A lens forms a sharp image on a screen. On inserting a parallel sided glass slab between the lens and the screen, it is found necessary to move the screen a distance `d` away the lens in order for the image to be sharp again. If the refractive index of the material of the slab is `mu`, the thickness of the slab is

Options

A. `mu d`
B. `(d)/(mu)`
C. `(mu - 1) (d)/(mu)`
D. `(mu d)/(mu - 1)`

Answer

To solve the problem, we need to determine the thickness of the glass slab that causes the screen to be moved a distance \( d \) away from the lens to regain a sharp image.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A lens forms a sharp image on a screen.
- When a parallel-sided glass slab is placed between the lens and the screen, the image shifts, and the screen must be moved to restore the sharp image.

2. **Identifying the Shift**:
- The distance the screen must be moved to regain focus is given as \( d \).
- The thickness of the glass slab is denoted as \( t \).
- The refractive index of the glass slab is denoted as \( \mu \).

3. **Using the Formula for Shift**:
- When a glass slab is inserted, the effective distance light travels changes due to the refractive index. The formula for the shift in the position of the image caused by the slab is given by:
\[ d = t \left(1 - \frac{1}{\mu}\right) \]

4. **Rearranging the Formula**:
- We can rearrange the formula to express the thickness \( t \) in terms of the distance \( d \) and the refractive index \( \mu \):
\[ d = t \left(1 - \frac{1}{\mu}\right) \]
- Rearranging gives:
\[ t = \frac{d}{1 - \frac{1}{\mu}} = \frac{d \mu}{\mu - 1} \]

5. **Final Expression for Thickness**:
- Thus, the thickness of the slab is:
\[ t = \frac{d \mu}{\mu - 1} \]

### Conclusion:
The thickness of the glass slab is given by:
\[ t = \frac{d \mu}{\mu - 1} \]