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Chapter 34: Problem 27

Suppose the thickness of a thin soap film \((n=1.32)\) surrounded by air is nonuniform and gradually tapers. Monochromatic light of wavelength \(550 \mathrm{nm}\) illuminates the film. At the thinnest end, a dark band is observed. How thick is the film at the next two dark bands closest to the first dark band?

Short Answer

Expert verified
Based on the calculations, the thickness of the soap film at the second dark band is approximately \(2.63 \times 10^{-7} m\), and at the third dark band, it is approximately \(2.19 \times 10^{-7} m\).

Step by step solution

01

Calculate the thickness at the first dark band

As mentioned, we need to calculate the thickness of the film at the next two dark bands. We will start with the first dark band, which has an order \(m=0\). Plugging this into the equation: $$2*(1.32)*d*0 = (0+\frac{1}{2})*550\times10^{-9}$$ We need to solve for \(d\) here, which is the thickness at the first dark band: $$d_{1} = \frac{0+\frac{1}{2}}{2*(1.32)*0}*550\times10^{-9}"m"$$
02

Determine the thickness at the second and third dark bands

Now, we will determine the thickness at the second and third dark bands. To do this, we increase the order value \(m\) to 1 and 2, respectively: For the second dark band (\(m=1\)): $$2*(1.32)*d_2*1 = (1+\frac{1}{2})*550\times10^{-9}"m"$$ Solve for \(d_2\): $$d_{2} = \frac{1+\frac{1}{2}}{2*(1.32)*1}*550\times10^{-9}"m"$$ For the third dark band (\(m=2\)): $$2*(1.32)*d_3*2 = (2+\frac{1}{2})*550\times10^{-9}"m"$$ Solve for \(d_3\): $$d_{3} = \frac{2+\frac{1}{2}}{2*(1.32)*2}*550\times10^{-9}"m"$$
03

Calculate the thickness values

Using the equations for finding \(d_{2}\) and \(d_{3}\), we can calculate the required values: $$d_{2} = \frac{1.5}{2*(1.32)}*550\times10^{-9} = 2.63\times10^{-7}"m"$$ $$d_{3} = \frac{2.5}{2*(1.32)*2}*550\times10^{-9} = 2.19\times10^{-7}"m"$$ Thus, the thickness of the film at the second dark band is \(2.63 \times 10^{-7} m\), and the thickness at the third dark band is \(2.19 \times 10^{-7} m\).

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