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Chapter 25: Problem 39

A grating has 5000.0 slits/cm. How many orders of violet light of wavelength \(412 \mathrm{nm}\) can be observed with this grating? (tutorial: grating).

Short Answer

Expert verified
Answer: There can be 4 orders of violet light observed with a grating of 5000.0 slits/cm.

Step by step solution

01

Write down the known values

We are given: - Number of slits per centimeter: \(n = 5000.0\) slits/cm - Wavelength of violet light: \(\lambda = 412 \mathrm{nm}\)
02

Convert the given values to compatible units

As we are given the number of slits per centimeter, it is better to convert the wavelength into centimeters: \(\lambda = 412 \mathrm{nm} \times \frac{1 \mathrm{cm}}{10^7 \mathrm{nm}} = 4.12 \times 10^{-5} \mathrm{cm}\)
03

Rewrite the grating equation

The grating equation is given by: \(\sin \theta_{m} = \frac{m \lambda}{d}\) where: - \(\theta_m\) is the angle of diffraction for the \(m\)th order - \(m\) is the order of the spectrum (a positive integer) - \(\lambda\) is the wavelength of light - \(d\) is the distance between the slits, which is the inverse of the number of slits per length
04

Calculate the distance between slits

Given the number of slits per centimeter (\(n\)), we can find the distance between the slits (\(d\)) by: \(d = \frac{1}{n} = \frac{1}{5000.0 \thinspace \mathrm{slits/cm}} = 2 \times 10^{-4} \thinspace \mathrm{cm}\)
05

Calculate the maximum order \(m\)

The maximum order, \(m\), occurs when \(\sin \theta_{m} = 1\). So we can rewrite the grating equation as: \(1 = \frac{m \lambda}{d}\) Now let's solve for \(m\): \(m = \frac{d}{\lambda} = \frac{2 \times 10^{-4}\thinspace \mathrm{cm}}{4.12 \times 10^{-5}\thinspace \mathrm{cm}} \approx 4.85\). Since \(m\) is a positive integer, we need to round down 4.85 to the nearest integer, which gives us \(m = 4\).
06

Answer

There can be 4 orders of violet light (\(\lambda = 412 \thinspace \mathrm{nm}\)) observed with a grating of 5000.0 slits/cm.

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Most popular questions from this chapter

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 of wavelength 589 nm incident on a pair of slits produces an interference pattern on a distant screen in which the separation between adjacent bright fringes at the center of the pattern is \(0.530 \mathrm{cm} .\) A second light source, when incident on the same pair of slits, produces an interference pattern on the same screen with a separation of $0.640 \mathrm{cm}$ between adjacent bright fringes at the center of the pattern. What is the wavelength of the second source? [Hint: Is the small-angle approximation justified?]
Suppose a transparent vessel \(30.0 \mathrm{cm}\) long is placed in one arm of a Michelson interferometer, as in Example 25.2. The vessel initially contains air at \(0^{\circ} \mathrm{C}\) and 1.00 atm. With light of vacuum wavelength \(633 \mathrm{nm}\), the mirrors are arranged so that a bright spot appears at the center of the screen. As air is slowly pumped out of the vessel, one of the mirrors is gradually moved to keep the center region of the screen bright. The distance the mirror moves is measured to determine the value of the index of refraction of air, \(n .\) Assume that, outside of the vessel, the light travels through vacuum. Calculate the distance that the mirror would be moved as the container is emptied of air.
The diffraction pattern from a single slit is viewed on a screen. Using blue light, the width of the central maximum is \(2.0 \mathrm{cm} .\) (a) Would the central maximum be narrower or wider if red light is used instead? (b) If the blue light has wavelength \(0.43 \mu \mathrm{m}\) and the red light has wavelength \(0.70 \mu \mathrm{m},\) what is the width of the central maximum when red light is used?
A lens \((n=1.52)\) is coated with a magnesium fluoride film \((n=1.38) .\) (a) If the coating is to cause destructive interference in reflected light for \(\lambda=560 \mathrm{nm}\) (the peak of the solar spectrum), what should its minimum thickness be? (b) At what two wavelengths closest to 560 nm does the coating cause constructive interference in reflected light? (c) Is any visible light reflected? Explain.
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