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Chapter 34: Problem 47

A double slit is opposite the center of a 1.8 -m wide screen \(2.0 \mathrm{~m}\) from the slits. The slit separation is \(24 \mu \mathrm{m}\) and the width of each slit is \(7.2 \mu \mathrm{m}\). How many fringes are visible on the screen if the slit is illuminated by \(600 .-\mathrm{nm}\) light?

Short Answer

Expert verified
The width of the screen is 1.8 m. Answer: 36 fringes are visible on the screen.

Step by step solution

01

Calculate the angular separation between bright fringes

We can calculate the angular separation between the bright fringes (maxima) using the double-slit interference formula: \(\theta = \arcsin(\frac{m\lambda}{d})\) Where: - \(\theta\) represents the angular separation between adjacent maxima, - \(m\) represents the order of the maxima, - \(\lambda\) represents the wavelength of the light (600 nm), - and \(d\) represents the slit separation (24 µm). Since we are interested in the angle between adjacent maxima, we will use \(m = 1\): \(\theta = \arcsin(\frac{1 \times 600\,\mathrm{nm}}{24\,\mu\mathrm{m}})\) \(\theta \approx 0.025\) radians
02

Calculate the distance between adjacent fringes on the screen

We can now convert the angular separation to a linear distance on the screen using the following formula: \(y = L \times \tan(\theta)\) Where: - \(y\) represents the distance between adjacent fringes, - \(L\) represents the distance between the slits and the screen (2.0 m), - and \(\theta\) represents the angular separation between adjacent maxima. Substitute the values and calculate \(y\): \(y = 2.0\,\mathrm{m} \times \tan(0.025) \approx 0.05\,\mathrm{m}\)
03

Determine the total number of fringes visible on the screen

To calculate the total number of fringes visible on the screen, we will divide the width of the screen (1.8 m) by the distance between adjacent fringes (\(y\)): Total fringes = \(\frac{1.8\,\mathrm{m}}{0.05\,\mathrm{m}} = 36\) There are 36 fringes visible on the screen.

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Most popular questions from this chapter

White light shines on a sheet of mica that has a uniform thickness of \(1.30 \mu \mathrm{m} .\) When the reflected light is viewed using a spectrometer, it is noted that light with wavelengths of \(433.3 \mathrm{nm}, 487.5 \mathrm{nm}, 557.1 \mathrm{nm}, 650.0 \mathrm{nm}\), and \(780.0 \mathrm{nm}\) is not present in the reflected light. What is the index of refraction of the mica?

Many times, radio antennas occur in pairs. The effect is that they will then produce constructive interference in one direction while producing destructive interference in another direction - a directional antenna-so that their emissions don't overlap with nearby stations. How far apart at a minimum should a local radio station, operating at \(88.1 \mathrm{MHz},\) place its pair of antennae operating in phase such that no emission occurs along a line \(45.0^{\circ}\) from the line joining the antennae?

Which of the following light types on a grating with 1000 rulings with a spacing of \(2.00 \mu \mathrm{m}\) would produce the largest number of maxima on a screen \(5.00 \mathrm{~m}\) away? a) blue light of wavelength \(450 \mathrm{nm}\) b) green light of wavelength \(550 \mathrm{nm}\) c) yellow light of wavelength \(575 \mathrm{nm}\) d) red light of wavelength \(625 \mathrm{nm}\) e) need more information

It is common knowledge that the visible light spectrum extends approximately from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\). Roughly, \(400 \mathrm{nm}\) to \(500 \mathrm{nm}\) corresponds to blue light, \(500 \mathrm{nm}\) to \(550 \mathrm{nm}\) corresponds to green, \(550 \mathrm{nm}\) to \(600 \mathrm{nm}\) to yelloworange, and above \(600 \mathrm{nm}\) to red. In an experiment, red light with a wavelength of \(632.8 \mathrm{nm}\) from a HeNe laser is refracted into a fish tank filled with water with index of refraction 1.33. What is the wavelength of the same laser beam in water, and what color will the laser beam have in water?

An instructor uses light of wavelength \(633 \mathrm{nm}\) to create a diffraction pattern with a slit of width \(0.135 \mathrm{~mm} .\) How far away from the slit must the instructor place the screen in order for the full width of the central maximum to be \(5.00 \mathrm{~cm} ?\)
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