A converging lens forms a real image `I` on its optic axis. A rectangular galss slab of refractive index `mu` and thickness `t` is introduced between the lens and `I`. `I` will move

Options

A. towards the lens ` (mu-1)x `
B. towards the lens by `(1-1/mu)x `
C. anway from the lens by `(mu- 1)`
D. away from the lens by `(1-1/mu )x `

Explanation

To solve the question regarding the movement of the image `I` when a rectangular glass slab of refractive index `μ` and thickness `t` is introduced between a converging lens and the image, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- A converging lens forms a real image `I` on its optic axis.
- The image is formed at a distance `v` from the lens, where `v` is the image distance.

2. **Effect of Introducing the Glass Slab**:
- When a glass slab is introduced, it alters the effective optical path due to its refractive index `μ` and thickness `t`.
- The light travels slower in the glass slab compared to air, which affects the position of the image.

3. **Calculating the Shift in Image Position**:
- The optical path length (OPL) in air is given by the distance traveled multiplied by the refractive index of air (which is approximately 1).
- The OPL in the glass slab is given by `μ * t`.
- The effective shift in the image position due to the introduction of the slab can be calculated using the formula:
\[ \text{Shift} = t \left(1 - \frac{1}{\mu}\right) \]
- This formula accounts for the difference in the optical path length before and after the introduction of the slab.

4. **Determining the New Position of Image `I'`**:
- The new position of the image `I'` after the introduction of the slab will be:
\[ I' = I + \text{Shift} \]
- Therefore, substituting the shift:
\[ I' = I + t \left(1 - \frac{1}{\mu}\right) \]

5. **Conclusion**:
- The image `I` will move closer to the lens by the distance calculated above, resulting in a new image position `I'`.