In this case, assume that n2 > n1 = n3. For which film-thickness(es) will constructive interference of reflected light occur?

Answer 1

To determine the film thickness(es) for which constructive interference of reflected light occurs in a thin film with three different refractive indices, we can utilize the principle of constructive or destructive interference based on the phase change upon reflection.

Given:

• \( n_2 > n_1 = n_3 \)

Here's the general process for determining the conditions for constructive interference:

1. Identify the Phase Change:

   • When light reflects from a medium with a higher refractive index (\( n_2 \)) than the medium from which it originates (\( n_1 \) or \( n_3 \)), a phase change of \( \pi \) (or half a wavelength) occurs.
   • For this case, since light reflects from \( n_1 \) to \( n_2 \) and from \( n_3 \) to \( n_2 \), there is a \( \pi \) phase change for each reflection at the boundary where \( n_1 \) and \( n_3 \) meet \( n_2 \).

2. Constructive Interference Condition:

   • The condition for constructive interference in a thin film can be given by the formula: \[ 2t = m \frac{\lambda}{n_2} \] where:
     • \( t \) = film thickness
     • \( m \) = order of interference (an integer: 0, 1, 2, ...)
     • \( \lambda \) = wavelength of the light in a vacuum
     • \( n_2 \) = refractive index of the film

3. Considering the Phase Changes:

   • Since there is a \( \pi \) phase shift on the reflection at both interfaces (from \( n_1 \) to \( n_2 \) and from \( n_3 \) to \( n_2 \)), you effectively have one extra half-wavelength shift. Thus, for constructive interference, the condition modifies to: \[ 2t = (m + \frac{1}{2})\frac{\lambda}{n_2} \]
   • This can be rearranged to: \[ t = \frac{(m + \frac{1}{2}) \lambda}{2n_2} \]

4. Final Expression:

   • The thicknesses at which constructive interference occurs will be given by: \[ t = \frac{(m + \frac{1}{2}) \lambda}{2 n_2} \quad (m = 0, 1, 2, \ldots) \]

In summary, constructive interference of reflected light occurs at film thicknesses given by:

\[ t = \frac{(m + \frac{1}{2}) \lambda}{2 n_2} \quad \text{for } m = 0, 1, 2, \ldots \]

This accounts for the phase change upon reflection and ensures that the path difference leads to constructive interference due to the additional half-wavelength shift.