**Problem Statement:**Two wavelengths of light are incident upon a diffraction grating with a grating constant of 426 lines/mm. If the wavelengths of light are 631 nm and 470 nm, what is the linear separation between the second-order maxima of these two wavelengths on a screen that is a distance of 1.73 meters away from the grating?**Explanation:**To solve this problem, you need to use the grating equation:\[ d \sin \theta = m \lambda \]Where:- \( d \) is the distance between grating lines (which is the reciprocal of the grating constant).- \( \theta \) is the angle of diffraction.- \( m \) is the order of the maximum (in this case, \( m = 2 \)).- \( \lambda \) is the wavelength of the light.First, calculate the distance \( d \) between the lines:\[ d = \frac{1}{426 \, \text{lines/mm}} = \frac{1}{426,000 \, \text{lines/m}} \]Then, apply the grating equation for each wavelength to find the angles \( \theta_1 \) and \( \theta_2 \) for the second-order maxima. Use those angles to calculate the linear separation \( \Delta y \) on the screen using the formula:\[ \Delta y = L (\tan \theta_1 - \tan \theta_2) \]Where \( L = 1.73 \) meters is the distance from the grating to the screen. Calculate the actual values to find the linear separation between the maxima for the given wavelengths.

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Two waveslengths of light are incident upon a diffraction grating with a grating constant of 426 lines/mm. If the wavelengths of light are 631 nm and 470 nm, what is the linear separation between the 2 ondrder maxima of these two wavelengths on a screen that is a distance 1.73 away from the grating?

Two waveslengths of light are incident upon a diffraction grating with a grating constant of 426 lines/mm. If the wavelengths of light are 631 nm and 470 nm, what is the linear separation between the 2 ondrder maxima of these two wavelengths on a screen that is a distance 1.73 away from the grating?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Interference of Light

The word "constructive" signifies addition in interference. Constructive interference occurs when the path of the waves that start from the slit differs by 1 entire wavelength or in the intervals of the same. Hence, the waves that are presented on the screen are in phase (crest to crest/trough to trough). The waves interfering constructively forms bright fringes on the screen.

Question

**Problem Statement:**

Two wavelengths of light are incident upon a diffraction grating with a grating constant of 426 lines/mm. If the wavelengths of light are 631 nm and 470 nm, what is the linear separation between the second-order maxima of these two wavelengths on a screen that is a distance of 1.73 meters away from the grating?

**Explanation:**

To solve this problem, you need to use the grating equation:

\[ d \sin \theta = m \lambda \]

Where:
- \( d \) is the distance between grating lines (which is the reciprocal of the grating constant).
- \( \theta \) is the angle of diffraction.
- \( m \) is the order of the maximum (in this case, \( m = 2 \)).
- \( \lambda \) is the wavelength of the light.

First, calculate the distance \( d \) between the lines:

\[ d = \frac{1}{426 \, \text{lines/mm}} = \frac{1}{426,000 \, \text{lines/m}} \]

Then, apply the grating equation for each wavelength to find the angles \( \theta_1 \) and \( \theta_2 \) for the second-order maxima. Use those angles to calculate the linear separation \( \Delta y \) on the screen using the formula:

\[ \Delta y = L (\tan \theta_1 - \tan \theta_2) \]

Where \( L = 1.73 \) meters is the distance from the grating to the screen. Calculate the actual values to find the linear separation between the maxima for the given wavelengths.

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