Question

A point object is kept in front of a convex spherical surface of radius of curvature R. Draw the ray diagram to show the formation of image and derive the relation between the object and image distance (u and v) in terms of refractive index n of the medium and R.

Answer 1

For a point object placed in a rarer medium (n₁) in front of a convex spherical surface of a denser medium (n₂), the relationship between object distance (u), image distance (v), and radius of curvature (R) is given by:

`(n_2)/v - (n_1)/u = (n_2 - n_1)/R`

In ΔOMC, the exterior angle is equal to the sum of the interior opposite angles:

i = α + γ   ...(1)

the angle

In ΔMIC, is the exterior angle:

γ = r + β

r = γ − β    ...(2)

According to Snell’s Law:

n₁ sin i = n₂ sin r

For small angles, sin θ ≈ θ (in radians):

n₁ i = n₂ r

Substituting equations (1) and (2) into this:

n₁(α + γ) = n₂(γ − β)

n₁α + n₁γ = n₂γ − n₂β

n₁α + n₂β = (n₂ − n₁)γ   ...(3)

Since the aperture is small, we can approximate the angles using the tangent (where MP is the perpendicular arc/height):

α ≈ `(MP)/(PO)`

β ≈ `(MP)/(PI)`

γ ≈ `(MP)/(PC)`

Using the New Cartesian Sign Convention:

Object distance PO = −u

Image distance PI = +v

Radius of curvature PC = +R

Substituting these into Equation (3):

`n_1 ((MP)/(-u)) + n_2((MP)/(v)) = (n_2 − n_1)((MP)/(R)) `

Dividing throughout by MP, we get the final relation:

`(n_2)/v - (n_1)/u = (n_2 - n_1)/R`