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

Applying Refraction at Spherical Surface: 

Given Information: Radius of the glass sphere, \(R = 40 \text{ cm}\).
Refractive index of the glass, \(\mu = 1.5\).
Distance of the bubble from the center of the sphere, \(CB = 20 \text{ cm}\).
Therefore, the distance of the bubble from the refracting surface (point B), \(u = -20 \text{ cm}\). (Negative because it's opposite the direction of incident light).
Refraction at a Spherical Surface:
The formula for refraction at a spherical surface is:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] where:
\(\mu_1\) is the refractive index of the medium where the object is located (glass = 1.5)
\(\mu_2\) is the refractive index of the medium where the image is formed (air = 1)
\(u\) is the object distance
\(v\) is the image distance
\(R\) is the radius of curvature of the surface
Applying the Formula:
\(\mu_1 = 1.5\)
\(\mu_2 = 1\)
\(u = -20 \text{ cm}\)
\(R = -40 \text{ cm}\) (Negative because the center of curvature is on the same side as the incident light).
Substituting these values into the formula: \[ \frac{1}{v} - \frac{1.5}{-20} = \frac{1 - 1.5}{-40} \] \[ \frac{1}{v} + \frac{1.5}{20} = \frac{-0.5}{-40} \] \[ \frac{1}{v} = \frac{0.5}{40} - \frac{1.5}{20} \] \[ \frac{1}{v} = \frac{0.5 - 3}{40} \] \[ \frac{1}{v} = \frac{-2.5}{40} \] \[ v = \frac{40}{-2.5} = -16 \text{ cm} \] Interpretation:
The image distance \(v = -16 \text{ cm}\). The negative sign indicates that the image is formed on the same side of the refracting surface as the object (the air bubble). Therefore:
Position: The image is located 16 cm from point B. Nature: Since the image is formed on the same side as the object, it is a virtual image.
The image of the air bubble is virtual and is located 16 cm from point B on the same side as the bubble.