Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 73

An object \(8.0 \mathrm{cm}\) high forms a virtual image \(3.5 \mathrm{cm}\) high located \(4.0 \mathrm{cm}\) behind a mirror. (a) Find the object distance. (b) Describe the mirror: is it plane, convex, or concave? (c) What are its focal length and radius of curvature?

Short Answer

Expert verified
What are the object distance, focal length, and radius of curvature? This is a convex mirror with an object distance of 9.143 cm, a focal length of 12.308 cm, and a radius of curvature of 24.615 cm.

Step by step solution

01

Calculate the Magnification Ratio

The magnification ratio (M) is the ratio of the image height (hi) to the object height (ho). We can calculate it by: $$ M = \frac{h_i}{h_o} $$ For this problem, \(h_o = 8.0\,\text{cm}\) and \(h_i = 3.5\,\text{cm}\). Plugging the values we get: $$ M = \frac{3.5}{8.0} $$
02

Determine the type of mirror

If the magnification ratio is positive, then the mirror is a virtual mirror and consequently, it is a convex mirror. If it is negative, then it is a concave mirror. In our calculation: $$ M = \frac{3.5}{8.0} > 0 $$ Since the magnification ratio is positive, the mirror is a convex mirror.
03

Calculate the object distance (do)

Since we know the magnification ratio (M), we can rewrite the magnification formula for a mirror as: $$ M = \frac{-d_i}{d_o} $$ We are given the image distance (\(d_i = -4.0\,\text{cm}\), negative because it is behind the mirror). So we can solve for the object distance (do): $$ d_o = -\frac{d_i}{M} $$ Plugging the values, we get: $$ d_o = -\frac{-4.0}{\frac{3.5}{8.0}} $$
04

Calculate the focal length (f) using the mirror equation

The mirror equation can be written as: $$ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $$ We have calculated \(d_o\) and we are given \(d_i\). So we can solve for the focal length (f): $$ f = \frac{1}{\frac{1}{d_o} + \frac{1}{d_i}} $$ Plugging the values of \(d_o\) and \(d_i\), we can find the value of \(f\).
05

Calculate the radius of curvature (R)

The radius of curvature (R) is twice the focal length (f) for a mirror. So we can calculate it as: $$ R = 2f $$ Plugging in the value of focal length (f) that we calculated in step 4, we can find the radius of curvature. Now, you have the object distance (d_o), the type of mirror (convex), the focal length (f), and the radius of curvature (R).

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

Show that the deviation angle \(\delta\) for a ray striking a thin converging lens at a distance \(d\) from the principal axis is given by \(\delta=d l f .\) Therefore, a ray is bent through an angle \(\delta\) that is proportional to \(d\) and does not depend on the angle of the incident ray (as long as it is paraxial). [Hint: Look at the figure and use the small-angle approximation \(\sin \theta=\tan \theta \approx \theta(\text { in radians) } .]\)
An object \(2.00 \mathrm{cm}\) high is placed \(12.0 \mathrm{cm}\) in front of a convex mirror with radius of curvature of \(8.00 \mathrm{cm} .\) Where is the image formed? Draw a ray diagram to illustrate.
A defect in a diamond appears to be 2.0 mm below the surface when viewed from directly above that surface. How far beneath the surface is the defect?
A diamond in air is illuminated with white light. On one particular facet, the angle of incidence is \(26.00^{\circ} .\) Inside the diamond, red light \((\lambda=660.0 \mathrm{nm}\) in vacuum) is refracted at \(10.48^{\circ}\) with respect to the normal; blue light $(\lambda=470.0 \mathrm{nm} \text { in vacuum })\( is refracted at \)10.33^{\circ} .$ (a) What are the indices of refraction for red and blue light in diamond? (b) What is the ratio of the speed of red light to the speed of blue light in diamond? (c) How would a diamond look if there were no dispersion?
A laser beam is traveling through an unknown substance. When it encounters a boundary with air, the angle of reflection is \(25.0^{\circ}\) and the angle of refraction is \(37.0^{\circ} .\) (a) What is the index of refraction of the substance? (b) What is the speed of light in the substance? (c) At what minimum angle of incidence would the light be totally internally reflected?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Oscillations

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Modern Physics

Read Explanation

Fluids

Read Explanation

Astrophysics

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