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

Provided Data Summary

• Optical Media Sequence: Air (refractive index \(n_1=1\)) transitions to Glass (refractive index \(n_2=1.5\)), and then back to Air.
• Spherical Lens Geometry: Radius of curvature \(R = 15\) cm; Diameter \(D = 30\) cm.
• Incident Light: Parallel beam, implying the object is at an infinite distance.
• Coordinate System Convention: Distances are measured from each surface. Positive distances are directed to the right, aligning with the direction of light propagation.
• Fundamental Refraction Formula for a Spherical Surface:

\( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)

Detailed Calculation Process

Phase 1 — Initial Refraction (Air to Glass) at the Left Surface

An object at infinity corresponds to \( \frac{1}{u_1} \approx 0 \). For the left surface, the center of curvature is situated to the right, hence \( R_1 = +15 \) cm.

Applying the formula: \( \frac{1.5}{v_1} - 0 = \frac{1.5 - 1}{15} \). Solving yields \( v_1 = 45 \) cm. This indicates that the light rays would converge 45 cm to the right of the left surface (within the glass) if the second surface were absent.

Phase 2 — Geometric Configuration for the Second Surface

The axial thickness of the sphere is its diameter, which is 30 cm. Consequently, the virtual focal point from Phase 1 is located \( 45 - 30 = 15 \) cm to the right of the right surface. This point acts as a virtual object for the right surface, situated at \( u_2 = +15 \) cm to its right, maintaining the established positive direction.

Phase 3 — Subsequent Refraction (Glass to Air) at the Right Surface

For the right surface, the center of curvature is to the left, resulting in \( R_2 = -15 \) cm. The refractive indices are \( n_1 = 1.5 \) (inside the glass) and \( n_2 = 1 \) (outside in the air).

Using the refraction formula: \( \frac{1}{v_2} - \frac{1.5}{15} = \frac{1 - 1.5}{-15} \). This simplifies to \( \frac{1}{v_2} - 0.1 = \frac{1}{30} \). Solving for \( v_2 \) gives \( \frac{1}{v_2} = 0.1333... \), which means \( v_2 = 7.5 \) cm.

Conclusion: The exiting light rays form a real image on the optical axis, located 7.5 cm to the right of the sphere's right surface (i.e., external to the ball).

The ball lens can also be characterized by:

• Equivalent Focal Length (EFL), measured from the lens's principal plane: \( f = \frac{nD}{4(n-1)} = \frac{1.5 \times 30}{4 \times 0.5} = 22.5 \) cm.
• Back Focal Length (BFL), measured from the right surface: \( \text{BFL} = f - R = 22.5 - 15 = 7.5 \) cm.

Essential Points for Recall

• When incident light is parallel, apply the surface refraction formula \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) sequentially for each surface, using the image formed by one surface as the object for the next.
• A quick method for ball lenses: \( \text{EFL} = \frac{nD}{4(n-1)} \) and \( \text{BFL} = \text{EFL} - R \). For the given parameters (\(n=1.5, R=15\) cm), the BFL is 7.5 cm.
• The resulting image in this scenario is real, significantly reduced (forming a point focus), and positioned outside the spherical lens.