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 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:
Show Hint
- 0.24 m right to the spherical surface
0.4 m left to the spherical surface
- 0.24 m left to the spherical surface
- 0.4 m right to the spherical surface
The Correct Option is B
Solution and Explanation
The problem asks to find the distance of the image of an object 'O' formed by a spherical surface that separates two media with refractive indices \( n_1 = 1 \) and \( n_2 = 1.5 \). The object distance, radius of curvature, and the positions of the object and center of curvature are given in the figure.
Concept Used:
For refraction at a single spherical surface, the relationship between the object distance (\( u \)), image distance (\( v \)), refractive indices (\( n_1 \) and \( n_2 \)), and the radius of curvature (\( R \)) is given by the lens maker's formula for a single surface:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
We use the Cartesian sign convention:
- All distances are measured from the pole (the vertex) of the spherical surface.
- Distances measured in the direction of the incident light are taken as positive.
- Distances measured in the direction opposite to the incident light are taken as negative.
Step-by-Step Solution:
Step 1: Identify and list the given quantities from the figure, applying the Cartesian sign convention. Let's assume the light travels from left to right.
- The object 'O' is in Medium-1, so the refractive index of the object medium is \( n_1 = 1 \).
- The light refracts into Medium-2, so the refractive index of the image medium is \( n_2 = 1.5 \).
- The object 'O' is located 0.2 m to the left of the spherical surface. Since this is against the direction of incident light, the object distance is \( u = -0.2 \, \text{m} \).
- The center of curvature 'C' is located to the right of the surface. Therefore, the radius of curvature is measured in the direction of incident light and is positive, \( R = +0.4 \, \text{m} \).
Step 2: Substitute these values into the spherical surface refraction formula.
\[ \frac{1.5}{v} - \frac{1}{(-0.2 \, \text{m})} = \frac{1.5 - 1}{(+0.4 \, \text{m})} \]
Step 3: Simplify the equation.
\[ \frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4} \]
Evaluating the fractions:
\[ \frac{1}{0.2} = 5 \quad \text{and} \quad \frac{0.5}{0.4} = \frac{5}{4} = 1.25 \]
So the equation becomes:
\[ \frac{1.5}{v} + 5 = 1.25 \]
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: Perform the final calculation.
\[ v = - \frac{1.5}{3.75} = - \frac{150}{375} = - \frac{2 \times 75}{5 \times 75} = - \frac{2}{5} \] \[ v = -0.4 \, \text{m} \]
The negative sign for \( v \) indicates that the image is formed on the same side as the object, which is to the left of the spherical surface. The image is virtual.
Thus, the distance of the image is 0.4 m left to the spherical surface.
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