Question

An air bubble is trapped at point B (CB = 20 cm) in a glass sphere of radius 40 cm and refractive index 1.5 as shown in the figure. Find the nature and position of the image of the bubble as seen by an observer at point P. 

Hint:

For refraction at a spherical surface, use the formula \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) to locate the image position.

Answer 1

Refraction at a Spherical Surface:

Given:
Radius of glass sphere, \(R = 40 \text{ cm}\).
Refractive index of glass, \(\mu = 1.5\).
Distance of bubble from sphere's center, \(CB = 20 \text{ cm}\).
Object distance (bubble to surface B), \(u = -20 \text{ cm}\) (negative as it's against incident light direction).
Formula for Refraction at a Spherical Surface:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] where:
\(\mu_1\): Refractive index of object's medium (glass = 1.5).
\(\mu_2\): Refractive index of image's medium (air = 1).
\(u\): Object distance.
\(v\): Image distance.
\(R\): Radius of curvature.
Calculation:
\(\mu_1 = 1.5\)
\(\mu_2 = 1\)
\(u = -20 \text{ cm}\)
\(R = -40 \text{ cm}\) (negative as center of curvature is on incident light side).
Substitution into the formula yields:
\[ \frac{1}{v} - \frac{1.5}{-20} = \frac{1 - 1.5}{-40} \]
\[ \frac{1}{v} + \frac{1.5}{20} = \frac{-0.5}{-40} \]
\[ \frac{1}{v} = \frac{0.5}{40} - \frac{1.5}{20} \]
\[ \frac{1}{v} = \frac{0.5 - 3}{40} \]
\[ \frac{1}{v} = \frac{-2.5}{40} \]
\[ v = \frac{40}{-2.5} = -16 \text{ cm} \]
Result:
The image distance is \(v = -16 \text{ cm}\). The negative sign signifies the image forms on the same side as the object (the air bubble) relative to the refracting surface.
Conclusion: The image of the air bubble is virtual and is located 16 cm from point B, on the same side as the bubble.