/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q90P A single-slit diffraction experi... [FREE SOLUTION] | 91影视

91影视

A single-slit diffraction experiment is set up with light of wavelength 420 nm, incident perpendicularly on a slit of width 5.10 mm. The viewing screen is 3.20 m distant. On the screen, what is the distance between the center of the diffraction pattern and the second diffraction minimum?

Short Answer

Expert verified

The distance between the center of the diffraction pattern and the second diffraction minimum is 0.527 m.

Step by step solution

01

Identification of given data

The given data can be listed below,

  • The wavelength of the light is,=420nm
  • The value of slit width is,a=5.10m
  • The distance of screen from slits is, D=3.20m
02

Concept/Significance of single slit diffraction

Sending a beam of light, electrons, or other particles through a single slit results in single slit diffraction. The beam diffracts widely in all directions behind a single, extremely thin (equivalent to the beam's wavelength) slit and covers the entire downstream viewing screen.

03

Determination of the distance between the center of the diffraction pattern and the second diffraction minimum

The diffraction minima of the single slit experiment is satisfied by,

asin=m

Here, ais the slit width, is the wavelength of light and m is the order of diffraction.

The angle of diffraction is given by,

sin=yD

Here, y is the distance between central and minimum diffraction pattern, and D is the distance of screen and slit.

Substitute all the values in the above,the distance between the center of the diffraction pattern and the second diffraction minimum is calculated as,

ayD=my=mDa=242010-9m3.20m5.1010-6m=0.527m

Thus, the distance between the center of the diffraction pattern and the second diffraction minimum is 0.527m.

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 91影视!

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

Manufacturers of wire (and other objects of small dimension) sometimes use a laser to continually monitor the thickness of the product. The wire intercepts the laser beam, producing a diffraction pattern like that of a single slit of the same width as the wire diameter (Fig.36-37). Suppose a helium 鈥 neon laser, of wavelength 632.8nm, illuminates a wire, and the diffraction pattern appears on a screen at distance L=2.60m. If the desired wire diameter is 1.37mm, what is the observed distance between the two tenth-order minima (one on each side of the central maximum)?

A circular obstacle produces the same diffraction pattern as a circular hole of the same diameter (except very near u 0).Airborne water drops are examples of such obstacles. When you see the Moon through suspended water drops, such as in a fog, you intercept the diffraction pattern from many drops. The composite of the central diffraction maxima of those drops forms a white region that surrounds the Moon and may obscure it. Figure 36-43 is a photograph in which the Moon is obscured. There are two faint, colored rings around the Moon (the larger one may be too faint to be seen in your copy of the photograph). The smaller ring is on the outer edge of the central maxima from the drops; the somewhat larger ring is on the outer edge of the smallest of the secondary maxima from the drops (see Fig. 36-10).The color is visible because the rings are adjacent to the diffraction minima (dark rings) in the patterns. (Colors in other parts of the pattern overlap too much to be visible.) (a) What is the color of these rings on the outer edges of the diffraction maxima? (b) The colored ring around the central maxima in Fig. 36-43 has an angular diameter that is 1.35 times the angular diameter of the Moon, which is 0.50掳. Assume that the drops all have about the same diameter. Approximately what is that diameter?

A diffraction grating having is illuminated with a light signal containing only two wavelengths and . The signal in incident perpendicularly on the grating. (a) What is the angular separation between the second order maxima of these two wavelengths? (b) What is the smallest angle at which two of the resulting maxima are superimposed? (c) What is the highest order for which maxima of both wavelengths are present in the diffraction pattern?

(a) What is the angular separation of two stars if their images are barely resolved by the Thaw refracting telescope at the Allegheny Observatory in Pittsburgh? The lens diameter is 76 cm and its focal length is 14 m. Assume =550nm. (b) Find the distance between these barely resolved stars if each of them is 10 light-years distant from Earth. (c) For the image of a single star in this telescope, find the diameter of the first dark ring in the diffraction pattern, as measured on a photographic plate placed at the focal plane of the telescope lens. Assume that the structure of the image is associated entirely with diffraction at the lens aperture and not with lens 鈥渆rrors.鈥

An astronaut in a space shuttle claims she can just barely resolve two point sources on Earth鈥檚 surface, 160 km below. Calculate their (a) angular and (b) linear separation, assuming ideal conditions. Take 位 = 540 nm and the pupil diameter of the astronaut鈥檚 eye to be 5.0 mm.

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.