Question

A spherical surface of radius of curvature \( R \), separates air from glass (refractive index = 1.5). The center of curvature is in the glass medium. A point object \( O \) placed in air on the optic axis of the surface, so that its real image is formed at \( I \) inside glass. The line \( OI \) intersects the spherical surface at \( P \) and \( PO = PI \). The distance \( PO \) equals:

• A. 5R
• B. 3R
• C. 2R
• D. 1.5R

Hint:

In optical systems involving refraction at spherical surfaces, the relationship between the object distance, image distance, and the focal length can be used to solve for various distances. In this case, we applied the refraction equation to find the distance where the image and object coincide on the spherical surface.

Answer 1

The Correct Option is A

The problem is solved using the formula for refraction at a spherical surface:

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

where:

• \(n_1 = 1\) (refractive index of air)
• \(n_2 = 1.5\) (refractive index of glass)
• \(u\) is the object distance
• \(v\) is the image distance
• \(R\) is the radius of curvature

Given that the object is placed on the optical axis, a real image is formed, and \(PO = PI\), it follows that \(u = -v\) and \(PO = v\).

Substituting these values into the formula:

\(\frac{1.5}{v} - \frac{1}{-v} = \frac{1.5 - 1}{R}\)

Simplification yields:

\(\frac{1.5 + 1}{v} = \frac{0.5}{R}\)

\(\frac{2.5}{v} = \frac{0.5}{R}\)

Cross-multiplication results in:

\(2.5R = 0.5v\)\)

Solving for \( v \):

\(v = 5R\)\)

Therefore, the distance \( PO \), which equals \( v \), is \(5R\).

The correct answer is \(5R\).