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

To solve this problem, let's use 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

In this scenario, the object is placed on the optical axis such that its real image is formed. Also, given that \(PO = PI\), we have \(u = -v\) and \(PO = v\).

Substitute the values in the formula:

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

After simplifying:

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

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

Cross-multiplying gives:

\(2.5R = 0.5v\)\)

Therefore, solving for \( v \):

\(v = 5R\)\)

Thus, the distance \( PO \), which is equal to \( v \), is \(5R\).

Therefore, the correct answer is \(5R\).

Answer 2

To solve this problem, we use the lensmaker's formula for a spherical refractive surface, given by:

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

where:
\(n_1 = 1\) (refractive index of air),
\(n_2 = 1.5\) (refractive index of glass),
\(R\) is the radius of curvature,
\(u\) is the object distance (positive, as it is on the left side),
\(v\) is the image distance (negative, as it is on the right side inside the glass).

The condition \(PO = PI\) implies that the object distance \(u = v\). Let's denote \(PO = PI = x\).

Substitute \(u = x\) and \(v = -x\) into the lensmaker's equation:

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

Simplifying, we have:

\(-\frac{1.5}{x} - \frac{1}{x} = \frac{0.5}{R}\)

Combining terms gives:

\(-\frac{2.5}{x} = \frac{0.5}{R}\)

Solving for \(x\), we find:

\[x = -\frac{2.5R}{0.5} = -5R\]

Since \(x\) represents distance and must be positive, this means \(x = 5R\).

Thus, the distance \(PO = 5R\).