On face of a glass `(mu = 1.50)` lens is coated with a thin film of magnesium fluoride `MgF_(2)(mu = 1.38)` to reduce reflection from the lens surface. Assuming the incident light to be perpendicular to the lens surface. The least coating thickness that eliminates the reflection at the centre of the visible spectrum `(lamda = 550 nm)` is about

Options

A. `0.05 mu m`
B. `0.10 mu m`
C. `1.38 mu m`
D. `2.80 mu m`

Explanation

To solve the problem, we need to find the least thickness of the magnesium fluoride (MgF₂) coating that will eliminate reflection from the surface of the glass lens. This occurs due to destructive interference of light waves reflected from the air-film interface and the film-glass interface.

### Step-by-Step Solution:

1. **Understanding Destructive Interference**:
For destructive interference to occur, the path difference between the two reflected rays must be an odd multiple of half wavelengths. The condition for destructive interference is given by:
\[ \Delta x = (2n + 1) \frac{\lambda}{2} \]
where \( n \) is an integer (0, 1, 2,...), and \( \lambda \) is the wavelength of light.

2. **Choosing the Minimum Thickness**:
To find the least thickness that eliminates reflection, we set \( n = 0 \):
\[ \Delta x = \frac{\lambda}{2} \]

3. **Calculating the Path Difference**:
The path difference for light reflecting off the two interfaces (air-film and film-glass) is given by:
\[ \Delta x = 2t \]
where \( t \) is the thickness of the film.

4. **Setting Up the Equation**:
Equating the two expressions for path difference:
\[ 2t = \frac{\lambda}{2n_{\text{film}}} \]
where \( n_{\text{film}} \) is the refractive index of the film (MgF₂).

5. **Substituting Values**:
Given:
- Wavelength \( \lambda = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m} \)
- Refractive index of MgF₂ \( n_{\text{film}} = 1.38 \)

We can substitute these values into the equation:
\[ 2t = \frac{550 \times 10^{-9}}{2 \times 1.38} \]

6. **Solving for Thickness \( t \)**:
Rearranging gives:
\[ t = \frac{550 \times 10^{-9}}{4 \times 1.38} \]

Now calculating:
\[ t = \frac{550 \times 10^{-9}}{5.52} \approx 99.64 \times 10^{-9} \, \text{m} = 0.1 \, \mu\text{m} \]

### Final Answer:
The least coating thickness that eliminates reflection at the center of the visible spectrum is approximately \( 0.1 \, \mu\text{m} \).

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