Question

An anti-reflection film coating of thickness 0.1 \( \mu \)m is to be deposited on a glass plate for normal incidence of light of wavelength 0.5 \( \mu \)m. What should be the refractive index of the film?

Hint:

For anti-reflection coatings, the refractive index and thickness of the film are related by the condition for destructive interference between the reflected light.

Answer 1

Step 1: Understand the condition for minimum reflection.
The condition for minimum reflection in an anti-reflection film is given by: \[ 2 n_f t = \frac{\lambda}{2} \] where \( n_f \) is the refractive index of the film, \( t \) is the thickness of the film, and \( \lambda \) is the wavelength of light. This condition ensures that the light reflected from the two surfaces of the film interferes destructively.
Step 2: Apply the given values.
Substitute the given values: \[ 2 n_f (0.1 \, \mu \text{m}) = \frac{0.5 \, \mu \text{m}}{2} \] Solving for \( n_f \): \[ n_f = \frac{0.5 \, \mu \text{m}}{2 \times 0.1 \, \mu \text{m}} = 2.5 \] Step 3: Conclusion.
Thus, the refractive index of the film should be 2.5.