos^(2) alpha)/(h)` (1)
Similarly
`d theta = (MN cos^(2) theta)/(h')` (2)
From Eqn (1) and (2)
`(d alpha)/(d theta) = (h' cos^(2) alpha)/(h cos^(2) theta)` or, `h' = (h cos^(2) theta)/(cos^(2) alpha) (d alpha)/(d theta)`
From the law of refraction
`n sin alpha = sin theta` (A)
`sin alpha = (sin theta)/(n)`, so, `cos alpha = sqrt((n^(2) - sin^(2) theta)/n^(2))` (B)
Differentiating Eqn (A)
`n cos alpha d alpha = cos theta d theta` or, `(d alpha)/(d theta) = (cos theta)/(n cos theta)` (4)
Using (4) in (3), we get
`h' = (h cos^(3) theta)/(n cos^(3) alpha)` (5)
Hence `h' = (h cos^(3) theta)/(n((n^(2) - sin^(2) theta)/(n^(2)))^(3//2)) =(n^(2)h cos^(3) theta)/((n^(2)-sin^(2) theta)^(3//2))` [Using Eqn (B)]