A convex lens of focal length `f_(1)` is placed in front of a luminous point object. The separation between the object and the lens is `3f_(1)` . A glass slab of thikness t is placed between the object and the lens. A real image of the object is formed at the shortest possible distance from the object.
a. Find the refractive index of the slab.
b. If a concave lens of very large focal length `f_(2)` is placed in contact with the convex lens, find the shifting of the image.

Explanation

### Step-by-Step Solution

**Part (a): Finding the Refractive Index of the Slab**

1. **Understanding the Setup**:
- We have a convex lens with a focal length \( f_1 \).
- The object is placed at a distance of \( 3f_1 \) from the lens.
- A glass slab of thickness \( t \) is placed between the object and the lens.

2. **Finding the Apparent Shift**:
- The real image is formed at the shortest possible distance from the object, which occurs when the object is at the focal point of the lens. This means the image is formed at \( 2f_1 \).
- The apparent position of the object due to the slab is shifted. The distance from the lens to the object is originally \( 3f_1 \), but due to the slab, it appears to be at \( 2f_1 \).

3. **Calculating the Apparent Shift**:
- The apparent shift \( S \) can be expressed as:
\[ S = \text{original distance} - \text{apparent distance} = 3f_1 - 2f_1 = f_1 \]

4. **Using the Formula for Apparent Shift**:
- The formula for the apparent shift due to a slab is given by:
\[ S = t \left(1 - \frac{1}{\mu}\right) \]
- Setting the two expressions for the apparent shift equal gives:
\[ f_1 = t \left(1 - \frac{1}{\mu}\right) \]

5. **Rearranging to Find the Refractive Index**:
- Rearranging the equation:
\[ 1 - \frac{1}{\mu} = \frac{f_1}{t} \]
\[ \frac{1}{\mu} = 1 - \frac{f_1}{t} \]
\[ \mu = \frac{t}{t - f_1} \]

**Conclusion for Part (a)**:
The refractive index of the slab is:
\[ \mu = \frac{t}{t - f_1} \]

---

**Part (b): Finding the Shifting of the Image with a Concave Lens**

1. **Understanding the New Setup**:
- A concave lens with a very large focal length \( f_2 \) is placed in contact with the convex lens.
- The focal length of the concave lens is negative, so \( f_2 \) is treated as negative.

2. **Finding the Combined Focal Length**:
- The formula for the combined focal length \( f \) of two lenses in contact is:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
- Substituting \( f_2 = -f_2 \):
\[ \frac{1}{f} = \frac{1}{f_1} - \frac{1}{f_2} \]
- Rearranging gives:
\[ f = \frac{f_1 f_2}{f_2 - f_1} \]

3. **Finding the Image Position**:
- The object distance \( u \) is the apparent position after the slab, which is \( -2f_1 \).
- Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
- Substituting \( u = -2f_1 \):
\[ \frac{1}{v} + \frac{1}{2f_1} = \frac{1}{f} \]
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{2f_1} \]

4. **Finding the New Image Position**:
- Substitute \( f \) into the equation:
\[ \frac{1}{v} = \frac{f_2 - f_1}{f_1 f_2} - \frac{1}{2f_1} \]
- After simplification, we find:
\[ v = \frac{2f_1 f_2}{f_2 - 2f_1} \]

5. **Calculating the Shift of the Image**:
- The original image position was at \( 2f_1 \).
- The shift \( \Delta v \) is given by:
\[ \Delta v = v - 2f_1 \]
- Substituting the expression for \( v \):
\[ \Delta v = \frac{2f_1 f_2}{f_2 - 2f_1} - 2f_1 \]
- Simplifying gives:
\[ \Delta v = \frac{4f_1^2}{f_2 - 2f_1} \]

6. **Considering the Large Focal Length**:
- If \( f_2 \) is very large, we can ignore \( -2f_1 \):
\[ \Delta v \approx \frac{4f_1^2}{f_2} \]

**Conclusion for Part (b)**:
The shift of the image is:
\[ \Delta v = \frac{4f_1^2}{f_2} \]

---