# Multiple maxima during thin film interference

I am trying to work through a problem about thin film interference. In the problem a thin film ($$n_1 = 1.5$$) is deposited on a layer of silicon ($$n_2 = 3$$). I need to find the thickness of the film which will result in it having a purple colour when viewed at normal incidence.

My first thought would be to simply find the thickness ($$d$$) which would result in violet/purple constructive interference (I am assuming the incident light is white) using:

$$2n_1d = m\lambda$$

But instead I am supposed to try to find a thickness that results in constructive interference for both red light and blue light. I was under the (maybe wrong) impression that a particular thickness results in constructive interference for only a specific wavelength?

I assume there must be something I'm missing about the concept so I'd really appreciate it if someone explain it to me.

Suppose a particular thickness of the slab to be $$d$$. Then for two different wavelength, the condition for constructive interference:
$$2nd=m_1\lambda_1$$ $$2nd=m_2\lambda_2$$ $$\frac{m_2}{m_1}=\frac{\lambda_1}{\lambda_2}$$ That should be a condition for constructive interference to match up. Note that the order of interference must be different. Though it's not impossible that two-wavelength can not produce constructive interference for the same width.
$$\frac{m_2}{m_1}=\frac{480}{\lambda_2}\Rightarrow \lambda_2=\frac{m_1}{m_2}(480)$$
Let's take $$m_1=2$$ and $$m_2=1$$ then $$\lambda_2=960\ nm$$ That goes too far. Let's put $$m_1=3$$ and $$m_=2$$, We find $$\lambda_2=720\ nm$$ which is close to wavelength of red light. So it's possible!