Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Visible light passes through a diffraction grating that has 900 slits/cm, and the interference pattern is observed on a screen that is 2.50 m from the grating. (a) Is the angular position of the first-order spectrum small enough for sin θθ to be a good approximation? (b) In the first- order spectrum, the maxima for two different wavelengths are separated on the screen by 3.00 mm. What is the difference in these wavelengths?

Short Answer

Expert verified
(a) Yes, the angle is small enough. (b) The wavelength difference is around 20 nm.

Step by step solution

01

Understand the approximation condition

For the approximation sinθθ (in radians) to be valid, θ should be small, typically less than 10 degrees. First, calculate the angle θ for the first-order spectrum to check the approximation condition.
02

Compute diffraction grating spacing

Convert the given grating density of 900 slits/cm to grating spacing d in meters. d=1900slits/cm=190000slits/m
03

Determine the angle θ using the first-order approximation

The angle θ for the first-order maximum is given by the formula sinθ=mλdSince we are interested in the value of θ for small approximation validity, note that without a specific wavelength λ, we cannot directly calculate θ but will instead reason about the typical values in the visible light spectrum (around 500 nm) to do a comparative analysis.
04

Calculate angular separation for two wavelengths

Calculate the angular separation Δθ between two wavelengths using Δθ=ΔyLwhere Δy=3.00mm=0.003m (given separation on screen) and L=2.5m. Δθ=0.0032.5
05

Compute wavelength difference

Use the relation Δθ=ΔλdRearrange to find Δλ=dΔθSubstitute the values of d from Step 2 and Δθ from Step 4.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Position
Angular position in the context of a diffraction grating refers to the angle at which the different orders of maximum intensity—the bright fringes—appear on a viewing screen. This is directly related to how light waves interfere with each other after passing through the grating's slits.

To find the angular position, we use the formula sinθ=mλd, where m is the order number, λ is the wavelength of the light, and d is the distance between grating slits. If you have visible light around wavelengths of 500 nm, to find the first-order maximum, you would use m=1.

For small angles, which are often the case in practical scenarios, the approximation sinθθ (in radians) is valid. This makes computation simpler, as you can directly measure or calculate θ without complex trigonometric functions. Checking smallness can be done by comparing θ against 10 degrees (roughly 0.1745 radians). If θ is significantly less, the approximation holds well.
Interference Pattern
When light passes through a diffraction grating, it creates an interference pattern—a series of alternating bright and dark spots or lines on a screen. This occurs due to constructive and destructive interference of light waves.

In constructive interference, the path difference between waves is a multiple of the wavelength, causing bright spots, called maxima. For destructive interference, waves cancel each other out, leading to dark spots, called minima.

The spacing between these fringes (bright or dark spots) on the screen is related to both the wavelength of the light and the distance to the screen. Interference patterns are thus directly tied to the wavelength and are a visual representation of the wave properties of light.

These patterns are noticeable and precisely measurable, making diffraction gratings powerful tools in experiments where determining light properties is essential.
Wavelength Difference
The wavelength difference between two light sources in a diffraction grating setup determines how far apart their corresponding maxima will appear on the screen. To determine the separation, one needs to consider both the distance between the grating and the screen, as well as the specific wavelengths involved.

With a known distance to the screen and a measured separation Δy on the screen between maxima, the angular separation Δθ can be computed using Δθ=ΔyL, where L is the screen distance.

It's straightforward to relate this angular separation back to a difference in wavelength Δλ with Δθ=Δλd, fully utilizing the grating spacing d.

The resulting difference in wavelengths is given by Δλ=dΔθ, which allows determinations of wavelength differences from readily observable screen measurements—a useful method in spectroscopy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are asked to design a space telescope for earth orbit. When Jupiter is 5.93 × 108 km away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 km apart. What minimum-diameter mirror is required? Assume a wavelength of 500 nm.

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 nm; for deuterium, the corresponding wavelength is 656.27 nm. (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum (θ=0) is 6.00×106W/m2. (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

Monochromatic x rays are incident on a crystal for which the spacing of the atomic planes is 0.440 nm. The first-order maximum in the Bragg reflection occurs when the incident and reflected x rays make an angle of 39.4 with the crystal planes. What is the wavelength of the x rays?

On December 26, 2004, a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than 150,000 people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 h and the speed of the wave to be 800 km/h. Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about 4500 km, while the distance between the southern end of Australia and Antarctica is about 3700 km. As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free