A soap film has an index of refraction \(n=1.50 .\) The film is viewed in transmitted light. (a) At a spot where the film thickness is $910.0 \mathrm{nm},$ which wavelengths are weakest in the transmitted light? (b) Which wavelengths are strongest in transmitted light?

Thin film interference occurs when light waves that have traveled through a thin film of a specific thickness and refractive index interfere with each other. We can use the following formula to find the condition for constructive interference (brightest spots): $$2 \cdot n \cdot d = m \cdot \lambda \hspace{20pt} (1)$$ For destructive interference (darkest spots), we have to consider a \(\frac{\lambda}{2}\) phase shift, so the condition is: $$2 \cdot n \cdot d = \left(m + \frac{1}{2}\right) \cdot \lambda \hspace{20pt} (2)$$ Where: \(d\) - thickness of the film \(n\) - index of refraction \(m\) - integer (0,1,2,3...)

We are given a soap film with index of refraction \(n = 1.50\) and film thickness \(d = 910.0 nm\). We'll use equation (2) to find the weakest transmitted light (destructive interference). Let's solve for \(\lambda\): $$\lambda = \frac{2 \cdot n \cdot d}{m + \frac{1}{2}}$$ Plugging in \(n\) and \(d\) and trying different \(m\) values, we obtain the corresponding wavelengths for weakest transmitted light.

Now, we'll use equation (1) to find the strongest transmitted light (constructive interference). Solving for \(\lambda\): $$\lambda = \frac{2 \cdot n \cdot d}{m}$$ Plugging in \(n\) and \(d\) and trying different \(m\) values, we obtain the corresponding wavelengths for strongest transmitted light.

For weakest transmitted light (destructive interference), let's assume \(m\) starts from 0, then for the first three possible wavelengths: \(m = 0:\) $$\lambda = \frac{2 \cdot 1.50 \cdot 910.0}{0 + \frac{1}{2}} = 5460 \ nm$$ \(m = 1:\) $$\lambda = \frac{2 \cdot 1.50 \cdot 910.0}{1 + \frac{1}{2}} = 1820 \ nm$$ \(m = 2:\) $$\lambda = \frac{2 \cdot 1.50 \cdot 910.0}{2 + \frac{1}{2}} = 910 \ nm$$ For strongest transmitted light (constructive interference), let's assume \(m\) starts from 1 then for the first three possible wavelengths: \(m = 1:\) $$\lambda = \frac{2 \cdot 1.50 \cdot 910.0}{1} = 2730 \ nm$$ \(m = 2:\) $$\lambda = \frac{2 \cdot 1.50 \cdot 910.0}{2} = 1365 \ nm$$ \(m = 3:\) $$\lambda = \frac{2 \cdot 1.50 \cdot 910.0}{3} = 910 \ nm$$ To summarize: (a) The weakest transmitted wavelengths (destructive interference) are approximately 5460 nm, 1820 nm, and 910 nm. (b) The strongest transmitted wavelengths (constructive interference) are approximately 2730 nm, 1365 nm, and 910 nm.