/*! 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} Q97P A spy satellite orbiting at 160 ... [FREE SOLUTION] | 91影视

91影视

A spy satellite orbiting at 160 km above Earth鈥檚 surface has a lens with a focal length of 3.6 m and can resolve objects on the ground as small as 30 cm. For example, it can easily measure the size of an aircraft鈥檚 air intake port. What is the effective diameter of the lens as determined by diffraction consideration alone? Assume =550nm.

Short Answer

Expert verified

The effective diameter of the lens is 0.36 m.

Step by step solution

01

Describe the diffraction by a circular aperture or a lens

The expression to calculate the effective diameter of the lens is given by,

sin=1.22d

Here, is the wavelength, d is diameter, and is angle.

Let D be the distance between two objects, and L be the distance between Earth and satellite. Then,

L=D=DL

From the above equation,

=1.22dDL=1.22dd=1.22LD 鈥.(1)

02

Determine the effective diameter of the lens

Substitute 55010-9m for , 160103m for L, and 0.30m for D in equation (1).

d=1.2255010-9m160103m0.30m=0.36m

Therefore, the effective diameter of the lens is 0.36 m.

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

Light of wavelength 440 nm passes through a double slit, yielding a diffraction pattern whose graph of intensity I versus angular position is shown in Fig. 36-44. Calculate (a) the slit width and (b) the slit separation. (c) Verify the displayed intensities of the m=1and m=2 interference fringes.

(a) For a given diffraction grating, does the smallest difference in two wavelengths that can be resolved increase, decrease, or remain the same as the wavelength increases? (b) For a given wavelength region (say, around 500 nm), is greater in the first order or in the third order?

Monochromatic light (wavelength=450nm) is incident perpendicularly on a single slit (width=0.4mm). A screen is placed parallel to the slit plane, and on it the distance between the two minima on either side of the central maximum is 1.8mm.

(a) What is the distance from the slit to the screen? (Hint:The angle to either minimum is small enough thatsintan.)

(b) What is the distance on the screen between the first minimum and the third minimum on the same side of the central maximum?

Millimetre-wave radar generates a narrower beam than conventional microwave radar, making it less vulnerable to antiradar missiles than conventional radar. (a) Calculate the angular width 2of the central maximum, from first minimum to first minimum, produced by a 220 GHz radar beam emitted by a 55.0-cm diameter circular antenna. (The frequency is chosen to coincide with a low-absorption atmospheric 鈥渨indow.鈥) (b) What is 2 for a more conventional circular antenna that has a diameter of 2.3 m and emits at wavelength 1.6 cm?

In an experiment to monitor the Moon鈥檚 surface with a light beam, pulsed radiation from a ruby laser (位= 0.69 碌m) was directed to the Moon through a reflecting telescope with a mirror radius of 1.3 m. A reflector on the Moon behaved like a circular flat mirror with radius 10 cm, reflecting the light directly back toward the telescope on Earth. The reflected light was then detected after being brought to a focus by this telescope. Approximately what fraction of the original light energy was picked up by the detector? Assume that for each direction of travel all the energy is in the central diffraction peak.

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.