Q123 / 180 · KCET 2013 · Physics

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

  • A
    1
  • B
    2
  • Correct answer
    C
    3
  • D
    4

Solution

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 $2 \mu t=(2 n+1) \frac{\lambda}{2}$ $\begin{gathered} 2 \times 1.5 \times t=(2 \times 1+1) \frac{600}{2} \\ 3 t=\frac{3 \times 600}{2} \\ \Rightarrow \quad t=300 \mathrm{~nm}=3 \times 10^{-6} \mathrm{~m}=3 \text { units } \end{gathered}$

Hence, the correct answer is option: 'C'.

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