Question

The average human eye sees colors with wavelength between $430nm$ to $680nm$. For what visible wavelength(s) will a $350nm$ thick ($n=1.35$) soap film produce maximum destructive interference?

• A. $945nm$
• B. $473nm$
• C. $315nm$
• D. None of these

Answer 1

The correct option is B $473nm$

Wave $R_a$ is out of phase (${180}^o$) with the incident wave whereas wave $R_b$ is in phase with the incident wave , so $R_a$ and $R_b$ are out of phase with each other.

thus using $2t=m\lambda /n$ for destructive interference (i-e. no reflection)

Given ; $n=1.35t=350nm$

$2\times 350nm=\frac{m\times {\lambda }_m}{1.35}$

$m=1$ gives ${\lambda }_1=945nm$

$m=2$ gives ${\lambda }_2=472.5nm\approx 473nm$

$m=3$ gives ${\lambda }_3=315nm$