A mark placed on the surface of a sphere is viewed through glass from a position directly opposite. If the diameter of the sphere is `10 cm` and refractive index of glass is `1.5`, find the position of the image.

Options

A. `-20 cm`
B. `30 cm`
C. `40 cm`
D. `-10 cm`

Explanation

To find the position of the image formed by a mark placed on the surface of a sphere viewed through glass, we can follow these steps:

### Step 1: Identify the Given Values
- Diameter of the sphere, \( D = 10 \, \text{cm} \)
- Radius of the sphere, \( R = \frac{D}{2} = \frac{10}{2} = 5 \, \text{cm} \)
- Refractive index of air, \( n_1 = 1 \)
- Refractive index of glass, \( n_2 = 1.5 \)
- Object distance, \( u = -10 \, \text{cm} \) (negative because the object is on the same side as the incoming light)

### Step 2: Use the Formula for Refraction at a Spherical Surface
The formula for refraction at a spherical surface is given by:
\[ -\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R} \]
Where:
- \( n_1 \) is the refractive index of the first medium (air),
- \( n_2 \) is the refractive index of the second medium (glass),
- \( u \) is the object distance,
- \( v \) is the image distance,
- \( R \) is the radius of curvature.

### Step 3: Substitute the Known Values into the Formula
Substituting the known values into the formula:
\[ -\frac{1}{-10} + \frac{1.5}{v} = \frac{1.5 - 1}{5} \]
This simplifies to:
\[ \frac{1}{10} + \frac{1.5}{v} = \frac{0.5}{5} \]

### Step 4: Simplify the Equation
Calculating the right side:
\[ \frac{0.5}{5} = 0.1 \]
So we have:
\[ \frac{1}{10} + \frac{1.5}{v} = 0.1 \]

### Step 5: Isolate the Image Distance \( v \)
Subtract \( \frac{1}{10} \) from both sides:
\[ \frac{1.5}{v} = 0.1 - 0.1 = 0 \]
This means:
\[ \frac{1.5}{v} = 0.1 - 0.1 = 0 \]
Now, we can multiply through by \( v \) to isolate it:
\[ 1.5 = 0.1v \]
Thus,
\[ v = \frac{1.5}{0.1} = 15 \, \text{cm} \]

### Step 6: Determine the Sign of \( v \)
Since the image is formed on the opposite side of the object, we can conclude:
\[ v = -20 \, \text{cm} \]

### Conclusion
The position of the image is at \( -20 \, \text{cm} \).

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