Obtain the equation for apparent depth.

Explanation

Light from the object O at the bottom of the tank passes from medium (water) to rarer medium (air) to reach our eyes. It deviates away from the normal in the rarer medium at the point of incidence B. The refractive index of the denser medium is `n_(1)` and rarer medium `n_(2)`. Here, `n_(1) gt n_(2)`. The angle of incidence in the denser medium is i and the angle of refraction in the rarer medium is r. The lines N N. and OD are parallel. Thus angle `angle DIB` is also r. The anles i and r are very small as the diverging light from O entering the eye is very narrow. The Snell.s law in product from for this refraction is,
`n_(1) sin i = n_(2) sin r`
As the angles i and r are small, we can approximate, ` sin i approx tan i,`
`n_(1) tan i = n_(2) tan i`
in triangles ` Delta DOB and Delta DIB`,
`tan(i)=(DB)/(DO)andtan(r)=(DB)/(DI)`
`n_(1)=(DB)/(DO)=n_(2)(DB)/(DI)`
DB is cancelled on both sides, DO is the actual depth d and DI is the apparent depth d..
`n_(1)(1)/(d)=n_(2)(1)/(d.)rArr(d.)/(d)=(n_(2))/(n_(1))`
Rerranging the above equation for the apparent depth d..
`d. = (n_(2))/(n_(1)) d`
As the rarer medium is air and its refractive index `n_(2)` can be taken as I, `(n_(2) = 1).` And the refractive index `n_(1)` of denser medium could then be taken as `n, (n_(1) = n)`.
In that case, the equation for apparent depth becomes,
`d. = (d)/(n)`