Question

A thin walled glass sphere of radius 2.5 cm is filled with water. An object (O) is placed at 7.5 cm from the surface of the sphere. Neglecting the effect of glass wall, at what distance the image (I) of the object, measured from the centre of sphere is formed?

• A. 20 cm
• B. 15 cm
• C. 10 cm
• D. 7.5 cm

Hint:

For refraction through a spherical surface, use $\frac\mu₂v - \frac\mu₁u = \frac\mu₂ - \mu₁R$.

Answer 1

The Correct Option is C

The problem involves finding the image distance of an object placed outside a thin-walled glass sphere filled with water. We will use the formula for refraction at a spherical surface to solve this problem. Given:

• Radius of the sphere, \(R = 2.5 \, \text{cm}\)
• Object distance from the surface, \(u = -7.5 \, \text{cm}\) (negative as per the sign convention)
• Refractive index of water, \(\mu_2 = 1.33\)
• Refractive index of air, \(\mu_1 = 1\)

The formula for refraction at a spherical surface is:

\(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\)

Substituting the given values:

\(\frac{1.33}{v} - \frac{1}{-7.5} = \frac{1.33 - 1}{2.5}\)

Simplifying the equation:

\(\frac{1.33}{v} + \frac{1}{7.5} = \frac{0.33}{2.5}\)

\(\frac{1.33}{v} = \frac{0.33}{2.5} - \frac{1}{7.5} = \frac{0.33 \times 3}{7.5} - \frac{1}{7.5}\)

\(\frac{1.33}{v} = \frac{0.99 - 1}{7.5} = \frac{-0.01}{7.5}\)

\(v = \frac{1.33 \times 7.5}{-0.01}\)

\(v \approx -10 \, \text{cm}\)

The image distance \(v\) from the surface of the sphere is \(-10 \, \text{cm}\). Measured from the center of the sphere, this distance is:

\(2.5 \, \text{cm} - 10 \, \text{cm} = 10 \, \text{cm}\)

Therefore, the correct answer is 10 cm.