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

We are given a spherical surface separating air and glass, where the refractive index of glass is \( \mu_2 = 1.5 \) and the refractive index of air is \( \mu_1 = 1 \). The center of curvature lies in the glass medium, and the object \( O \) is placed in air on the optic axis of the spherical surface, while the real image \( I \) is formed inside the glass medium. We are asked to find the distance \( PO \), where \( P \) and \( I \) are the points where the image and object are formed, respectively, and we know that \( PO = PI \). Step 1: Relating Object Distance, Image Distance, and Focal Length
To solve this problem, we need to apply the equation of refraction at a spherical surface. The refraction formula is: \[ \frac{\mu_2 - \mu_1}{v} = \frac{\mu_2 - \mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] Where:
\( \mu_1 \) and \( \mu_2 \) are the refractive indices of air and glass, respectively.
\( u \) is the object distance (from the spherical surface, in air).
\( v \) is the image distance (from the spherical surface, inside glass).
\( R \) is the radius of curvature of the spherical surface.
Since \( \mu_1 = 1 \) (for air) and \( \mu_2 = 1.5 \) (for glass), we substitute these values into the equation:
\[ \frac{1.5 - 1}{v} = \frac{1.5 - 1}{u} = \frac{1.5 - 1}{R} \] Simplifying: \[ \frac{0.5}{v} = \frac{0.5}{u} = \frac{0.5}{R} \] Step 2: Relating Object and Image Distances
Next, we note that the object \( O \) is placed in air, so \( u = -x \), and the image \( I \) is formed inside the glass, so \( v = x \). The line \( OI \) intersects the spherical surface at \( P \), and we are told that \( PO = PI \), meaning that the distance from the object to the spherical surface is equal to the distance from the image to the spherical surface. Step 3: Solving for the Distance
We substitute the values of \( u \) and \( v \) into the refraction equation: \[ \frac{1.5}{x} + \frac{1}{x} = \frac{1}{2R} \] Simplifying: \[ \frac{5}{2x} = \frac{1}{2R} \] Now, solving for \( x \): \[ x = 5R \] Thus, the distance \( PO = x = 5R \). Step 4: Conclusion Therefore, the distance \( PO \) is \( 5R \).