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Chapter 25: Problem 16

A thin film of oil \((n=1.50)\) is spread over a puddle of water \((n=1.33) .\) In a region where the film looks red from directly above $(\lambda=630 \mathrm{nm}),$ what is the minimum possible thickness of the film? (tutorial: thin film).

Short Answer

Expert verified
Answer: The minimum thickness of the oil film is 210 nm.

Step by step solution

01

Understand thin film interference

Thin film interference occurs when light waves reflect off two surfaces separated by a thin film, causing constructive or destructive interference between the reflected waves. In this case, the film is the oil on top of the water. Constructive interference leads to a brighter color, while destructive interference results in a darker color.
02

Find the interference condition for a minimum thickness

We will use the following equation for the constructive interference of the thin film: $$2nt = m\lambda_{air}$$ where \(n\) is the refractive index of the oil, \(t\) is the film thickness, \(m\) is an integer representing the order of constructive interference, and \(\lambda_{air}\) is the wavelength of light in air. In our case, \(n=1.50\), \(\lambda_{air}=630\,\text{nm}\), and we want to find the smallest possible \(t\) for constructive interference, so we will consider the case when \(m=1\).
03

Solve for the minimum thickness of the film

Plug in the given values and solve for \(t\): $$2(1.50)t = 1(630\,\text{nm})$$ $$t = \frac{630\,\text{nm}}{2(1.50)}$$ $$t = 210\,\text{nm}$$ Therefore, the minimum possible thickness of the film for it to appear red when viewed from above is 210 nm.

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Most popular questions from this chapter

In bright light, the pupils of the eyes of a cat narrow to a vertical slit \(0.30 \mathrm{mm}\) across. Suppose that a cat is looking at two mice $18 \mathrm{m}$ away. What is the smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of $560 \mathrm{nm} ?$ Assume the resolution is limited by diffraction only.
Coherent light from a laser is split into two beams with intensities \(I_{0}\) and \(4 I_{0},\) respectively. What is the intensity of the light when the beams are recombined? If there is more than one possibility, give the range of possibilities. (tutorial: two waves).
Light of wavelength 589 nm incident on a pair of slits produces an interference pattern on a distant screen in which the separation between adjacent bright fringes at the center of the pattern is \(0.530 \mathrm{cm} .\) A second light source, when incident on the same pair of slits, produces an interference pattern on the same screen with a separation of $0.640 \mathrm{cm}$ between adjacent bright fringes at the center of the pattern. What is the wavelength of the second source? [Hint: Is the small-angle approximation justified?]
Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)
Red light of 650 nm can be seen in three orders in a particular grating. About how many slits per centimeter does this grating have?
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