Question

Obtain an expression for refraction at a single convex spherical surface, i.e., the relation between μ₁ (rarer medium), μ₂ (denser medium), object distance u, image distance v and the radius of curvature R.

Answer 1

In ΔCOA,

i = α + γ and r = γ - β

γ = r + β

Let O be a point object in the rarer-medium of refractive index μ₁ and lying on the principle axis. The image of this object, O is formed by refraction at the convex spherical surface of radius at the point 'I' into the medium 'B' of refractive index μ₂ of curvature, as shown in the figure.

The convex surface has a small aperture and the angles of incidence ‘i’ and refraction ‘r’ are small.

Let ∠AOP = α, ∠AIP = β, ∠ACP = γ from A, and draw AN ⊥^er to the principle axis for refraction at point A₁. By snell's law,

${\displaystyle \frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}}$

∵ i and r are small.

then ${\displaystyle \frac{i}{r}=\frac{{\mu }_2}{{\mu }_1}}$

${\displaystyle \Rightarrow {\mu }_{1}i={\mu }_{2}r}$

Substituting the value of i and r.

μ₁(α + γ) = μ₂(γ - β)

∵ α, β, γ are small, they can be replaced.

∴ μ₁(tan α + tan γ) = μ₂(tan γ - tan β) 

or ${\displaystyle {\mu }_1(\frac{\text{AN}}{\text{NO}}+\frac{\text{AN}}{\text{NC}})={\mu }_2(\frac{\text{AN}}{\text{NC}}+\frac{\text{AN}}{\text{NI}})}$

or ${\displaystyle \frac{{\mu }_1}{\text{NO}}+\frac{{\mu }_1}{\text{NC}}=\frac{{\mu }_2}{\text{NC}}-\frac{{\mu }_2}{\text{NI}}}$

Due to small aperture, the point N lies close to P. Also, applying sign convention, we get

NO ~ PO = - u, NC ~ PC = R

NI ~ PI = v

Putting the value in the above expression

${\displaystyle \frac{{\mu }_1}{-u}+\frac{{\mu }_1}{\text{R}}=\frac{{\mu }_2}{\text{R}}-\frac{{\mu }_2}{v}}$

OR

${\displaystyle \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{\text{R}}}$