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 find the least coating thickness of magnesium fluoride (MgF₂) that eliminates reflection at the center of the visible spectrum (λ = 550 nm), we can use the principle of destructive interference for thin films.

### Step-by-Step Solution:

1. **Understanding the Condition for Destructive Interference**:
For a thin film, destructive interference occurs when the path difference between the light reflected from the top and bottom surfaces of the film leads to a phase difference of an odd multiple of π (or λ/2). The condition for destructive interference is given by:
\[ 2nT = \left(m + \frac{1}{2}\right)\lambda \]
where:
- \( n \) is the refractive index of the film (MgF₂ in this case),
- \( T \) is the thickness of the film,
- \( m \) is an integer (0, 1, 2, ...),
- \( \lambda \) is the wavelength of light in vacuum.

2. **Using the First Order of Destructive Interference**:
For the least thickness that eliminates reflection, we can take \( m = 0 \):
\[ 2nT = \frac{1}{2}\lambda \]
Rearranging gives:
\[ T = \frac{\lambda}{4n} \]

3. **Substituting the Values**:
Given:
- \( \lambda = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m} \)
- \( n = 1.38 \) (for MgF₂)

Substituting these values into the equation:
\[ T = \frac{550 \times 10^{-9}}{4 \times 1.38} \]

4. **Calculating the Thickness**:
First, calculate \( 4 \times 1.38 \):
\[ 4 \times 1.38 = 5.52 \]
Now substitute this back into the equation for \( T \):
\[ T = \frac{550 \times 10^{-9}}{5.52} \approx 99.64 \times 10^{-9} \, \text{m} \]
Converting this to micrometers:
\[ T \approx 0.1 \, \mu m \]

5. **Final Answer**:
The least coating thickness that eliminates reflection at the center of the visible spectrum is approximately:
\[ T \approx 0.1 \, \mu m \]