/*! 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} Q27P If Superman really had x-ray vis... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Q27P (page 1110)

If Superman really had x-ray vision at 0.10nm wavelength and a 4.0mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.0cm to do this?

Short Answer

Expert verified

The altitude using the angular separation formula is 1.6脳103km.

Step by step solution

01

Given data

Wavelength of x-Ray,位=0.10nm

The pupil diameter d=4.00mm

The resolving separation D=5.0cm

02

Definition of X-rays

X-rays is defined as the form of electromagnetic radiation, similar to light but of shorter wavelength and capable of penetrating solids and of ionizing gases. Such radiation having wavelengths in the range of approximately 0.1-10Nanometers.

03

Determining the altitude using the angular separation formula

Let the altitude be 'L'

Since angular separation 蠒R= 1.22位d

But 蠒R=D/L

So D=1.22位Ld

Or,

L=Dd1.22位=5.0脳10-2m脳4.0脳10-3m1.22脳0.10脳10-9m=1.639脳106mL=1.6脳103km

Hence, the altitude using the angular separation formula is 1.6脳103km.

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

In a double-slit experiment, the slit separation d is 2.00 times the slit width w. How many bright interference fringes are in the central diffraction envelope?

Light of wavelength 633nmis incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.20掳. What is the width of the slit?

Figure36-38gives 伪versus the sine of the angle 胃in a single-slit diffraction experiment using light of wavelength 610nm. The vertical axis scale is set by as role="math" localid="1663169810058" 伪s=12rad. What are (a) the slit width, (b) the total number of diffraction minima in the pattern (count them on both sides of the center of the diffraction pattern), (c) the least angle for a minimum, and (d) the greatest angle for a minimum?

Light of wavelength 600 nm is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by sin胃=0.2andsin胃=0.3 .The fourth order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width , what are the (c) largest, (d) second largest, and (e) third largest values of the order of the maxima produced by the grating?

(a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if 位=550nm, d=0.150mm, and a=30渭m? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Astrophysics

Read Explanation

Solid State Physics

Read Explanation

Scientific Method Physics

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.