/*! 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} Q1DQ A spherical mirror is cut in hal... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A spherical mirror is cut in half horizontally. Will an image be formed by the bottom half of the mirror? If so, where will the image be formed?

Short Answer

Expert verified

Yes, the image will be formed by the bottom half of the mirror.

Step by step solution

01

Radius of curvature

The radius of curvature of a spherical mirror is actually the radius of the circle which the spherical mirror is a part. It can also be defined as the distance between the centre of the curvature and the pole of the mirror which is on the principal axis.

02

Explanation

Now, using the concept which says that the radius of the curvature of the mirror will not be altered when the spherical mirror is cut horizontally that is along the optical axis. Also, the use of the relation between the radius of curvature and the focal length of the mirror to be describe the location of the image formed by the lower half of the mirror. The image will be formed by the lower half of the mirror.

Now, the radius of curvature of the spherical mirror is twice the focal length of the mirror. The focal length of the lower part of the mirror will be same as initial case because the radius of curvature will bot be changed when it cut horizontally. Hence by mirror equation location of the image formed by the lower part of the mirror is also as earlier case.

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

Devise straightforward experiments to measure the speed of light in a given glass using (a) Snell’s law; (b) total internal reflection; (c) Brewster’s law

Young’s experiment is performed with light from excited helium atoms (λ= 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

The Lens of the Eye: The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 cm tall is placed 30.0 cm from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)

(a) Prove that when two thin lenses with focal lengths f1and f2are placed in contact, the focal length Æ’ of the combination is given by the relationship 1f=1f1+1f2 (b) A converging meniscus lens (see Fig. 34.32a) has an index of refraction of 1.55 and radii of curvature for its surfaces of magnitudes 4.50 cm and 9.00 cm. The concave surface is placed upward and filled with carbon tetrachloride (CCI4), which has n = 1.46. What is the focal length of the CCI4-glass combination?

Curvature of the Cornea: In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40 , and all the bending occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40 cm from the cornea’s vertex is focused on the retina?

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.