Question

A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed.

Hint:

For refraction at a curved surface, use the lens-maker’s equation, keeping sign conventions in mind.

Answer 1

Given Data (for quick reference)

• Mediums: Air (n1=1) → Glass (n2=1.5) → Air.
• Sphere (ball lens) radius: R = 15 cm; diameter: D = 30 cm.
• Incoming beam: parallel (object at infinity).
• Sign convention (Cartesian): distances measured from each surface; positive to the right (direction of light).
• Refracting-surface formula:

n₂/v − n₁/u = (n₂ − n₁)/R

Step-by-Step Derivation

Step 1 — First refraction (air → glass) at the left surface

Object at infinity ⇒ 1/u₁ ≈ 0. For the left surface, the center of curvature is to the right, so R₁ = +15 cm.

n₂/v₁ − n₁/u₁ = (n₂ − n₁)/R₁ ⇒ (1.5)/v₁ − 0 = (1.5 − 1)/15 ⇒ v₁ = 45 cm.

This means the rays would meet 45 cm to the right of the left surface (inside the glass) if there were no second surface.

Step 2 — Geometry for the second surface

The sphere’s thickness along the axis is its diameter: 30 cm. The “would-be” focus of Step 1 is therefore 45 − 30 = 15 cm to the right of the right surface. For the right surface, this is a virtual object located u₂ = +15 cm to its right (our positive direction remains to the right).

Step 3 — Second refraction (glass → air) at the right surface

Here, the center of curvature is to the left, so R₂ = −15 cm, with n₁ = 1.5 (inside) and n₂ = 1 (outside).

n₂/v₂ − n₁/u₂ = (n₂ − n₁)/R₂ 
1/v₂ − (1.5/15) = (1 − 1.5)/(−15) ⇒ 1/v₂ − 0.1 = 1/30 ⇒ 1/v₂ = 0.1333… ⇒ v₂ = 7.5 cm.

Final answer: The emergent rays form a real image on the axis at a distance 7.5 cm to the right of the sphere’s right surface (i.e., outside the ball).
Equivalently, the ball lens has:

• Effective focal length (EFL) from the center: f = nD / (4(n−1)) = (1.5×30)/(4×0.5) = 22.5 cm.
• Back focal length (BFL) from the right surface: BFL = f − R = 22.5 − 15 = 7.5 cm.

Key Takeaways (Exam-ready)

• For parallel light, treat each surface with n₂/v − n₁/u = (n₂ − n₁)/R and carry the intermediate image to the next surface.
• Ball-lens quick check: EFL = nD / (4(n−1)), BFL = EFL − R. For n=1.5, R=15 cm, BFL=7.5 cm.
• The final image here is real, highly reduced (a point focus), located outside the sphere.