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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

Sonya is designing a diffraction experiment for her students. She has a laser that emits light of wavelength \(627 \mathrm{nm}\) and a grating with a distance of \(2.40 \times 10^{-3} \mathrm{mm}\) between slits. She hopes to shine the light through the grating and display a total of nine interference maxima on a screen. She finds that no matter how she arranges her setup, she can see only seven maxima. Assuming that the intensity of the light is not the problem, why can't Sonya display the \(m=4\) interference maxima on either side?
Diffraction by a single Slit The central bright fringe in a single-slit diffraction pattern from light of wavelength 476 nm is \(2.0 \mathrm{cm}\) wide on a screen that is $1.05 \mathrm{m}$ from the slit. (a) How wide is the slit? (b) How wide are the first two bright fringes on either side of the central bright fringe? (Define the width of a bright fringe as the linear distance from minimum to minimum.)
Gratings A grating has exactly 8000 slits uniformly spaced over \(2.54 \mathrm{cm}\) and is illuminated by light from a mercury vapor discharge lamp. What is the expected angle for the third-order maximum of the green line $(\lambda=546 \mathrm{nm}) ?$
The photosensitive cells (rods and cones) in the retina are most densely packed in the fovea- the part of the retina used to see straight ahead. In the fovea, the cells are all cones spaced about \(1 \mu \mathrm{m}\) apart. Would our vision have much better resolution if they were closer together? To answer this question, assume two light sources are just far enough apart to be resolvable according to Rayleigh's criterion. Assume an average pupil diameter of \(5 \mathrm{mm}\) and an eye diameter of \(25 \mathrm{mm}\). Also assume that the index of refraction of the vitreous fluid in the eye is \(1 ;\) in other words, treat the pupil as a circular aperture with air on both sides. What is the spacing of the cones if the centers of the diffraction maxima fall on two nonadjacent cones with a single intervening cone? (There must be an intervening dark cone in order to resolve the two sources; if two adjacent cones are stimulated, the brain assumes a single source.)
A camera lens \((n=1.50)\) is coated with a thin film of magnesium fluoride \((n=1.38)\) of thickness \(90.0 \mathrm{nm}\) What wavelength in the visible spectrum is most strongly transmitted through the film?
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