Answer:
7.22*10⁻7m
Explanation:
To solve this problem we have to take into account that the phase difference is given by
[tex]\delta=(\frac{2\pi}{\lambda})2nd[/tex]
where lambda is the wavelength, n is the index of refraction and d is the thickness of the film. For the condition of a maximum we have
[tex](\frac{2\pi}{\lambda})2nd=2\pi m\\[/tex]
Hence, the thickness is
[tex]d=\frac{m \lambda}{2n}[/tex]
in this case the region correspond to the third maximum m=3
[tex]d=\frac{3(645*10^{-9}m)}{2(1.34)}=7.22*10^{-7}m[/tex]
the thickness of the film is 7.22*10⁻7m
hope this helps!
