Question

A spherical surface separates two media of refractive indices \( n_1 = 1 \) and \( n_2 = 1.5 \) as shown in the figure. Distance of the image of an object \( O \), if \( C \) is the center of curvature of the spherical surface and \( R \) is the radius of curvature, is:

• A. 0.24 m right to the spherical surface
• B. 0.4 m left to the spherical surface
• C. 0.24 m left to the spherical surface
• D. 0.4 m right to the spherical surface

Hint:

When solving for the position of an image in spherical surfaces, always use the appropriate sign convention for distances and refractive indices.

Answer 1

The Correct Option is B

The objective is to determine the image distance for an object 'O' positioned at a spherical interface separating two media with refractive indices \( n_1 = 1 \) and \( n_2 = 1.5 \). The figure provides the object distance, radius of curvature, and the locations of the object and center of curvature.

Concept Used:

The formula governing refraction at a single spherical surface, relating object distance (\( u \)), image distance (\( v \)), refractive indices (\( n_1 \) and \( n_2 \)), and radius of curvature (\( R \)), is:

\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

The following Cartesian sign convention is applied:

1. Measurements originate from the pole (vertex) of the spherical surface.
2. Distances in the direction of incident light are positive.
3. Distances against the direction of incident light are negative.

Step-by-Step Solution:

Step 1: Extract and list the given parameters from the figure, adhering to the Cartesian sign convention, assuming light propagation from left to right.

• Object Medium Refractive Index: \( n_1 = 1 \).
• Image Medium Refractive Index: \( n_2 = 1.5 \).
• Object Distance: The object 'O' is 0.2 m to the left of the surface, opposing the incident light, hence \( u = -0.2 \, \text{m} \).
• Radius of Curvature: The center of curvature 'C' is to the right of the surface, aligning with the incident light direction, thus \( R = +0.4 \, \text{m} \).

Step 2: Input these values into the spherical surface refraction equation.

\[ \frac{1.5}{v} - \frac{1}{(-0.2 \, \text{m})} = \frac{1.5 - 1}{(+0.4 \, \text{m})} \]

Step 3: Simplify the equation.

\[ \frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4} \]

Calculate the fractional values:

\[ \frac{1}{0.2} = 5 \quad \text{and} \quad \frac{0.5}{0.4} = 1.25 \]

The equation is now:

\[ \frac{1.5}{v} + 5 = 1.25 \]

Step 4: Solve for the image distance \( v \).

\[ \frac{1.5}{v} = 1.25 - 5 \] \[ \frac{1.5}{v} = -3.75 \] \[ v = \frac{1.5}{-3.75} \]

Step 5: Execute the final computation.

\[ v = - \frac{1.5}{3.75} = - \frac{150}{375} = - \frac{2 \times 75}{5 \times 75} = - \frac{2}{5} \] \[ v = -0.4 \, \text{m} \]

The negative value of \( v \) signifies that the image is formed on the same side as the object, to the left of the spherical surface, and is virtual. Therefore, the image is located 0.4 m to the left of the spherical surface.