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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

A steep cliff west of Lydia's home reflects a \(1020-\mathrm{kHz}\) radio signal from a station that is \(74 \mathrm{km}\) due east of her home. If there is destructive interference, what is the minimum distance of the cliff from her home? Assume there is a \(180^{\circ}\) phase shift when the wave reflects from the cliff.
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?
Sonya is designing a diffraction experiment for her students. She has a laser that emits light of wavelength \(627 \mathrm{nm}\) and a grating with a distance of \(2.40 \times 10^{-3} \mathrm{mm}\) between slits. She hopes to shine the light through the grating and display a total of nine interference maxima on a screen. She finds that no matter how she arranges her setup, she can see only seven maxima. Assuming that the intensity of the light is not the problem, why can't Sonya display the \(m=4\) interference maxima on either side?
In bright light, the pupils of the eyes of a cat narrow to a vertical slit \(0.30 \mathrm{mm}\) across. Suppose that a cat is looking at two mice $18 \mathrm{m}$ away. What is the smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of $560 \mathrm{nm} ?$ Assume the resolution is limited by diffraction only.
Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]
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