​A silicon monoxide ($$n$$​ = 1.45) film of 100 nm thickness is used to coat a glass camera lens ($$n$$​ = 1.56). What wavelength of light in the visible region will be most efficiently transmitted by this system?​

Correct option is C

​$$\begin{aligned}&\textbf{For maximum transmission (constructive interference):} \\&\text{We use the formula for constructive interference:} \\&\quad 2nd = m\lambda \\\\&\text{Substitute values:} \\&\quad \lambda = \frac{2 \times 1.45 \times 100 \, \text{nm}}{m} = \frac{290}{m} \, \text{nm} \\\\&\text{This gives possible wavelengths:} \\&\quad \lambda = 290 \, \text{nm}, \, 145 \, \text{nm}, \, 96.67 \, \text{nm} \ldots \\\\&\text{All outside the visible range.} \\\\&\textbf{For destructive interference (maximum transmission through anti-reflection):} \\&\quad 2nd = \left(m + \frac{1}{2} \right) \lambda \Rightarrow \lambda = \frac{2nd}{\left(m + \frac{1}{2} \right)} \\\\&\text{Let's try values:} \\&\quad \lambda = \frac{2 \times 1.45 \times 100}{1.5} = \frac{290}{1.5} = 193.33 \, \text{nm} \quad \text{(not visible)} \\&\quad \lambda = \frac{290}{0.5} = 580 \, \text{nm} \quad \text{(visible)} \\&\quad \lambda = \frac{290}{2.5} = 116 \, \text{nm} \quad \text{(not visible)}\end{aligned}$$​​