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Chapter 3: Problem 38

What is the angular width of the central fringe of the interference pattern of (a) 20 slits separated by \(d=2.0 \times 10^{-3} \mathrm{mm} ?\) (b) 50 slits with the same separation? Assume that \(\lambda=600 \mathrm{nm}\).

Short Answer

Expert verified
The angular width of the central fringe for the 20-slits scenario is \(0.015 \mathrm{rad}\) and for the 50-slits scenario is \(0.006 \mathrm{rad}\).

Step by step solution

01

Understand the relationship between angular width, wavelength and slit separation

The spread (angular width) of the central fringe in a multiple-slit interference pattern is directly proportional to the wavelength of the light and inversely proportional to the number of slits (N) and the slit separation. This is captured by the formula \(\theta = \lambda / (Nd)\) where \(\theta\) is the angular width, \(\lambda\) is the wavelength of the incident light, \(N\) is the number of slits and \(d\) is the slit separation.
02

Calculate the angular width for the 20-slits scenario

Substitute the given values into the formula: \(N=20\), \(d=2.0 \times 10^{-3} \mathrm{mm} = 2.0 \times 10^{-6} \mathrm{m}\), \(\lambda=600 \mathrm{nm} = 600 \times 10^{-9} \mathrm{m}\). The angular width \( \theta_{20} \) is now calculated by plugging these values into the formula to get: \( \theta_{20} = 600 \times 10^{-9} \mathrm{m} / (20 \times 2.0 \times 10^{-6} \mathrm{m}) = 0.015 \mathrm{rad}\).
03

Calculate the angular width for the 50-slits scenario

Repeat step 2 but now using the given value for the number of slits: \(N=50\). We get: \( \theta_{50} = 600 \times 10^{-9} \mathrm{m} / (50 \times 2.0 \times 10^{-6} \mathrm{m}) = 0.006 \mathrm{rad}\).

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Most popular questions from this chapter

Find the distance between two slits that produces the first minimum for 410 -nm violet light at an angle of \(45.0^{\circ}\).

A thin wedge filled with air is produced when two flat glass plates are placed on top of one another and a slip of paper is inserted between them at one edge. Interference fringes are observed when monochromatic light falling vertically on the plates are seen in reflection. Is the first fringe near the edge where the plates are in contact a bright fringe or a dark fringe? Explain.

Young's double-slit experiment is performed immersed in water \((n=1.333) .\) The light source is a HeNe laser, \(\lambda=632.9 \mathrm{nm}\) in vacuum. (a) What is the wavelength of this light in water? (b) What is the angle for the third order maximum for two slits separated by 0.100 mm.

In a thermally stabilized lab, a Michelson interferometer is used to monitor the temperature to ensure it stays constant. The movable mirror is mounted on the end of a 1.00 -m-long aluminum rod, held fixed at the other end. The light source is a He Ne laser, \(\lambda=632.8 \mathrm{nm}\). The resolution of this apparatus corresponds to the temperature difference when a change of just one fringe is observed. What is this temperature difference?

Is it possible to create a experimental setup in which there is only destructive interference? Explain.
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