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Chapter 36: Problem 16

Immersed in Water A double-slit arrangement produces interference fringes for sodium light \((\lambda=589 \mathrm{~nm})\) that are \(0.20^{\circ}\) apart. What is the angular fringe separation if the entire arrangement is immersed in water \((\bar{n}=1.33)\) ?

Short Answer

Expert verified
The new angular fringe separation when immersed in water is approximately \( 0.15^{\circ} \).

Step by step solution

01

- Review the Original Angular Separation

The initial angular separation of the interference fringes in air is given as \( 0.20^{\circ} \). This separation corresponds to the sodium light wavelength \( \lambda = 589 \, \text{nm} \).
02

- Understand the Effect of Medium

When the arrangement is immersed in water, the wavelength of the sodium light will change due to the refractive index of water. The refractive index \( \bar{n} = 1.33 \).
03

- Calculate the New Wavelength in Water

The wavelength of light in a medium is given by \( \lambda' = \frac{\lambda}{\bar{n}} \). Substituting \( \lambda = 589 \, \text{nm} \) and \( \bar{n} = 1.33 \): \[ \lambda' = \frac{589 \, \text{nm}}{1.33} \approx 442.86 \, \text{nm} \]
04

- Relate Angular Separation to Wavelength

The angular separation \( \theta \) of interference fringes is directly proportional to the wavelength of the light used. Therefore, the new angular separation \( \theta' \) in water can be found using: \[ \theta' = \theta \cdot \frac{\lambda'}{\lambda} \]
05

- Apply the Wavelength Change to Angular Separation

Substitute the known values: \[ \theta' = 0.20^{\circ} \cdot \frac{442.86 \, \text{nm}}{589 \, \text{nm}} \approx 0.15^{\circ} \]

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Key Concepts

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

Double-Slit Experiment
The double-slit experiment is a fundamental demonstration of the wave-like nature of light. It involves a light source, typically a monochromatic one, which passes through two closely spaced slits. The light waves emerging from these slits interfere with each other, creating a series of bright and dark fringes on a screen placed behind the slits.
These fringes are known as interference fringes.
The distance between adjacent bright or dark fringes, termed as fringe separation, depends on the wavelength of light and the distance between the slits. This pattern of interference is the result of constructive and destructive interference, demonstrating the wave behavior of light.

- **Constructive Interference:** Occurs when the peaks of two waves coincide, resulting in increased amplitude, causing bright fringes.
- **Destructive Interference:** Occurs when a peak of one wave coincides with the trough of another, resulting in reduced amplitude, causing dark fringes.
Wavelength Change in Medium
When light travels through different media, its speed and wavelength change, though its frequency remains constant.
The speed of light in a medium is given by the equation:

This equation demonstrates that the wavelength (位) gets reduced when light enters a medium with a refractive index.
In the given problem, sodium light with a wavelength of 位 = 589 nm in air is used. In water, which has a refractive index (虆n = 1.33), the new wavelength can be found using :

位' = 位/虆n
Substituting the known values, we get:
鈥�
位' = 442.86 nm (approximated)
This reduction in wavelength affects various optical phenomena, as seen in the change in angular fringe separation in the double-slit experiment.
Refractive Index Effect on Light
The refractive index (虆n) of a material is a measure of how much the speed of light is reduced inside the material compared to its speed in a vacuum.
It also affects the wavelength of light passing through it, altering the interference pattern observed in experiments like the double-slit experiment.

The refractive index is defined as:
This equation states that the speed of light in the medium (c ) is c=谓 , where 谓 is the speed of light in a vacuum.

The higher the refractive index, the slower the light travels through the medium.
In the given exercise, when the double-slit arrangement is immersed in water (虆n = 1.33), the light鈥檚 wavelength changes according to:
位' = 位/虆n.
This change in wavelength leads to a new angular fringe separation:胃' = 胃 * (位'/位). The result of this change, in this case, is that the angular fringe separation becomes approximately **0.15 degrees**.

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Most popular questions from this chapter

Coat Glass We wish to coat flat glass \((n=1.50)\) with a transparent material \((n=1.25)\) so that reflection of light at wavelength \(600 \mathrm{~nm}\) is eliminated by interference. What minimum thickness can the coating have to do this?

Camera Lens A camera lens with index of refraction greater than \(1.30\) is coated with a thin transparent film of index of refraction \(1.25\) to eliminate by interference the reflection of light at wavelength \(\lambda\) that is incident perpendicularly on the lens. In terms of \(\lambda\), what minimum film thickness is needed?

Two Interference Patterns In a double-slit experiment the distance between slits is \(5.0 \mathrm{~nm}\) and the slits are \(1.0 \mathrm{~m}\) from the screen. Two interference patterns can be seen on the screen: one due to light with wavelength \(480 \mathrm{~nm}\), and the other due to light with wave length \(600 \mathrm{~nm}\). What is the separation on the screen between the third-order \((m=3)\) bright fringes of the two interference patterns?

Mica Flake A thin flake of mica \((n=1.58)\) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe \((m=7)\) before the mica was used. If \(\lambda=550 \mathrm{~nm}\), what is the thickness of the mica? (Hint: Consider the wavelength of the light within the mica.)

Broad Beam A broad beam of monochromatic light is directed perpendicularly through two glass plates that are held together at one end to create a wedge of air between them. An observer intercepting light reflected from the wedge of air, which acts as a thin film, sees 4001 dark fringes along the length of the wedge. When the air between the plates is evacuated, only 4000 dark fringes are seen. Calculate the index of refraction of air from these data.
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