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Coherent light with wavelength \(450 \mathrm{nm}\) falls on a pair of slits. On a screen \(1.80 \mathrm{~m}\) away, the distance between dark fringes is \(3.90 \mathrm{~mm} .\) What is the slit separation?

Short Answer

Expert verified
The slit separation, d, is approximately 207 micrometers or 0.207 millimeters.

Step by step solution

01

Identify Knowns

The exercise has provided the following values: wavelength (λ) of \(450 nm = 450 \times 10^{-9} m\), the distance between the screen and the slits (L) of \(1.80 m\), and the distance between consecutive dark fringes (y) of \(3.90 mm = 3.90 \times 10^{-3} m\). Note that n=0 as we consider the first dark fringe.
02

Rearrange The Equation To Solve For d

The given formula can be rearranged from \(y = \frac{(2n+1) \lambda L}{2d}\) to solve for d as \(d = \frac{ (2n+1) \lambda L}{2y}\). This becomes \(d = \frac{\lambda L}{y}\) since \(n=0\).
03

Substitute The Knowns Into The Equation

Substitute the known values into the formula for d. Thus, \(d = \frac{450 \times 10^{-9} m \times 1.80 m }{3.90 \times 10^{-3} m}.\)
04

Calculate The Answer

On performing the calculation, by multiplying the numerator, dividing by the denominator, d results to be \(d \approx 0.207 mm = 207 \mu m\).

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Key Concepts

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

Interference Pattern
When we talk about an interference pattern, we're delving into the fascinating world of wave behavior, which is a foundational concept in physics.

Think of throwing two stones into a pond at nearby points. You'll see ripples, or 'waves', emanating from the points where the stones hit the water. Now, notice how these ripples interact; at some points, they'll merge and produce a larger ripple, while at other spots, they seem to cancel each other out, leaving calm water. This phenomenon, seen with water, light, or any other wave-like entity, is called interference.

The pattern created by this merger, the series of bright and dark bands, is specifically referred to as the 'interference pattern'. In our exercise, coherent light is used, which means all light waves are in phase and have a constant phase difference, producing a very distinct and stable interference pattern on the screen.
Double-Slit Experiment
The double-slit experiment is a pivotal demonstration of the dual nature of light, revealing its wave-like characteristics. Conducted first by Thomas Young in 1801, this experiment involves coherent light, such as a laser, passing through two closely spaced slits. What emerges on the other side is fascinating – instead of seeing two separate bands of light on a screen behind the slits, we see several bands of varying brightness.

This phenomenon occurs because the light waves spread out after passing through the slits and then interfere with each other. Bright bands, or 'fringes', form where the waves constructively interfere, and dark ones form when they destructively interfere. Understanding the principles behind this experiment aids in grasping how light behaves in certain conditions and is fundamental to the topics of diffraction and interference.
Wavelength of Light
Every color of light has a corresponding wavelength, which is the distance between successive peaks or troughs of a wave. It's a crucial concept in understanding how light interacts with objects and how we perceive various colors. In the visible spectrum, violet has the shortest wavelength and red has the longest.

In our exercise, the wavelength given is 450 nm, which falls within the blue-violet region of the visible spectrum. When dealing with interference and diffraction, the wavelength is a key factor in determining the pattern produced. For instance, different wavelengths will produce interference patterns with differing spacing between the fringes, thus affecting the calculations for determining other parameters like slit separation in the context of our textbook problem.
Dark Fringes
In the context of the double-slit experiment and interference patterns, dark fringes represent points on the screen where destructive interference has occurred. This means the crest of one wave meets the trough of another, and as a result, they cancel each other out.

The position of these dark fringes follows a predictable formula and is related to the wavelength of the light used, as well as the separation of the slits, among other factors. The 'formula' mentioned in the exercise, which was manipulated to find the slit separation, directly arises from the mathematics describing these dark fringes. The placement and spacing of the dark fringes provide insightful information enabling the calculation of various properties of the wave and the experimental setup, such as the slit separation, which in the context of the given exercise ultimately was found to be approximately 207 micrometers.

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Most popular questions from this chapter

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm 19.0^{\circ}\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(\frac{1}{10}\) the maximum intensity on the screen?

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) wavelengths of the light in the \(\mathrm{TiO}_{2}\) film.

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?

Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is \(120 \mathrm{~m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of \(40 \mathrm{~m}\) to the right of antenna \(B .\) The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wavelength for which there will be constructive interference at point \(Q ?\)

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference 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 \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

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