Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q45E (page 1214)

You are asked to design a space telescope for earth orbit. When Jupiter is 5.93×108kmaway (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.

Short Answer

Expert verified

The minimum diameter of the mirror is 1.45m.

Step by step solution

01

Rayleigh’s Criterion

According to the Rayleigh criterion for the diffraction limit to resolution, two images are only resolvable when the center of one's diffraction pattern is directly over the first minimum of the other's diffraction pattern.

02

Minimum-diameter mirror

Putting the values;

Hence, the minimum diameter of the mirror is 1.45m.

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.

Start your free trial

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

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

Monochromatic coherent light passing through two thin slits is viewed on a distant screen. Are the bright fringes equally spaced on the screen? If so, why? If not, which ones are closest to being equally spaced?

According to the discussion in Section 34.2, light rays are reversible. Are the formulas in the table in this chapter’s Summary still valid if object and image are interchanged? What does reversibility imply with respect to the forms of the various formulas?

When light is incident on an interface between two materials, the angle of the refracted ray depends on the wavelength, but the angle of the reflected ray does not. Why should this be?

An interference pattern is produced by eight equally spaced narrow slits. The caption for Fig. 36.14 claims that minima occur for f =3π/4, π/4, 3π/2 and7π/4. Draw the phasor diagram for each of these four cases, and explain why each diagram proves that there is in fact a minimum. In each case, for which pairs of slits are there totally destructive interference?

There appears to be anambiguity for the case s = 10cm to whether s′ is s'=+∞and-∞whether the image is erect or inverted. How is this resolved? Or is it?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Electrostatics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Famous Physicists

Read Explanation
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