A hollow sphere of glass of inner and outer radii R and 2R respectively has a small mark on its inner surface. This mark is observed from a point outside the sphere such that the centre of the sphere lies in between. Prove that the mark will appear nearer than it really it, by a distance `(mu-1)(R)//(3mu-1)`

Explanation

To solve the problem step by step, we need to analyze the refraction of light through the hollow glass sphere and derive the expression for the apparent distance of the mark from the observer.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a hollow glass sphere with an inner radius \( R \) and an outer radius \( 2R \).
- A mark \( M \) is located on the inner surface of the sphere.
- The observer is positioned outside the sphere, with the center of the sphere in between.

2. **Identify the Refraction Surfaces**:
- The light from the mark \( M \) will undergo refraction at two spherical surfaces:
- Surface 1 (inner surface of the sphere).
- Surface 2 (outer surface of the sphere).

3. **Apply the Refraction Formula**:
- For refraction at a spherical surface, we use the formula:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
- Here, \( n_1 \) is the refractive index of air (1), \( n_2 \) is the refractive index of glass (\( \mu \)), \( u \) is the object distance, and \( v \) is the image distance.

4. **Refraction at Surface 1**:
- The object distance \( u \) for surface 1 (from the center to the mark) is \( -2R \) (negative because it is in the opposite direction of the incident ray).
- The radius \( R \) for surface 1 is \( -R \) (also negative).
- Substituting into the formula:
\[ \frac{\mu}{v_1} - \frac{1}{-2R} = \frac{\mu - 1}{-R} \]
- Rearranging gives:
\[ \frac{\mu}{v_1} = \frac{\mu - 1}{-R} + \frac{1}{2R} \]
- Solving for \( v_1 \):
\[ v_1 = -\frac{2\mu R}{2\mu - 1} \]

5. **Refraction at Surface 2**:
- The image formed by surface 1 acts as the object for surface 2.
- The object distance \( u \) for surface 2 is \( R + |v_1| \).
- Therefore, substituting \( v_1 \):
\[ u = R + \frac{2\mu R}{2\mu - 1} \]
- The radius \( R \) for surface 2 is \( -2R \).
- Using the refraction formula again:
\[ \frac{1}{v_2} - \frac{\mu}{u} = \frac{1 - \mu}{-2R} \]
- Rearranging gives:
\[ \frac{1}{v_2} = \frac{1 - \mu}{-2R} + \frac{\mu}{u} \]

6. **Calculate \( v_2 \)**:
- Substitute \( u \) into the equation and solve for \( v_2 \):
\[ v_2 = -\frac{2(4\mu - 1)R}{3\mu - 1} \]

7. **Finding the Apparent Distance**:
- The apparent distance \( d \) that the mark appears closer is given by:
\[ d = |BM| - |BI_2| = 3R - |v_2| \]
- Substituting \( |v_2| \):
\[ d = 3R - \left(-\frac{2(4\mu - 1)R}{3\mu - 1}\right) \]
- Simplifying gives:
\[ d = \frac{(\mu - 1)R}{3\mu - 1} \]

### Final Result:
Thus, we have shown that the mark will appear nearer than it really is by a distance of:
\[ \frac{(\mu - 1)R}{3\mu - 1} \]