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Chapter 4: Problem 84

Red light (wavelength 632.8 nm in air) from a HeliumNeon laser is incident on a single slit of width \(0.05 \mathrm{mm}\). The entire apparatus is immersed in water of refractive index \(1.333 .\) Determine the angular width of the central peak.

Short Answer

Expert verified
The angular width of the central peak in water is approximately \(1.898 \% \).

Step by step solution

01

Find the wavelength of light in water

First, we need to find the wavelength of the light in water using the given refractive index and the wavelength in air. The wavelength in water is related to the wavelength in air by the formula: \[\lambda_{water} = \frac{\lambda_{air}}{n}\] where \(n\) is the refractive index of water. Substituting the given values, we get: \[\lambda_{water} = \frac{632.8 \times 10^{-9} m}{1.333}\]
02

Calculate the wavelength of light in water

Now, we calculate the wavelength of light in water. \[\lambda_{water} = 474.522 \times 10^{-9} m\]
03

Find the angular width of the central peak

Next, we will use the formula for the angular width of the central peak of a single-slit interference pattern: \[\theta = \frac{2 \lambda}{a}\] where \(\theta\) is the angular width of the central peak, \(\lambda\) is the wavelength of light in water, and \(a\) is the width of the slit. Substituting the values, we get: \[\theta = \frac{2 (474.522 \times 10^{-9} m)}{0.05 \times 10^{-3} m}\]
04

Calculate the angular width

Now, we calculate the angular width. \[\theta = 0.01898088 = 1.898 \% \] Therefore, the angular width of the central peak in water is approximately \(1.898 \% \).

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Most popular questions from this chapter

How narrow is a slit that produces a diffraction pattem on a screen \(1.8 \mathrm{m}\) away whose central peak is \(1.0 \mathrm{m}\) wide? Assume \(\lambda=589 \mathrm{nm}\)

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