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Chapter 25: Problem 40

A grating is made of exactly 8000 slits; the slit spacing is $1.50 \mu \mathrm{m} .\( Light of wavelength \)0.600 \mu \mathrm{m}$ is incident normally on the grating. (a) How many maxima are seen in the pattern on the screen? (b) Sketch the pattern that would appear on a screen \(3.0 \mathrm{m}\) from the grating. Label distances from the central maximum to the other maxima.

Short Answer

Expert verified
Answer: There are a total of 5 maxima seen in the diffraction pattern. The distances of the maxima from the central maximum on the screen are 1.32 m for the maxima at m=1 and 4.0 m for the maxima at m=2.

Step by step solution

01

Calculate the maximum order m

Use the grating equation: \(d \cdot sin(\theta) = m \cdot \lambda\), where d is the slit spacing, \(\theta\) is the diffraction angle, m is the order, and \(\lambda\) is the wavelength of the incident light. To find the maximum order m, set \(\theta\) at \(90^{\circ}\), i.e., \(sin(90^{\circ}) = 1\). Then, the equation becomes \(d = m \cdot \lambda\). Solve for m: \(m = \frac{d}{\lambda}\). Plug in d = \(1.50 \mu\mathrm{m}\) and \(\lambda = 0.600 \mu\mathrm{m}\): \(m = \frac{1.50}{0.600} = 2.5\). As m must be an integer, the maximum order m is 2.
02

Calculate the number of maxima

Including the central maximum (m = 0), we have orders 0, 1, and 2 on both sides of the central maximum. Therefore, the total number of maxima seen in the pattern is \((2 \times 2) + 1 = 5\).
03

Calculate the angular separation of maxima

For each order m, find the angular separation \(\theta\) using the grating equation. For m=1: \(sin(\theta_1) = \frac{1 \cdot 0.600 \mu\mathrm{m}}{1.50 \mu\mathrm{m}}\), so \(\theta_1 = sin^{-1}(0.4) \approx 23.58^{\circ}\). For m=2: \(sin(\theta_2) = \frac{2 \cdot 0.600 \mu\mathrm{m}}{1.50 \mu\mathrm{m}}\), so \(\theta_2 = sin^{-1}(0.8) \approx 53.13^{\circ}\).
04

Calculate distances of maxima from the central maximum on the screen

Using the angular separation, we can calculate the distances of these maxima from the central maximum on a screen placed 3.0 m away from the grating. We do this by using the formula \(y = L \cdot tan(\theta)\), where y is the distance from the central maximum, L is the distance between grating and screen, and \(\theta\) is the angular separation. For the maxima at m=1, the distance is \(y_1 = 3.0 \cdot tan(23.58^{\circ}) \approx 1.32 \mathrm{m}\). For the maxima at m=2, the distance is \(y_2 = 3.0 \cdot tan(53.13^{\circ}) \approx 4.0 \mathrm{m}\).
05

Sketch the pattern and label the distances

Sketch the pattern with the central maximum at the center and two maxima on each side. Label the distances of the maxima at m=1 as \(1.32 \mathrm{m}\), and the maxima at m=2 as \(4.0 \mathrm{m}\).

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Most popular questions from this chapter

About how close to each other are two objects on the Moon that can just barely be resolved by the \(5.08-\mathrm{m}-\) (200-in.)-diameter Mount Palomar reflecting telescope? (Use a wavelength of \(520 \mathrm{nm}\).)
Light from a red laser passes through a single slit to form a diffraction pattern on a distant screen. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum on the screen?
The Michelson Interferometer A Michelson interferometer is adjusted so that a bright fringe appears on the screen. As one of the mirrors is moved \(25.8 \mu \mathrm{m}, 92\) bright fringes are counted on the screen. What is the wavelength of the light used in the interferometer?
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.
A thin film of oil \((n=1.50)\) of thickness \(0.40 \mu \mathrm{m}\) is spread over a puddle of water \((n=1.33) .\) For which wavelength in the visible spectrum do you expect constructive interference for reflection at normal incidence?
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