Question

A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at

• A. 30 cm left
• B. infinity
• C. 1 cm to the right
• D. 18 cm to the left

Hint:

Negative \(v\) indicates virtual image on same side as object.

Answer 1

The Correct Option is A

To determine the position of the image formed by the glass rod with a spherical end, we can use the lens maker's formula for a spherical surface:

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

Where:

• \(n_1\) is the refractive index of the medium from which the object is placed (air), typically approximated as 1.
• \(n_2\) is the refractive index of the glass, generally more than 1 (e.g., 1.5 for glass).
• \(u\) and \(v\) are the object and image distances from the spherical surface, respectively.
• \(R\) is the radius of curvature of the spherical surface.

Let's apply the values:

• \(u = -15 \, \text{cm}\) (object distance, negative as per convention)
• \(R = 30 \, \text{cm}\)
• \(n_1 = 1\), \(n_2 = 1.5\) (approximate values for air and glass)

Substituting these values into the formula, we get:

\(\frac{1.5}{v} + \frac{1}{15} = \frac{1.5 - 1}{30}\)

This simplifies to:

\(\frac{1.5}{v} = \frac{0.5}{30} - \frac{1}{15}\)

Solving further:

\(\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}\)

\(\frac{1.5}{v} = \frac{1 - 4}{60}\)

\(\frac{1.5}{v} = \frac{-3}{60}\)

\(v = \frac{-1.5 \times 60}{3}\)

\(v = 30 \, \text{cm}\)

The negative sign indicates that the image is formed on the same side as the object. Therefore, the image is formed 30 cm to the left of the spherical surface.

Thus, the correct answer is: 30 cm left.