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When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 \(\mathrm{nm}\) . What is the vacuum wavelength of the light in \(\mathrm{nm}\) ?

Short Answer

Expert verified
The vacuum wavelength is approximately 625.6 nm.

Step by step solution

01

Understanding Destructive Interference

For destructive interference in a thin film, the path difference needs to equal an odd multiple of half wavelengths (i.e., \( (2m+1) \frac{\lambda}{2} \)). In our case, the interference is due to reflections on both surfaces of the film, with light refracting through the film.
02

Determine the Condition for the Given Thickness

Since we are given that the second smallest nonzero thickness occurs, we can denote this as \( m = 1 \) for destructive interference, meaning: \[ 2t = (m+0.5) \frac{\lambda}{n} \rightarrow 2t = (2 \cdot 1+1) \frac{\lambda}{2n} \rightarrow 2t = \frac{3\lambda}{2n} \].
03

Substitution of Known Values

Given \( t = 296 \, nm \) and \( n = 1.33 \), substitute these values into the expression: \[ 2 \times 296 = \frac{3\lambda}{2 \cdot 1.33} \].
04

Solve for \( \lambda \)

Rearrange and solve for \( \lambda \): \[ \lambda = \frac{2 \times 1.33 \times 2 \times 296}{3} = \frac{313.44}{3} \]. When you calculate this, you get: \ \approx 625.6 \, nm\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Destructive Interference
Destructive interference is a fascinating phenomenon where waves combine to produce a reduced or nullified effect. This occurs when two waves are out of phase, meaning their peaks and troughs do not line up. In optics, this often involves light waves that reflect from different interfaces.

For destructive interference in a thin film, like a soap bubble, the path difference between the two reflected light waves must be an odd multiple of half wavelengths, mathematically expressed as \((2m+1)\frac{\lambda}{2}\). This condition ensures the crest of one wave aligns with the trough of another, resulting in cancellation. Valuing the path difference is key to predicting interference outcomes in various applications, like anti-reflective coatings or noise-cancelling headphones.
Thin Film Interference
Thin film interference occurs when light waves reflect off the different layers of a thin film, creating interference patterns. These patterns arise because the light reflects from both the top and bottom surfaces of the film.

In a simple setup, like a soap film with air on each side, light reflects on the film's surfaces and travels different path lengths. When these reflected waves meet, they cause interference - constructive or destructive, depending on the thickness of the film and the light's wavelength. The thickness required for a specific interference pattern is calculated using its refractive index. By understanding these interactions, scientists and engineers can harness these effects to design coatings and devices with specific optical properties.
Monochromatic Light
Monochromatic light is light of a single wavelength or color, making it very pure. This type of light is vital for experiments and technologies that rely on interference, like holography and spectroscopy.

Lasers are a common source of monochromatic light due to their precise and focused beams. In experiments involving interference, using monochromatic light avoids the complexity of different wavelengths interfering differently, simplifying analysis and ensuring more reliable results. Understanding how to work with monochromatic light is essential in the fields of physics and engineering.
Wavelength Calculation
Wavelength calculation, especially in the context of thin films, requires careful consideration of several factors. To determine the wavelength of light causing interference in a film, you need the film's thickness and refractive index.

The calculation involves the formula for destructive interference in the film: \(2t = \frac{3\lambda}{2n}\), where \(t\) is the thickness, \(n\) is the refractive index, and \(\lambda\) is the wavelength. Rearranging this equation allows for solving the unknown wavelength, crucial for applications ranging from microscopic imaging to telecommunications. This process highlights the need for precise measurement and understanding of optical principles in practical technology development.

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Most popular questions from this chapter

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure 27.6\(a .\) The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times 10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm} )\) of the monochromatic light?

Violet light (wavelength \(=410 \mathrm{nm} )\) and red light (wave length \(=660 \mathrm{nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for \((\mathbf{b})\) the second order maximum and \((\mathbf{c})\) the third-order maximum. (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength 625 \(\mathrm{nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{m} ?\)

Two concentric circles of light emit light whose wavelength is 555 nm. The larger circle has a radius of 4.0 cm, and the smaller circle has a radius of 1.0 cm. When taking a picture of these lighted circles, a camera admits light through an aperture whose diameter is 12.5 mm. What is the maximum distance at which the camera can (a) distinguish one circle from the other and (b) reveal that the inner circle is a circle of light rather than a solid disk of light?

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00) .\) A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 625 \(\mathrm{nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 242 \(\mathrm{nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

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