/*! 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} Problem 42 For a concave mirror of focal le... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 42

For a concave mirror of focal length \(20 \mathrm{~cm}\), if the object is at a distance of \(30 \mathrm{~cm}\) from the pole, then the nature of the image and its magnification will be (A) real and \(-2\) (B) virtual and \(-2\) (C) real and \(+2\) (D) virtual and \(+2\)

Short Answer

Expert verified
The correct option is (C) real and \(+2\). To obtain this answer, we first use the mirror formula and given values to calculate the image distance. We find the image distance to be \(d_i = -60 \mathrm{~cm}\), indicating a real image. Then, we use the magnification formula to find the magnification \(M = 2\), which is positive, meaning the image is erect and twice the size of the object.

Step by step solution

01

Recall the mirror formula and magnification formula

The mirror formula is given by the equation \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] where - \(f\) is the focal length of the mirror - \(d_o\) is the distance of the object from the mirror's pole - \(d_i\) is the distance of the image from the mirror's pole The magnification formula for a mirror is given by the equation: \[M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\] where - \(h_i\) is the height of the image - \(h_o\) is the height of the object Now, let's use these formulas to solve the exercise.
02

Plug in the known values

We are given the focal length, \(f = -20 \mathrm{~cm}\) (negative because it's a concave mirror) and the object distance, \(d_o = 30 \mathrm{~cm}\). Let's plug these values into the mirror formula and solve for the image distance, \(d_i\): \[\frac{1}{-20} = \frac{1}{30} + \frac{1}{d_i}\]
03

Solve for the image distance (d_i)

Now, we solve for the image distance, \(d_i\): \[\frac{1}{d_i} = \frac{1}{-20} - \frac{1}{30}\] \[\frac{1}{d_i} = \frac{30 - 20}{(-20)(30)}\] \[\frac{1}{d_i} = \frac{10}{(-20)(30)}\] \[d_i = -60 \mathrm{~cm}\] The image distance, \(d_i = -60 \mathrm{~cm}\), is negative, indicating that the image is formed on the same side as the object, which means it's a real image.
04

Calculate the magnification (M)

Now, let's use the magnification formula to find the magnification \(M\) using the given object distance \(d_o\) and the calculated image distance \(d_i\): \[M = -\frac{d_i}{d_o}\] \[M = -\frac{-60}{30}\] \[M = 2\] The magnification is positive, which means the image is erect and has twice the size of the object.
05

Choose the correct option

Since the image is real and the magnification is 2, the correct choice is: (C) real and \(+2\)

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

An object \(2.4 \mathrm{~m}\) in front of a lens forms a sharp image on a film \(12 \mathrm{~cm}\) behind the lens. A glass plate \(1 \mathrm{~cm}\) thick, of refractive index \(1.50\) is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film? (A) \(7.2 \mathrm{~m}\) (B) \(2.4 \mathrm{~m}\) (C) \(3.2 \mathrm{~m}\) (D) \(5.6 \mathrm{~m}\)

An astronomical telescope has magnifying power 10 . The focal length of the eyepiece is \(20 \mathrm{~cm}\). The focal length of the objective is (A) \(\frac{1}{200} \mathrm{~cm}\) (B) \(\frac{1}{2} \mathrm{~cm}\) (C) \(2 \mathrm{~cm}\) (D) \(200 \mathrm{~cm}\)

In a Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}(\lambda\) being the wavelength of light used) is \(I\). If \(I_{0}\) denotes the maximum intensity, \(\frac{I}{I_{0}}\) is equal to [2007] (A) \(\frac{3}{4}\) (B) \(\frac{1}{\sqrt{2}}\) (C) \(\frac{\sqrt{3}}{2}\) (D) \(\frac{1}{2}\)

A plane mirror is placed at the bottom of a tank containing a liquid of refractive index \(\mu, P\) is a small object at a height \(h\) above the plane mirror. An observer \(O\), vertically above \(P\), outside the liquid, observers \(P\) and its image in the mirror. The apparent distance between object and its image will be Plane mi (A) \(2 \mu h\) (B) \(\frac{2 h}{\mu}\) (C) \(\frac{2 h}{\mu-1}\) (D) \(h\left(\frac{1}{\mu}\right)\)

A ray incident at a point at an angle of incidence of \(60^{\circ}\) enters a glass sphere of R.1. \(n=\sqrt{3}\) and gets reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is (A) \(50^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(40^{\circ}\)
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Mechanics and Materials

Read Explanation

Rotational Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Medical 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.