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Chapter 35: Problem 57

35.57. Two thin parallel slits are made in an opaque sheet of film. When a monochromatic beam of light is shone through them at normal incidence, the first bright fringes in the transmitted light occur in air at \(\pm 18.0^{\circ}\) with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringes now occur at \(\pm 12.6^{\circ} .\) Find the index of refraction of the liquid.

Short Answer

Expert verified
The index of refraction of the liquid is approximately 1.42.

Step by step solution

01

Understand the Condition for Bright Fringes

The condition for bright fringes in a double-slit experiment is given by the equation \( d \sin \theta = m \lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( m \) is the order number (an integer), and \( \lambda \) is the wavelength of light in the medium. This is true for m = 卤1, 2, 3... Here, we have the first-order bright fringe (\( m = 1 \)).
02

Calculate the Wavelength in Air

Since the bright fringe angle in air is \( \theta = 18.0^{\circ} \), use the equation \( \sin \theta_1 = \frac{m \lambda_0}{d} \). The wavelength of light in air, \( \lambda_0 \), does not change, and we use \( \theta_1 = 18.0^{\circ} \) to calculate \( \sin 18.0^{\circ} = \frac{m \lambda_0}{d} \). We do not need to know individual \( d \) or \( \lambda_0 \) because they will be canceled out in the next steps.
03

Calculate the Wavelength in Liquid

Similarly, when the apparatus is immersed in a liquid, use the same fringe condition for the angle \( \theta = 12.6^{\circ} \): \( \sin \theta_2 = \frac{m \lambda_l}{d} \), where \( \lambda_l \) is the wavelength in the liquid and \( \theta_2 = 12.6^{\circ} \). So, \( \sin 12.6^{\circ} = \frac{m \lambda_l}{d} \).
04

Relate Wavelengths to Indices of Refraction

The wavelength of the light in a medium is related to its wavelength in vacuum and the refractive index by \( \lambda_l = \frac{\lambda_0}{n} \). Thus, \( \sin 12.6^{\circ} \cdot n = \sin 18.0^{\circ} \) because \( m \lambda_0/d = m \lambda_l \cdot n/d \).
05

Solve for the Index of Refraction

Finally, \[n = \frac{\sin 18.0^{\circ}}{\sin 12.6^{\circ}}\]Substituting the values:\[n = \frac{\sin 18.0^{\circ}}{\sin 12.6^{\circ}} \approx \frac{0.309}{0.218} \approx 1.42\]The index of refraction of the liquid is approximately 1.42.

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Key Concepts

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

Interference of Light
In the double-slit experiment, interference of light creates patterns of bright and dark fringes on a screen. This phenomenon occurs because light waves overlap and interact with one another. When two or more light waves meet, they can add up to form a brighter light or cancel each other out to create a darker spot. This process is known as interference.

There are two types of interference: constructive and destructive. Constructive interference happens when the crests of two waves align, resulting in increased amplitude and brightness. This is what forms bright fringes in the double-slit experiment. Meanwhile, destructive interference occurs when the crest of one wave aligns with the trough of another, causing them to cancel out and form dark spots. Understanding how light waves interfere with each other is crucial in predicting and explaining the patterns observed in experiments like the double-slit test.

In essence, interference demonstrates the wave nature of light, showcasing how light can behave in ways similar to water waves.
Refractive Index
The refractive index, often denoted by the symbol \( n \), is a measure of how much the speed of light reduces when it travels through a medium. It is a critical factor when understanding the behavior of light as it moves from one medium to another, such as air to water or glass. The refractive index can be calculated using the formula: \[ n = \frac{c}{v} \], where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.

In the context of the double-slit experiment, the refractive index affects the angle at which bright fringes appear. This index change modifies the speed of light, leading to a change in its wavelength in that medium. For instance, if an apparatus operating in air is submerged in a liquid, the refractive index of the liquid will determine how much the path of light bends and therefore alter the interference pattern.

Calculating the refractive index helps explain these changes, providing insight into how light behaves in different environments.
Wavelength in Medium
The wavelength of light in a medium is shorter than its wavelength in a vacuum or air due to the change in speed. In any medium other than a vacuum, light slows down, causing its wavelength to decrease. This relationship between light speed, wavelength, and refractive index is mathematically expressed as \( \lambda_l = \frac{\lambda_0}{n} \), where \( \lambda_l \) is the wavelength in the medium, \( \lambda_0 \) is the wavelength in a vacuum, and \( n \) is the refractive index.

Understanding this concept is essential in the double-slit experiment. When observing shifts in fringe patterns upon submersion of the apparatus in another medium, it's directly tied to changes in wavelength. Since the refractive index influences the speed of light in a medium, it consequently alters the wavelength.

By knowing the refractive index, you can determine how the wavelength of light changes inside any given medium. This understanding is crucial for precise calculations and interpretation of light behavior in various experimental and real-world applications.
Bright Fringes
Bright fringes are the result of constructive interference, occurring at specific angles where the path difference between light waves corresponds to whole multiples of the wavelength. In the formula for bright fringes, \( d \sin \theta = m \lambda \), \( m \) represents the order of the fringe, and \( \theta \) is the angle where the fringe appears. These bright spots signify positions on a screen where the light waves have combined constructively after passing through the slits.

In the original exercise, the relationship between angle changes and bright fringes when the apparatus is submerged in a liquid highlights how refractive index can influence these patterns. Identifying the shift in angle gives clues about the medium鈥檚 refractive index.

Bright fringes are not just important for understanding simple interference patterns. They serve as practical benchmarks for testing and exploring properties of light in physics experiments, shedding light on wave dynamics and material properties.

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Most popular questions from this chapter

35.15. Coherent light with wavelength 600 \(\mathrm{nm}\) passes through two very narrow slits and the interference pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slits. The first-order bright fringe is at 4.84 \(\mathrm{mm}\) from the center of the central bright fringe. For what wavelength of light will the first order dark fringe be observed at this same point on the screen?

35.31. A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 \(\mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO_ 2 film.

35.46. A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wavelength 525 \(\mathrm{nm}\) that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 \(\mathrm{nm}\) per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

35.30. A plate of glass 9.00 \(\mathrm{cm}\) long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 \(\mathrm{mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 \(\mathrm{nm}\) . How many interference fringes are observed per centimeter in the reflected light?

35.54. White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?
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