Find the positions of the perinciple planes, the focal and model points of a thin biconvex symmetric glass lens with curvature radius of its surfaces equal to `R = 7.50 cm`. There is air on one side of the lens and water on the other.

Explanation

The primary and secondary focal length of a thick lens are given as,
`f =-(n//Phi) {1-(d//n') Phi_(2)}`
and `f' =+ (n''//Phi) {1 -(d//n') Phi_(1)}`,
where `Phi` is the lens power `n, n'` and `n''` are the refractive indices of first medium, lens material and the second medium beyond the lens. `Phi_(1)` and `Phi_(2)` are the powers of first second spherical surfcae of the lens.
Here, `n = 1`, for lens, `n' = n`, fro air
and `n'' = n_(0)`, for water.
So, `{:(f,=,-,1,//,Phi_(1)),(andf',=,+,n_(0),//,Phi):}},` as `d ~~0`, (1)
Now, power of a thin lens,
`Phi = Phi_(1) + Phi_(2)`,
where, `Phi_(1) = ((n - 1))/(R )`
and `Phi_(2) = ((n_(0) - n))/(-R)`
So, `Phi = (2n - n_(0))//R`
From equations (1) and (2), we get (2)
`f = (-R)/((2n - n - 1)) =-11.2 cm`
and `f = (n_(0)R)/((2n - n_(0) - 1)) =+ 14.9 cm`.
Since the distance between the primary principle point and primary nodal point is given as,
`x = f{(n'' - n)//n''}`
So, in this case, `x = (n_(0)//Phi) (n_(0) - 1)//n_(0) = (n_(0) - 1)//Phi`
`= (n_(0))/(Phi) - (1)/(Phi) = f' + f = 3.7 cm`.