Interference from Thin Films — Question 9

Question 9

One film can reflect a maximum of one color and a minimum of another

A nondispersive film of index 1.50 is surrounded by air. At normal incidence it must give constructive reflection at $600\,\mathrm{nm}$ and destructive reflection at $450\,\mathrm{nm}$. Use the single relative reflection reversal and require the smallest positive thickness.

1. Write the simultaneous integer conditions for $D=2nt$ and solve for its smallest positive value.

2. Find the thickness and verify the total relative phases at both wavelengths modulo $2\pi$.

Question 9 - Solution

One film can reflect a maximum of one color and a minimum of another

(a) Method and calculation

$$D=(m+1/2)(600\,\mathrm{nm})=q(450\,\mathrm{nm})\Longrightarrow4m+2=3q.$$$$m=1,\ q=2\ \text{is the first nonnegative solution}\Longrightarrow\boxed{D=900\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\boxed{t=D/(2n)=300\,\mathrm{nm}}.$$$$\boxed{\delta_{600}=2\pi(900/600)+\pi=4\pi\equiv0;\quad\delta_{450}=5\pi\equiv\pi}.$$

Physical check

The boundary phase must be included for both colors. Equal integer propagation cycles alone would describe the opposite reflected condition.