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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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Most popular questions from this chapter

A double slit is opposite the center of a 1.8 -m wide screen \(2.0 \mathrm{~m}\) from the slits. The slit separation is \(24 \mu \mathrm{m}\) and the width of each slit is \(7.2 \mu \mathrm{m}\). How many fringes are visible on the screen if the slit is illuminated by \(600 .-\mathrm{nm}\) light?

A Newton's ring apparatus consists of a convex lens with a large radius of curvature \(R\) placed on a flat glass disc. (a) Show that the distance \(x\) from the center to the air, thickness \(d,\) and the radius of curvature \(R\) are given by \(x^{2}=2 R d\) (b) Show that the radius of nth constructive interference is given by \(x_{\mathrm{n}}=\left[\left(n+\frac{1}{2}\right) \lambda R\right]^{1 / 2} .\) (c) How many bright fringes may be seen if it is viewed by red light of wavelength 700\. \(\mathrm{nm}\) for \(R=10.0 \mathrm{~m},\) and the plane glass disc diameter is \(5.00 \mathrm{~cm} ?\)

White light is shone on a very thin layer of mica \((n=1.57),\) and above the mica layer, interference maxima for two wavelengths (and no other in between) are seen: one blue wavelength of \(480 \mathrm{nm},\) and one yellow wavelength of \(560 \mathrm{nm} .\) What is the thickness of the mica layer?

One type of hologram consists of bright and dark fringes produced on photographic film by interfering laser beams. If this is illuminated with white light, the image will appear reproduced multiple times, in different pure colors at different sizes. a) Explain why. b) Which colors correspond to the largest and smallest images, and why?

A Michelson interferometer is illuminated with a 600.-nm light source. How many fringes are observed if one of the mirrors of the interferometer is moved a distance of 200. \(\mu \mathrm{m} ?\)
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