To ensure almost $100$ per cent transmittivity,photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediate between that of air and glass (which makes the optical element of the lens). $A$ typically used dielectric film is $MgF_2$ $(n = 1.38)$. What should the thickness of the film be so that at the center of the visible spectrum $5500 \, \mathring{A}$ there is maximum transmission?

(N/A) For maximum transmission,the reflected rays must interfere destructively. This occurs when the path difference between the two reflected rays is an odd multiple of $\frac{\lambda}{2}$.
For normal incidence $(i = 0)$,the path difference is $\Delta x = 2nd$.
Since both reflections occur at a rarer to denser medium interface,there is no additional phase shift due to reflection.
Thus,the condition for destructive interference (minimum reflection,maximum transmission) is $2nd = (m + \frac{1}{2}) \lambda$,where $m = 0, 1, 2, \dots$.
For minimum thickness,we take $m = 0$,so $2nd = \frac{\lambda}{2}$.
$d = \frac{\lambda}{4n} = \frac{5500 \times 10^{-10} \, \text{m}}{4 \times 1.38} \approx 996.4 \times 10^{-10} \, \text{m} \approx 996.4 \, \mathring{A}$.