Consider the point object O placed in the medium with refractive index equal to n1 .
As n1 sin i = n2 sin r
For small aperture i , r → 0
⇒ n1 i = n2 r
In ΔOAC , i = α + γ
In ΔIAC , γ = r + β ⇒ r = γ – β
⇒ n1(α + γ) = n2( γ – β)
As aperture is small, α ≈ tan α , β ≈ tan β , for paraxial rays
γ ≈ tan γ
⇒ n1(tan α + tanγ) = n2(tan γ – tanβ)
$ \displaystyle n_1 (\frac{AM}{MO} + \frac{AM}{MC} ) = n_2 ( \frac{AM}{MC} – \frac{AM}{MI}) $
$ \displaystyle n_1 (\frac{1}{MO} + \frac{1}{MC} ) = n_2 ( \frac{1}{MC} – \frac{1}{MI}) $
As M is close to P ,
$ \displaystyle n_1 (\frac{1}{PO} + \frac{1}{PC} ) = n_2 ( \frac{1}{PC} – \frac{1}{PI}) $
Applying sign convention,
i.e., PO = -u , PI= v & PC = R
∴For a spherical surface,
$ \displaystyle \frac{n_2}{v}-\frac{n_1}{u} = \frac{n_2 -n_1}{R} $
The symbols should be carefully remembered as : n2 → refractive index of the medium into which light rays are entering ; n1 → refractive index of the medium from which light rays are coming. Care should also be taken while applying the sign convention to R.
Illustration : Shown in the figure is a spherical surface of radius of curvature R & R.I. (n = 1.5). Find the distance of the silvering of the plane surface so as to form an image at the pole due to a very distant object.
Solution:
Since light comes from very distant object
u = ∞
$ \displaystyle \frac{n}{v}-\frac{1}{\infty} = \frac{n-1}{R} $
$ \displaystyle v = \frac{nR}{n-1} $
$ \displaystyle v = \frac{1.5 R}{1.5-1} $
v = 3 R
Due to presence of the plane mirror, the image formed at I behaves as a virtual object for the plane mirror & a real image I’ is formed in front of the plane mirror, at the pole P.
⇒ x + x = v
⇒ x = v/2
x = 1.5R