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\(\bullet$$\bullet\) Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first inter- ference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(19.46^{\circ}\) instead. What is the index of refrac- tion of this liquid?

Short Answer

Expert verified
The index of refraction of the liquid is approximately 1.73.

Step by step solution

01

Understanding the Problem

The problem involves thin slits illuminated by monochromatic coherent light. Initially, the light is in air and produces the first interference minima at angles \( \pm 35.20^{\circ} \). When immersed in a liquid, the minima occur at \( \pm 19.46^{\circ} \). We need to find the index of refraction of the liquid.
02

Applying the Condition for Minima

In a double-slit experiment, the condition for interference minima is given by \(d \sin \theta_m = (m + 1/2)\lambda\), where \(d\) is the slit separation, \(\theta_m\) is the angle for the \(m\)-th minimum, and \(\lambda\) is the wavelength of light. For the first minimum, \(m = 0\), so the condition simplifies to \(d \sin \theta_1 = \frac{1}{2}\lambda\).
03

Assessing Initial Conditions

With air, the angle \(\theta = 35.20^{\circ}\), so \(d \sin 35.20^{\circ} = \frac{1}{2}\lambda_{air}\). This equation links \(d\), \(\lambda_{air}\), and the condition for the minima.
04

Applying Conditions in the Liquid

In the liquid, the angle for the first minimum is \(\theta = 19.46^{\circ}\), hence \(d \sin 19.46^{\circ} = \frac{1}{2}\lambda_{liquid}\).
05

Relating Wavelengths through Refraction

The wavelength in the liquid \(\lambda_{liquid}\) is related to the wavelength in air \(\lambda_{air}\) by \(\lambda_{liquid} = \frac{\lambda_{air}}{n}\), where \(n\) is the refractive index.
06

Equating the Conditions

From Step 3: \(d \sin 35.20^{\circ} = \frac{1}{2}\lambda_{air}\). From Step 4, substituting from Step 5: \(d \sin 19.46^{\circ} = \frac{1}{2}\frac{\lambda_{air}}{n}\). Equate these to solve for \(n\):\(\sin 35.20^{\circ} = n \sin 19.46^{\circ}\).
07

Solving for the Index of Refraction

Rearrange to solve for \(n\): \[ n = \frac{\sin 35.20^{\circ}}{\sin 19.46^{\circ}} \]. Calculate using trigonometric values: \(\sin 35.20^{\circ} = 0.576\) (approx.) and \(\sin 19.46^{\circ} = 0.333\) (approx.), yielding: \[ n \approx \frac{0.576}{0.333} \approx 1.73 \].

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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 in optics that explains the phenomenon of interference of light. This experiment highlights how light behaves like a wave, showcasing patterns of bright and dark areas known as the interference pattern. It involves shining light through two narrow, closely spaced slits. When this monochromatic light, meaning light of a single wavelength, passes through the slits, it diffracts. This means it spreads out rather than traveling in straight lines.
As a result, the light waves overlap and interact in regions of constructive interference, leading to bright fringes, and destructive interference, resulting in dark fringes. The pattern can be seen on a screen positioned behind the slits. The specific conditions for these bright and dark spots involve the slit separation, wavelength of the light, and the distance from the slits to the screen, as described by the interference equations.
Index of Refraction
The index of refraction, usually represented by the letter \(n\), is a measure of how much light slows down as it travels through a medium compared to its speed in a vacuum. In simpler terms, when light enters a different medium, it changes speed and direction. The index of refraction quantifies this change. It's calculated using Snell's Law, which relates the angles and indices of refraction of two different media.
In the context of this problem, when the double-slit setup is immersed in a liquid, the light still produces an interference pattern, but the angles at which the minima occur change due to the light's altered speed. The index of refraction tells us how much the light is altered. To find this number, we use the angles of the interference minima in air and in the liquid to relate back to the change in light's path using trigonometric calculations. Knowing these two angles allows us to determine the refractive index of the liquid accurately.
Monochromatic Light
Monochromatic light is light consisting of a single wavelength or color. Unlike white light, which contains all the colors of the visible spectrum, monochromatic light provides a uniform source for conducting interference experiments. This simplifies calculations as there is only one wavelength to consider. It ensures the interference pattern is clear and not confused with overlapping wavelengths.
In interference experiments such as the double-slit setup, using monochromatic light like in laser beams increases accuracy. This is because having only one wavelength allows for consistent constructive and destructive interference, as multiple wavelengths could produce a complex and muddled pattern. In our exercise, monochromatic light enables us to apply the interference equations without the added complication of multiple wavelengths, allowing for simpler deductions, such as finding the refractive index using trigonometric calculations.
Trigonometric Calculation
Trigonometric calculations play a pivotal role in solving problems related to interference patterns created by the double-slit experiment. The relationship between the angles of interference patterns and the physical dimensions of the experimental setup is described using trigonometric functions. Specifically, the sine function is used to relate the angles of the light rays to the distance between the slits and the wavelength.
The formula used for calculating the index of refraction when light passes from air into a liquid relies on trigonometric functions. We know that the ratio of the wavelengths in different mediums can be expressed through the sines of the angles where minima occur. Hence, employing trigonometric identities allows us to solve for unknowns like the index of refraction easily. These calculations are central to understanding how light behaves differently when it moves from one medium to another, as light change direction based on these trigonometric principles. In this exercise, it is crucial to correctly calculate these angles to accurately determine the refractive index.

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Most popular questions from this chapter

\(\bullet\) Monochromatic \(x\) rays are incident on a crystal for which the spacing of the atomic planes is 0.440 nm. The first-order maximum in the Bragg reflection occurs when the incident and reflected rays make an angle of \(39.4^{\circ}\) with the crystal planes. What is the wavelength of the rays?

\(\bullet$$\bullet\) Sensitive eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 \(\mathrm{nm}\) has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

\(\bullet\) Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge, so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546 nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

\(\bullet$$\bullet\) A wildlife photographer uses a moderate telephoto lens of focal length 135 \(\mathrm{mm}\) and maximum aperture \(f / 4.00\) to photo- graph a bear that is 11.5 \(\mathrm{m}\) away. Assume the wavelength is 550 \(\mathrm{nm}\) . (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0,\) what would be the width of the small- est resolvable feature on the bear"?

\( (f… # \)\bullet$$\bullet\( Two small loudspeakers that are 5.50 \)\mathrm{m}\( apart are emitting sound in phase. From both of them, you hear a singer singing \)\mathrm{C} \\#\( (frequency 277 \)\mathrm{Hz} )\( , while the speed of sound in the room is 340 \)\mathrm{m} / \mathrm{s}\( . Assuming that you are rather far from these speak- ers, if you start out at point \)P\( equidistant from both of them and walk around the room in front of them, at what angles (measured relative to the fine from \)P$ to the midpoint between the speakers) will you hear the sound (a) maximally enhanced, (b) cancelled? Neglect any reflections from the walls.

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