Refraction at Curved Surface μ
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.
Transverse Magnification for Refracting Spherical Surface
Transverse Magnification $\large m = \frac{h_I}{h_O} = \frac{-A’B’}{AB}$
$\large \frac{Sini}{Sinr} = \frac{\mu_2}{\mu_1}$
For small angle of incidence & refraction Sini = tani & Sinr = tanr
$\large \frac{tani}{tanr} = \frac{\mu_2}{\mu_1}$
In triangle ABP & A’B’P ,
$\large \frac{AB/PB}{A’B’/PB’} = \frac{\mu_2}{\mu_1}$
$\large -\frac{A’B’}{AB} = \frac{-PB’/\mu_2}{PB/\mu_1}$
Hence $\large m = \frac{v/\mu_2}{u/\mu_1} $
Solved Example : 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