Q. A thin film of soap solution $(n = 1.4)$ lies on the top of a glass plate $(n = 1.5).$ When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 400 and 630 nm. The minimum thickness of the soap solution is
A) 420 nm
B) 450 nm
C) 630 nm
D) 1260 nm
Explanation
(b) $\ \ \ \ \ \ \ \ \ \ \ \ n_{1}\lambda_{1} = n_{2}\lambda_{2}$
$\therefore n_{1} \times 420 = n_{2} \times 630$
Or $\ \ \ \ \ \ \ 2n_{1} = 3n_{2}$
If $n_{2} = 2,\ $ then $n_{1} = 3\ $
Therefore, thickness of soap solution is given by
$\mu_{1}t = n_{1}\frac{\lambda_{1}}{2}$
Or $t = \frac{3\ \times \ 420}{1.4\ \times \ 2} = 450\ nm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.