Consider the point object O placed in the medium with refractive index equal to n_{1} .

As n_{1} sin i = n_{2} sin r

For small aperture i , r → 0

⇒ n_{1} i = n_{2} r

In ΔOAC , i = α + γ

In ΔIAC , γ = r + β ⇒ r = γ – β

⇒ n_{1}(α + γ) = n_{2}( γ – β)

As aperture is small, α ≈ tan α , β ≈ tan β , for paraxial rays

γ ≈ tan γ

⇒ n_{1}(tan α + tanγ) = n_{2}(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 : n_{2} → refractive index of the medium into which light rays are entering ; n_{1} → 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