White light reflected from a soap film (Refractive Index \(=1.5\) ) has a maxima at \(600 \mathrm{~nm}\) and a minima at \(450 \mathrm{~nm}\) with no minimum in between. Then the thickness of the film is

We know for maxima
\(2 \mu t \cos r=\frac{(2 n+1) \lambda_{1}}{2}...(i)\)
For minima \(2 \mu t \cos r=(n+1) \lambda_{2}...(ii)\)
Given here,
\(\lambda_{1}=600 \mathrm{~nm}\)
and
\(\lambda_{2}=450 \mathrm{~nm}\)
Dividing Eq. (i) by (ii) and substituting values of \(\lambda_{1}\) and \(\lambda_{2}\), we get
\(\begin{aligned} \frac{(2 n+1) \lambda_{1}}{2(n+1) \lambda_{2}} &=1 \\ 90 m+90 &=60 n+60 \\ 30 n &=30 \\ n &=1 \end{aligned}\)
Using one of the equation, we get
In given condition \(2 \mu t=(2 n+1) \frac{\lambda}{2}\)
\(\begin{gathered}
2 \times 1.5 \times t=(2 \times 1+1) \frac{600}{2} \\
3 t=\frac{3 \times 600}{2} \\
\Rightarrow \quad t=300 \mathrm{~nm}=3 \times 10^{-6} \mathrm{~m}=3 \text { units }
\end{gathered}\)