Question

A thin film of soap solution ${\mu _s} = 1.4$ lies on the top of a glass plate ${\mu _g} = 1.5$ When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths $420nm$ and $630nm$. The minimum thickness of the soap solution is:
(A) $420nm$
(B) $500nm$
(C) $450nm$
(D) $490nm$

Answer 1

Hint: In order to solve this question, we will use the maximum intensity condition for reflection due to thin-film, and using the given information we will form the equation and will solve for the value of the thickness of the soap film.

Formula used:
If n is the order of maximum intensity, ‘t’ is the thickness of the thin film and $\lambda ,\mu $ be the wavelength and refractive index then, the condition of maximum intensity is:
$2\mu t = n\lambda $

Complete answer:
We have given that for two wavelengths $420nm$ and $630nm$, two adjacent maxima occur which means if $n$ is the order of wavelength $420$ then $n - 1$ will be the order for wavelength $630nm$. so, using the maxima condition formula we have equations for both wavelengths as
$
  2{\mu _s}t = n(420) \to (i) \\
  2{\mu _s}t = n - 1(630) \to (ii) \\
 $

So, equating equations (i) and (ii) to find n we get,
$
  420n = 630n - 630 \\
  210n = 630 \\
  n = 3 \\
 $

So, $n = 3$ is order of maxima when two adjacent maxima occurs now, using maxima formula at $n = 3$ , $\lambda = 420nm$ and ${\mu _s} = 1.4$ we get,
$
  2\mu t = n\lambda \\
  t = \dfrac{{3 \times 420}}{{2 \times 1.4}} \\
  t = 450nm \\
 $
so, the thickness of the soap film is $450nm$

Hence, the correct answer is option (C) $450nm$

Note:It should be remembered that, the wavelength is directly proportional to the maxima order that’s why for the larger wavelength of $630nm$ we take the order as $n - 1$ whereas for the smaller wavelength of $420nm$ we take the order as $n$ and nanometre is the unit of length denoted by nm and its related with meter as $1nm = {10^{ - 9}}m$