/*! 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} Q. 29 Excercises And Problems  An air bubble inside an 8.0-cm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Chapter 34: Q. 29 Excercises And Problems (page 991)

An air bubble inside an 8.0-cm-diameter plastic ball is localid="1651400303536" 2.0cmfrom the surface. As you look at the ball with the bubble turned toward you, how far beneath the surface does the bubble appear to be?

Short Answer

Expert verified

As a result, the picture of the bubble appears to be 1.5cmbeneath the surface.

Step by step solution

01

Relationship between the refractive indices 

To locate the picture, use the relationship between the refractive indices of the two media and the radius of the lens. An image will be created at distance sfrom a spherical refracting surface if an object is positioned at distances'from it.

n1s+n2s'=n2-n1R

n1is refractive index of air,n2is refractive index of plastic, and Ris radius of curvature.

02

Expression for s'

Radius of plastic ball,

R=-Diameter of plastic ball2

=-8.0cm2

=-4.0cm

n1s+n2sl=n2-n1Rrearrange for s'

s'=n2n2-n1R-n1s

03

Calculation for  image of the bubble's position 

The image of the bubble's position is calculated as follows:

s'=n2n2-n1R-n1s

Substitute1for role="math" localid="1651400853986" n2,1.59forrole="math" localid="1651400866937" n1,-4.0cm for Rand2.0cm fors

s'=11-1.59-4.0cm-1.592.0cm

=10.147cm-1-0.795cm-1

=-1.5cm

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 91Ó°ÊÓ!

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

When you look at your reflection in the bowl of a spoon, it is upside down. Why?

A student has built a 15−cm-long pinhole camera for a science fair project. She wants to photograph her 180−cm-tall friend and have the image on the film be0.5cm high. How far should the front of the camera be from her friend?

A 1.0-cm-tall object is 20cmin front of a concave mirror that has a60cmfocal length. Calculate the position and height of the image. State whether the image is in front of or behind the mirror, and whether the image is upright or inverted.

A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be 30cmoutside the aquarium. What is the actual distance between the can and the aquarium? (You can ignore the thin glass wall of the aquarium.)

A keratometer is an optical device used to measure the radius of curvature of the eye’s cornea—its entrance surface. This measurement is especially important when fitting contact lenses, which must match the cornea’s curvature. Most light incident on the eye is transmitted into the eye, but some light reflects from the cornea, which, due to its curvature, acts like a convex mirror. The keratometer places a small, illuminated ring of known diameter 7.5 cm in front of the eye. The optometrist, using an eyepiece, looks through the center of this ring and sees a small virtual image of the ring that appears to be behind the cornea. The optometrist uses a scale inside the eyepiece to measure the diameter of the image and calculate its magnification. Suppose the optometrist finds that the magnification for one patient is 0.049. What is the absolute value of the radius of curvature of her cornea?

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.