Chapter 24 - The Wave Nature of Light - Search and Learn - Page 712: 6

See the detailed answer below.

We have here two general cases: 1) The first case, as we see in the first figure below, is when the index of refraction of the coating is greater than that of air and greater than that of glass $( n_{\rm coating}\gt n_{glass})$. So, we have two reflected rays with one $\pi$-phase change. a) For minimum interference, the bath length difference between the two rays is given by $$2t= \dfrac{m\lambda}{n_{\rm coating}}$$ Thus, the thickness needed for the coating is given by $$\boxed{t=\dfrac{m\lambda}{2n_{\rm coating}}}$$ b) For maximum interference, the bath length difference between the two rays is given by $$2t= \dfrac{\left(m+\frac{1}{2}\right)\lambda}{n_{\rm coating}}$$ Thus, the thickness needed for the coating is given by $$\boxed{t=\dfrac{\left(m+\frac{1}{2}\right)\lambda}{2n_{\rm coating}}}$$ ---- 2) The second case, as we see in the second figure below, is when the index of refraction of the coating is greater than that of air but is less than that of glass $( n_{\rm coating}\lt n_{glass})$. So, we have two reflected rays with two $\pi$-phase changes. a) For minimum interference, the bath length difference between the two rays is given by $$2t= \dfrac{\left(m+\frac{1}{2}\right)\lambda}{n_{\rm coating}}$$ Thus, the thickness needed for the coating is given by $$\boxed{t=\dfrac{\left(m+\frac{1}{2}\right)\lambda}{2n_{\rm coating}}}$$ b) For maximum interference, the bath length difference between the two rays is given by $$2t= \dfrac{m\lambda}{n_{\rm coating}}$$ Thus, the thickness needed for the coating is given by $$\boxed{t=\dfrac{m\lambda}{2n_{\rm coating}}}$$ ------------- We know that a film coating can reduce reflection from $4\%$ to $1\%$. This means that the transmitted light will be increased from $96\%$ to $99\%$.