High-quality camera lenses are often coated to prevent reflection. A lens has an optical index of refraction of 1.72 and a coating with an optical index os refraction 1.31. For near normal incidence, the minimum thickness of the coating to prevent reflection for wavelength of `5.3 xx 10^(-7) m` is

Options

A. `0.75 mu m`
B. 0.2 mm
C. `0.1 mu m`
D. 1.75 mm

Explanation

To find the minimum thickness of the coating that prevents reflection for a lens with a given refractive index, we can use the formula for destructive interference. The condition for destructive interference at normal incidence is given by:

\[ d = \frac{\lambda}{4n} \]

where:
- \( d \) is the thickness of the coating,
- \( \lambda \) is the wavelength of light in vacuum,
- \( n \) is the refractive index of the coating.

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength of light, \( \lambda = 5.3 \times 10^{-7} \, \text{m} \)
- Refractive index of the coating, \( n = 1.31 \)

2. **Use the formula for minimum thickness:**
\[ d = \frac{\lambda}{4n} \]

3. **Substitute the known values into the formula:**
\[ d = \frac{5.3 \times 10^{-7}}{4 \times 1.31} \]

4. **Calculate the denominator:**
\[ 4 \times 1.31 = 5.24 \]

5. **Now calculate \( d \):**
\[ d = \frac{5.3 \times 10^{-7}}{5.24} \approx 1.01 \times 10^{-7} \, \text{m} \]

6. **Convert the thickness into micrometers:**
\[ d \approx 0.1 \, \mu m \]

### Final Answer:
The minimum thickness of the coating to prevent reflection is approximately \( 0.1 \, \mu m \).

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