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Chapter 34: Problem 23

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?

Short Answer

Expert verified
Answer: The distance between the slits and the screen is approximately 1.0 m.

Step by step solution

01

Write the formula for the position of the maxima in an interference pattern

Write the formula as \(y_n = \frac{n\lambda L}{d}\).
02

Plug in the given values

We know the following values: - Wavelength \(\lambda = 540 \,\text{nm} = 540 \times 10^{-9}\, \text{m}\) (converting to meters) - Separation between the slits \(d = 0.100\, \text{mm} = 0.100 \times 10^{-3} \,\text{m}\) (converting to meters) - Distance between the first side maximum and the center of the pattern \(y_1 = 5.40\, \text{mm} = 5.40 \times 10^{-3}\, \text{m}\) (converting to meters) - Order of the maximum \(n = 1\) Now plug these values into the formula: \(5.40\times 10^{-3}=\frac{1\times 540\times 10^{-9} L}{0.100\times 10^{-3}}\)
03

Solve for L

To find the value of \(L\), we will first multiply both sides of the equation by \(0.100\times 10^{-3}\): \(5.40\times 10^{-3} \cdot 0.100 \times 10^{-3} = 540 \times 10^{-9} L\) Now, divide both sides by the remaining factor on the right side of the equation, which in this case is \(540 \times 10^{-9}\): \(L = \frac{5.40 \times 10^{-3} \cdot 0.100\times 10^{-3}}{540\times 10^{-9}}\) Upon calculating, we get: \(L \approx 1.0\,\text{m}\) Thus, the distance between the slits and the screen is approximately 1 meter.

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Most popular questions from this chapter

Some mirrors for infrared lasers are constructed wit alternating layers of hafnia and silica. Suppose you want to produce constructive interference from a thin film of hafnia \((n=1.90)\) on BK-7 glass \((n=1.51)\) when infrared radiation of wavelength \(1.06 \mu \mathrm{m}\) is used. What is the smallest film thickness that would be appropriate, assuming the laser beam is oriented at right angles to the film?

An airplane is made invisible to radar by coating it with a 5.00 -mm-thick layer of an antireflective polymer with the index of refraction \(n=1.50 .\) What is the wavelength of radar waves for which the plane is made invisible?

Plane light waves are incident on a single slit of width \(2.00 \mathrm{~cm} .\) The second dark fringe is observed at \(43.0^{\circ}\) from the central axis. What is the wavelength of the light?

Why are radio telescopes so much larger than optical telescopes? Would an \(X\) -ray telescope also have to be larger than an optical telescope?

What is the wavelength of the X-rays if the first-order Bragg diffraction is observed at \(23.0^{\circ}\) related to the crystal surface, with inter atomic distance of \(0.256 \mathrm{nm} ?\)
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