# Refraction at Curved Surface

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

i = α + γ  &   γ =  r + 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