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Chapter 25: Problem 75

Two slits separated by \(20.0 \mu \mathrm{m}\) are illuminated by light of wavelength \(0.50 \mu \mathrm{m} .\) If the screen is \(8.0 \mathrm{m}\) from the slits, what is the distance between the \(m=0\) and \(m=1\) bright fringes?

Short Answer

Expert verified
Answer: The distance between the central bright fringe (m=0) and the first bright fringe adjacent to the central fringe (m=1) is 0.20 m.

Step by step solution

01

Write down the double-slit interference formula

The interference formula for a double-slit experiment is given by: $$ \Delta y = \dfrac{m \lambda d}{a} $$ where \(\Delta y\) is the distance between the adjacent bright fringes, \(m\) is the fringe order (\(0,1,2,\dots\)), \(\lambda\) is the wavelength of the light, \(d\) is the distance between the screen and the slits, \(a\) is the distance between the slits.
02

Plug in the given values

We are given the distance between the slits (\(a = 20.0\,\mu m\)), the wavelength of the light (\(\lambda = 0.50\,\mu m\)), and the distance between the screen and the slits (\(d = 8.0\,m\)). We are looking for the distance between the \(m=0\) and \(m=1\) bright fringes.
03

Calculate the distance between the bright fringes

We need to find the difference in the fringe distances for \(m=0\) and \(m=1\). In other words, we want to calculate \(\Delta y(m=1) - \Delta y (m=0)\). Using the given values, we can plug them into the formula: $$ \begin{aligned} \Delta y (1) &= \dfrac{1 \times 0.50\,\mu m \times 8.0\,m}{20.0\,\mu m} \\ \Delta y (0) &= \dfrac{0 \times 0.50\,\mu m \times 8.0\,m}{20.0\,\mu m} \\ \end{aligned} $$
04

Solve for the distance between the bright fringes

Solving for the distances, we get: $$ \begin{aligned} \Delta y (1) &= \dfrac{4.0}{20.0\,\mu m} m = 0.20\, m \\ \Delta y (0) &= 0\, m \end{aligned} $$ Since the distance between the \(m=0\) and \(m=1\) bright fringes corresponds to \(\Delta y(1) - \Delta y(0)\), we find: $$ \Delta y = 0.20\, m - 0\, m = 0.20\, m $$ Hence, the distance between the \(m=0\) and \(m=1\) bright fringes is \(0.20\,m\).

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Most popular questions from this chapter

You are given a slide with two slits cut into it and asked how far apart the slits are. You shine white light on the slide and notice the first-order color spectrum that is created on a screen \(3.40 \mathrm{m}\) away. On the screen, the red light with a wavelength of 700 nm is separated from the violet light with a wavelength of 400 nm by 7.00 mm. What is the separation of the two slits?
The photosensitive cells (rods and cones) in the retina are most densely packed in the fovea- the part of the retina used to see straight ahead. In the fovea, the cells are all cones spaced about \(1 \mu \mathrm{m}\) apart. Would our vision have much better resolution if they were closer together? To answer this question, assume two light sources are just far enough apart to be resolvable according to Rayleigh's criterion. Assume an average pupil diameter of \(5 \mathrm{mm}\) and an eye diameter of \(25 \mathrm{mm}\). Also assume that the index of refraction of the vitreous fluid in the eye is \(1 ;\) in other words, treat the pupil as a circular aperture with air on both sides. What is the spacing of the cones if the centers of the diffraction maxima fall on two nonadjacent cones with a single intervening cone? (There must be an intervening dark cone in order to resolve the two sources; if two adjacent cones are stimulated, the brain assumes a single source.)
Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)
The diffraction pattern from a single slit is viewed on a screen. Using blue light, the width of the central maximum is \(2.0 \mathrm{cm} .\) (a) Would the central maximum be narrower or wider if red light is used instead? (b) If the blue light has wavelength \(0.43 \mu \mathrm{m}\) and the red light has wavelength \(0.70 \mu \mathrm{m},\) what is the width of the central maximum when red light is used?
A red line (wavelength \(630 \mathrm{nm}\) ) in the third order overlaps with a blue line in the fourth order for a particular grating. What is the wavelength of the blue line?
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