A glass sphere with `10 cm` radius has a `5 cm` radius spherical hole at its centre. A narraow beam of parallel light is directed into the sphere. Where, if anywhere, will the sphere produce an image? The index of refraction of the glass is `1.50.`

Explanation

To solve the problem step by step, we will analyze the refraction of light through the glass sphere with a central spherical hole. The goal is to determine where the images are formed as the light passes through the sphere.

### Step 1: Understand the Geometry of the Problem
We have a glass sphere with a radius of 10 cm and a central spherical hole with a radius of 5 cm. The refractive index of glass is given as \( n_2 = 1.5 \) and the refractive index of air is \( n_1 = 1 \).

### Step 2: Identify the Surfaces
The sphere has four surfaces where refraction occurs:
1. Surface 1 (outer surface of the sphere)
2. Surface 2 (inner surface of the hole)
3. Surface 3 (inner surface of the sphere)
4. Surface 4 (outer surface of the hole)

### Step 3: Refraction at Surface 1
Using the formula for refraction at a spherical surface:
\[ \frac{n_2}{V} - \frac{n_1}{U} = \frac{n_2 - n_1}{R} \]
Here, \( U = -\infty \) (object at infinity), \( R = 10 \) cm (radius of the sphere), \( n_2 = 1.5 \), and \( n_1 = 1 \).

Substituting the values:
\[ \frac{1.5}{V} - \frac{1}{-\infty} = \frac{1.5 - 1}{10} \]
This simplifies to:
\[ \frac{1.5}{V} = \frac{0.5}{10} \]
\[ V = \frac{1.5 \times 10}{0.5} = 30 \text{ cm} \]
This means the first image \( I_1 \) is formed 30 cm from the surface 1 in the direction of the incident light.

### Step 4: Refraction at Surface 2
The image \( I_1 \) acts as an object for surface 2. The distance from \( I_1 \) to surface 2 is \( 30 - 5 = 25 \) cm (positive since it is in the direction of the incident light).

Using the same refraction formula:
\[ U = 25 \text{ cm}, \quad R = -5 \text{ cm}, \quad n_2 = 1, \quad n_1 = 1.5 \]
Substituting:
\[ \frac{1}{V_2} - \frac{1.5}{25} = \frac{1 - 1.5}{-5} \]
This simplifies to:
\[ \frac{1}{V_2} = \frac{1.5}{25} - \frac{0.5}{5} \]
Calculating:
\[ \frac{1}{V_2} = \frac{1.5}{25} - \frac{2.5}{25} = -\frac{1}{25} \]
Thus:
\[ V_2 = -25 \text{ cm} \]
This means the second image \( I_2 \) is formed 25 cm to the left of surface 2.

### Step 5: Refraction at Surface 3
The distance from \( I_2 \) to surface 3 is \( 25 + 10 = 35 \) cm (negative since it is in the opposite direction).

Using the refraction formula:
\[ U = -35 \text{ cm}, \quad R = -5 \text{ cm}, \quad n_2 = 1.5, \quad n_1 = 1 \]
Substituting:
\[ \frac{1.5}{V_3} - \frac{1}{-35} = \frac{1 - 1.5}{-5} \]
This simplifies to:
\[ \frac{1.5}{V_3} + \frac{1}{35} = \frac{-0.5}{-5} \]
Calculating:
\[ \frac{1.5}{V_3} + \frac{1}{35} = 0.1 \]
Solving for \( V_3 \):
\[ \frac{1.5}{V_3} = 0.1 - \frac{1}{35} \]
Calculating:
\[ \frac{1.5}{V_3} = \frac{3.5 - 1}{35} = \frac{2.5}{35} \]
Thus:
\[ V_3 = \frac{1.5 \times 35}{2.5} = 21 \text{ cm} \]

### Step 6: Refraction at Surface 4
The distance from \( I_3 \) to surface 4 is \( 21 + 5 = 26 \) cm (negative since it is in the opposite direction).

Using the refraction formula:
\[ U = -26 \text{ cm}, \quad R = -10 \text{ cm}, \quad n_2 = 1, \quad n_1 = 1.5 \]
Substituting:
\[ \frac{1}{V_4} - \frac{1.5}{-26} = \frac{1 - 1.5}{-10} \]
This simplifies to:
\[ \frac{1}{V_4} + \frac{1.5}{26} = \frac{-0.5}{-10} \]
Calculating:
\[ \frac{1}{V_4} + \frac{1.5}{26} = 0.05 \]
Solving for \( V_4 \):
\[ \frac{1}{V_4} = 0.05 - \frac{1.5}{26} \]
Calculating:
\[ \frac{1}{V_4} = \frac{1.3 - 1.5}{26} = -\frac{0.2}{26} \]
Thus:
\[ V_4 = -130 \text{ cm} \]

### Conclusion
The sphere produces four images at the following distances:
1. \( I_1 \) at \( +30 \) cm
2. \( I_2 \) at \( -25 \) cm
3. \( I_3 \) at \( -21 \) cm
4. \( I_4 \) at \( -130 \) cm