Using formula of spherical surface or otherwise, find the apparent depth of an object placed `10cm` below the water surface, if seen near normally form air.

Explanation

Put `R = oo` in the formula of the Refraction at Spherical Surfaces we get,
`v = (un_(2))/(n_(1))`
`u = -10cm`
`n_(1) = (4)/(3)`
`n_(2) = 1`
`v = (10xx1)/(4//3) = 7.5cm`

negative sign imples that the images is formed in water.
`d_(app) = (d_("real"))/(mu_("rel"))`
`= (10)/(4//3) = (30)/(4) = 7.5cm`