Answer
Please see the work below.
Work Step by Step
(a) The thickness of the film is given as
$ t_{min}=\frac{\lambda}{4n}$
We plug in the known values to obtain:
$ t_{min}=\frac{652\times 10^{-9}m}{4(1.33)}$
$ t_{min}=123\times 10^{-9}m=123nm $
(b) We know that
$\lambda=\frac{2(1.33)(123\times 10^{-9}m)}{m}=\frac{327.18nm}{m}$
$\lambda_1=\frac{227.18nm}{1}$
for $ m=2$
$\lambda_2=\frac{327.18nm}{2}=163.59nm $
for $ m=3$
$\lambda_3=\frac{327.18nm}{3}=109.06nm $
Thus, no visible wavelengths destructively interfere as we have obtained the wavelengths which are less than $400nm $.
